Textbook of Mathematics Grade 10 (FBISE / NBF)
Class 10 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Textbook of Mathematics Grade 10 (FBISE / NBF)

Mastery Guide: Vectors in Plane, Direct Line Segments, Component Forms & Geometrical Proofs

📖 Chapter 7: Vectors in Plane 📅 Updated: Sep 25, 2026
FBISE Class 10 • Single National Curriculum (SNC)

Chapter 7: Vectors in Plane

Complete Student Mastery Guide: Direct Line Segments, Component Forms, Head-to-Tail Rule, Ratio Formula & Geometrical Proofs

📑 Quick Navigation & Chapter Index

📖 Unit Overview & Target Learning Outcomes

In this chapter, students explore the mathematical bridge between geometry and physics: Vectors in Plane. While scalars measure magnitude alone (such as mass, temperature, or speed), vectors embody both magnitude and spatial direction (such as displacement, force, and velocity). Mastering vector algebra and coordinate geometry empowers students to model spatial navigation, aeronautics, forces, and rigorous geometrical proofs.

  • Scalars vs. Vectors: Rigorously distinguish between scalar quantities (magnitude only) and vector quantities (magnitude + direction).
  • Geometrical Representation: Represent vectors as directed line segments $\vec{AB}$ featuring an initial point (tail), a terminal point (head), length (norm), and direction angle $\theta$.
  • Cartesian & Component Forms: Express vectors in ordered pair form $[x, y]$ and standard basis unit vector form $x\hat{i} + y\hat{j}$.
  • Position Vectors: Define position vector $\vec{OP}$ relative to the origin $(0, 0)$ and compute displacement $\vec{AB} = \vec{OB} - \\vec{OA} = \vec{b} - \vec{a}$.
  • Vector Arithmetic: Add, subtract, and scale vectors algebraically and geometrically using the Head-to-Tail Rule (Triangle Law), Parallelogram Law, and Polygon Law.
  • Unit Vectors & Normalization: Normalize any non-zero vector to obtain a unit vector $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$ of unit magnitude in the specified direction.
  • Parallel & Anti-Parallel Vectors: Apply the scalar proportionality test $\vec{u} = k\vec{v} \iff \frac{x_1}{x_2} = \frac{y_1}{y_2} = k$ to determine parallelism and collinearity.
  • Rectangular Components: Resolve vectors into horizontal ($F_x = |\vec{F}|\cos\theta$) and vertical ($F_y = |\vec{F}|\sin\theta$) components, and compute direction $\theta = \tan^{-1}(F_y / F_x)$.
  • Ratio Formula: Calculate the position vector of internal ($\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$) and external division points, midpoint ($\vec{r} = \frac{\vec{a} + \vec{b}}{2}$), and points of trisection.
  • Geometrical Proofs: Execute formal vector proofs for the Triangle Midpoint Theorem, Trapezium Midline Theorem, Varignon's Theorem, and parallelogram diagonal properties.
Figure 7.1: Anatomy of a Geometrical Vector (Directed Line Segment)
A (Initial Point / Tail) (x₁, y₁) B (Terminal Point / Head) (x₂, y₂) Magnitude |v| = Length θ (Direction) Vector Notation: v = AB = [x, y]

💡 Kid-Friendly Tips for Success

🧠 Memory Hooks & Golden Rules

  • The GPS Metaphor (Scalar vs. Vector): If your speedometer says $60\text{ km/h}$, that is a Scalar (Speed). But if your GPS tells you to drive $60\text{ km/h}$ due North towards Islamabad, that is a Vector (Velocity)! Direction turns speed into velocity.
  • Vector AB = Head minus Tail: To find the vector from point $A$ to point $B$, always subtract the tail coordinates from the head coordinates: $\vec{AB} = B - A = \vec{b} - \vec{a}$. Never do $A - B$ unless you want the opposite vector $\vec{BA}$!
  • Head-to-Tail Train Rule: When adding vectors, imagine train carriages coupling together: Connect the Tail of the 2nd carriage to the Head of the 1st carriage. The resultant is the express shortcut connecting the very start (first tail) to the final stop (last head).
  • Unit Vector = "Shrinking to Size 1": A unit vector $\hat{u}$ is simply your vector with its direction completely intact, but shrunk down to have an exact length of $1$. You get it by dividing the vector by its own length: $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$.
  • The Ratio Formula Cross-Multiplication Trick: In $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$, remember that ratio weight $m$ (distance from $A$) multiplies the opposite vector $\vec{b}$, while ratio weight $n$ (distance from $B$) multiplies $\vec{a}$. They criss-cross!
Figure 7.2: Vector Resolution into Horizontal & Vertical Components
x-axis y-axis O (0,0) P (x, y) r = OP = xi + yj x = |r| cos θ (Horizontal) y = |r| sin θ (Vertical) θ Key Resolution Formulas: |r| = √(x² + y²) θ = tan⁻¹(y / x) r = [|r|cosθ, |r|sinθ]

🌍 Real-World Connections & Engineering

✈️ Aviation & Crosswind Navigation

Commercial aircraft fly through moving air masses. Pilots use vector addition ($\vec{v}_{\text{ground}} = \vec{v}_{\text{airspeed}} + \vec{v}_{\text{wind}}$) to calculate crosswind crab angles so the plane stays exactly on its runway approach trajectory without drifting off-course.

🎮 3D Video Game Engines & Physics

Game engines (like Unreal Engine and Unity) use 2D/3D vectors for everything: player movement, camera rotations, gravity vectors, projectile trajectories, and collision normals. Normalizing vectors ensures characters walk at uniform speeds in all diagonal directions.

🏗️ Structural Civil Engineering & Trusses

Bridge trusses and crane cables balance tension and compression forces. Civil engineers resolve each cable tension into horizontal and vertical components ($\sum F_x = 0$, $\sum F_y = 0$) to guarantee structural equilibrium and prevent bridge collapse.

🚢 Marine Navigation & River Crossings

Ferry captains crossing fast-flowing rivers like Indus or Chenab must angle their boats upstream. By adding the boat's engine vector to the river current vector, the ferry achieves a net resultant velocity directed straight across to the target dock.

Figure 7.3: Vector Addition via Head-to-Tail Rule (Triangle Law)
O (Tail of a) A (Head of a / Tail of b) Vector a = [x₁, y₁] B (Head of b) Vector b = [x₂, y₂] Resultant R = a + b Rule of Addition: 1. Tail of b joins Head of a 2. Resultant joins Tail a → Head b R = [x₁+x₂, y₁+y₂]

🔑 Essential Inquiries & Conceptual Questions

🤔 Provocative Mathematical Thinking Cues

  • Can two vectors have equal magnitude but NOT be equal?
    Yes! Consider $\vec{u} = [3, 4]$ and $\vec{v} = [-3, 4]$. Both have magnitude $\sqrt{3^2 + 4^2} = 5$, but $\vec{u} \neq \vec{v}$ because their horizontal directions point in opposite directions.
  • Why does a zero vector have an arbitrary direction?
    A zero vector starts and ends at the exact same point ($\vec{0} = [0, 0]$), so its length is zero. Because there is no line segment spanning two distinct points, no specific direction angle $\theta$ can be uniquely fixed, allowing it to adapt mathematically to any directional context.
  • Why is the magnitude of a vector always non-negative ($|\vec{u}| \ge 0$)?
    Geometrically, magnitude represents physical Euclidean length. Algebraically, it is the principal positive square root $\sqrt{x^2 + y^2}$, which is always $\ge 0$. It equals zero if and only if $x = 0$ and $y = 0$ (the null vector).
  • What is the geometric significance of subtracting vectors $\vec{a} - \vec{b}$?
    While $\vec{a} + \vec{b}$ gives the main diagonal of a parallelogram (originating from the common tail), $\vec{a} - \vec{b}$ represents the other diagonal connecting the head of $\vec{b}$ to the head of $\vec{a}$.
Figure 7.4: Parallelogram Law of Addition & Commutativity (a + b = b + a)
O (Origin) P Q R Vector a Vector b Vector a Vector b OR = a + b = b + a Commutative Law: ΔOPR: a + b ΔOQR: b + a ∴ a + b = b + a

🌟 Section-by-Section Explanations & Reference Tables

Table 7.1: Comprehensive Physical Quantities Classification
Physical Quantity Symbol / Units Classification Key Determining Attribute
Mass $m$ (kg, g) Scalar Possesses magnitude only; independent of spatial direction.
Time $t$ (s, min, h) Scalar One-dimensional forward progression; magnitude only.
Distance / Length $d, l$ (m, km) Scalar Total path length traversed regardless of direction.
Speed $v$ (m/s, km/h) Scalar Scalar rate of motion: $\text{Distance} / \text{Time}$.
Displacement $\vec{d}$ (m, km) Vector Shortest directed straight-line vector from start to finish.
Velocity $\vec{v}$ (m/s) Vector Rate of change of displacement in a specified direction.
Acceleration $\vec{a}$ (m/s²) Vector Rate of change of velocity vector over time.
Force / Weight $\vec{F}, \vec{W}$ (N) Vector Push or pull possessing magnitude, line of action, and direction.
Table 7.2: Vector Operations & Formulas Summary
Vector Operation Algebraic Component Rule ($[x, y]$) Standard Basis ($\hat{i}, \hat{j}$) Geometrical Interpretation
Vector Addition $[x_1 + x_2, y_1 + y_2]$ $(x_1+x_2)\hat{i} + (y_1+y_2)\hat{j}$ Head-to-tail triangle law / Parallelogram diagonal.
Vector Subtraction $[x_1 - x_2, y_1 - y_2]$ $(x_1-x_2)\hat{i} + (y_1-y_2)\hat{j}$ Connecting heads of $\vec{v}$ and $\vec{u}$ ($ riangle$ second diagonal).
Scalar Multiple $[kx, ky]$ $kx\hat{i} + ky\hat{j}$ Scales length by $|k|$; reverses direction if $k < 0$.
Magnitude (Norm) $\sqrt{x^2 + y^2}$ $\sqrt{x^2 + y^2}$ Straight-line length of directed segment via Pythagoras.
Unit Vector $\hat{u}$ $\left[\frac{x}{|\vec{u}|}, \frac{y}{|\vec{u}|}\right]$ $\frac{x\hat{i} + y\hat{j}}{\sqrt{x^2+y^2}}$ Normalized vector of length 1 pointing along $\vec{u}$.
Position Vector $\vec{AB}$ $[x_2 - x_1, y_2 - y_1]$ $(x_2-x_1)\hat{i} + (y_2-y_1)\hat{j}$ Displacement vector from tail $A(x_1, y_1)$ to head $B(x_2, y_2)$.
Internal Ratio Formula $\frac{m[x_2, y_2] + n[x_1, y_1]}{m + n}$ $\frac{m\vec{b} + n\vec{a}}{m + n}$ Position vector of point dividing $AB$ internally in ratio $m:n$.
Figure 7.5: Ratio Formula for Position Vector of Internal Division Point P (m : n)
Origin O A (a) OA = a B (b) OB = b P (r) OP = r m n Ratio Formula (Internal Division): r = (m·b + n·a) / (m + n) Midpoint (m=n=1): r = (a + b) / 2

🎯 Unit Synthesis Summary

Core Concepts Retained for Lifetime Mastery:

1. Dual Nature of Vectors: Every vector in $\mathbb{R}^2$ is completely characterized by two independent pieces of information: its magnitude (length $|\vec{u}| = \sqrt{x^2 + y^2}$) and its direction angle ($\theta = \tan^{-1}(y/x)$). In component form, $\vec{u} = [x, y] = x\hat{i} + y\hat{j}$.

2. Vector Invariance under Translation: Vectors are free entities; translating a vector anywhere in the plane without altering its length or angle produces an identical equal vector.

3. Head-to-Tail & Parallelogram Equivalence: Vector addition is commutative ($\vec{a} + \vec{b} = \vec{b} + \vec{a}$) and associative. Geometrically, the diagonal of a parallelogram formed by coterminal vectors $\vec{a}$ and $\vec{b}$ gives their resultant sum.

4. The Power of Position Vectors in Geometry: Expressing geometric vertices via position vectors $\vec{a}, \vec{b}, \vec{c}$ turns complex geometric proofs (such as Varignon's Theorem and the Triangle Midpoint Theorem) into clean, elegant linear algebraic equations.

Exercise 7.1 • Complete Step-by-Step Solutions

Ex 7.1 • Q1 Vector Graphical Representation 4 Marks
Draw and represent the following vectors geometrically with appropriate scale and direction:
(i) $10\text{ N}$ force along the positive x-axis
(ii) $50\text{ m/s}$ velocity at $150^\circ$ with the positive x-axis
(iii) $220\text{ m}$ displacement towards North
(iv) $24\text{ m/s}^2$ acceleration at $45^\circ$ with the positive x-axis.
Step-by-Step Geometrical Construction & Resolution:
(i) $10\text{ N}$ force along positive x-axis:
• Scale: Let $1\text{ cm} = 2\text{ N} \implies \text{Length} = \frac{10}{2} = 5\text{ cm}$.
• Direction: Directed along positive x-axis (Angle $\theta = 0^\circ$).
• Component Form: $\vec{F} = [10\cos 0^\circ, 10\sin 0^\circ] = [10, 0] = 10\hat{i}\text{ N}$.

(ii) $50\text{ m/s}$ velocity at $150^\circ$ with positive x-axis:
• Scale: Let $1\text{ cm} = 10\text{ m/s} \implies \text{Length} = \frac{50}{10} = 5\text{ cm}$.
• Direction: In Quadrant II at $\theta = 150^\circ$.
• Component Form: $\vec{v} = [50\cos 150^\circ, 50\sin 150^\circ] = [50(-\frac{\sqrt{3}}{2}), 50(\frac{1}{2})] = [-25\sqrt{3}, 25] \approx [-43.3, 25]\text{ m/s} = -25\sqrt{3}\hat{i} + 25\hat{j}\text{ m/s}$.

(iii) $220\text{ m}$ displacement towards North:
• Scale: Let $1\text{ cm} = 50\text{ m} \implies \text{Length} = \frac{220}{50} = 4.4\text{ cm}$.
• Direction: Due North corresponds to positive y-axis (Angle $\theta = 90^\circ$).
• Component Form: $\vec{d} = [0, 220] = 220\hat{j}\text{ m}$.

(iv) $24\text{ m/s}^2$ acceleration at $45^\circ$ with positive x-axis:
• Scale: Let $1\text{ cm} = 6\text{ m/s}^2 \implies \text{Length} = \frac{24}{6} = 4\text{ cm}$.
• Direction: In Quadrant I making an angle of $45^\circ$ with the horizontal.
• Component Form: $\vec{a} = [24\cos 45^\circ, 24\sin 45^\circ] = [24(\frac{1}{\sqrt{2}}), 24(\frac{1}{\sqrt{2}})] = [12\sqrt{2}, 12\sqrt{2}] \approx [16.97, 16.97]\text{ m/s}^2 = 12\sqrt{2}\hat{i} + 12\sqrt{2}\hat{j}\text{ m/s}^2$.
✓ Final Result: (i) 10i N, (ii) -25√3 i + 25j m/s, (iii) 220j m, (iv) 12√2 i + 12√2 j m/s²
Ex 7.1 • Q2 Scalar Multiples of Vectors 4 Marks
Given vector $\vec{F}$ is $4\text{ cm}$ long making an angle of $45^\circ$ with the positive x-axis. Find the length and orientation of:
(i) $2\vec{F}$
(ii) $-\vec{F}$
(iii) $0.5\vec{F}$
(iv) $-1.5\vec{F}$
(v) $-0.5\vec{F}$.
Step-by-Step Geometrical & Scalar Properties:
Given: $|\vec{F}| = 4\text{ cm}$ and direction $\theta = 45^\circ$.
• If scalar $k > 0$, the vector maintains direction $\theta = 45^\circ$ and length becomes $k|\vec{F}|$.
• If scalar $k < 0$, the direction reverses by $180^\circ$ (new direction $\theta = 45^\circ + 180^\circ = 225^\circ$) and length is $|k||\vec{F}|$.

(i) $2\vec{F}$:
• Length $= 2 \times 4\text{ cm} = 8\text{ cm}$.
• Direction $= 45^\circ$ (Same direction as $\vec{F}$).

(ii) $-\vec{F}$:
• Length $= |-1| \times 4\text{ cm} = 4\text{ cm}$.
• Direction $= 45^\circ + 180^\circ = 225^\circ$ (Directly opposite to $\vec{F}$).

(iii) $0.5\vec{F}$:
• Length $= 0.5 \times 4\text{ cm} = 2\text{ cm}$.
• Direction $= 45^\circ$ (Same direction as $\vec{F}$).

(iv) $-1.5\vec{F}$:
• Length $= |-1.5| \times 4\text{ cm} = 6\text{ cm}$.
• Direction $= 45^\circ + 180^\circ = 225^\circ$ (Opposite direction).

(v) $-0.5\vec{F}$:
• Length $= |-0.5| \times 4\text{ cm} = 2\text{ cm}$.
• Direction $= 45^\circ + 180^\circ = 225^\circ$ (Opposite direction).
✓ Final Result: (i) 8 cm at 45°, (ii) 4 cm at 225°, (iii) 2 cm at 45°, (iv) 6 cm at 225°, (v) 2 cm at 225°
Ex 7.1 • Q3 Vector Arithmetic on Compass Bearings 8 Marks
Given vectors $\vec{a} = 3\text{ units West}$ ($[-3, 0]$) and $\vec{b} = 3\text{ units North}$ ($[0, 3]$). Compute the component form and magnitude of the following combinations:
(i) $2\vec{a} + \vec{b}$
(ii) $\vec{a} - 2\vec{b}$
(iii) $3\vec{a} + 1.5\vec{b}$
(iv) $(2\vec{a} + \vec{b}) + (\vec{a} - 2\vec{b})$
(v) $0.5(\vec{a} + \vec{b})$
(vi) $3\vec{a} - 2\vec{b}$
(vii) $2\vec{a} - 2.5\vec{b}$
(viii) $-(\vec{a} - 2\vec{b})$.
Step-by-Step Algebraic Evaluation:
Given: $\vec{a} = [-3, 0] = -3\hat{i}$ and $\vec{b} = [0, 3] = 3\hat{j}$.

(i) $2\vec{a} + \vec{b}$:
• $2[-3, 0] + [0, 3] = [-6, 0] + [0, 3] = [-6, 3] = -6\hat{i} + 3\hat{j}$.
• Magnitude $= \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}\text{ units} \approx 6.71$.

(ii) $\vec{a} - 2\vec{b}$:
• $[-3, 0] - 2[0, 3] = [-3, 0] - [0, 6] = [-3, -6] = -3\hat{i} - 6\hat{j}$.
• Magnitude $= \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\text{ units} \approx 6.71$.

(iii) $3\vec{a} + 1.5\vec{b}$:
• $3[-3, 0] + 1.5[0, 3] = [-9, 0] + [0, 4.5] = [-9, 4.5] = -9\hat{i} + 4.5\hat{j}$.
• Magnitude $= \sqrt{(-9)^2 + (4.5)^2} = \sqrt{81 + 20.25} = \sqrt{101.25} = 4.5\sqrt{5}\text{ units} \approx 10.06$.

(iv) $(2\vec{a} + \vec{b}) + (\vec{a} - 2\vec{b})$:
• Combining like terms $= 3\vec{a} - \vec{b} = 3[-3, 0] - [0, 3] = [-9, -3] = -9\hat{i} - 3\hat{j}$.
• Magnitude $= \sqrt{(-9)^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}\text{ units} \approx 9.49$.

(v) $0.5(\vec{a} + \vec{b})$:
• $0.5([-3, 0] + [0, 3]) = 0.5[-3, 3] = [-1.5, 1.5] = -1.5\hat{i} + 1.5\hat{j}$.
• Magnitude $= \sqrt{(-1.5)^2 + (1.5)^2} = \sqrt{2.25 + 2.25} = \sqrt{4.5} = 1.5\sqrt{2}\text{ units} \approx 2.12$.

(vi) $3\vec{a} - 2\vec{b}$:
• $3[-3, 0] - 2[0, 3] = [-9, 0] - [0, 6] = [-9, -6] = -9\hat{i} - 6\hat{j}$.
• Magnitude $= \sqrt{(-9)^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117} = 3\sqrt{13}\text{ units} \approx 10.82$.

(vii) $2\vec{a} - 2.5\vec{b}$:
• $2[-3, 0] - 2.5[0, 3] = [-6, 0] - [0, 7.5] = [-6, -7.5] = -6\hat{i} - 7.5\hat{j}$.
• Magnitude $= \sqrt{(-6)^2 + (-7.5)^2} = \sqrt{36 + 56.25} = \sqrt{92.25} \approx 9.60\text{ units}$.

(viii) $-(\vec{a} - 2\vec{b})$:
• $-[-3, -6] = [3, 6] = 3\hat{i} + 6\hat{j}$.
• Magnitude $= \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = 3\sqrt{5}\text{ units} \approx 6.71$.
✓ Final Result: (i) [-6, 3], (ii) [-3, -6], (iii) [-9, 4.5], (iv) [-9, -3], (v) [-1.5, 1.5], (vi) [-9, -6], (vii) [-6, -7.5], (viii) [3, 6]
Ex 7.1 • Q4 Vector Scalar Proportions 4 Marks
In the coordinate grid, the direction and displacement of vector $\vec{a}$ is $4$ square units to the right and $3$ square units up (i.e., $\vec{a} = [4, 3]$). Determine the scalar relationship of the following vectors with $\vec{a}$:
(i) Vector $\vec{p} = [8, 6]$
(ii) Vector $\vec{q} = [-4, -3]$
(iii) Vector $\vec{r} = [-8, -6]$
(iv) Vector $\vec{s} = [2, 1.5]$
(v) Vector $\vec{t} = [-2, -1.5]$.
Step-by-Step Scalar Relationship Analysis:
Given reference vector: $\vec{a} = [4, 3] = 4\hat{i} + 3\hat{j}$.
We test for scalar multiple $k$ such that $\vec{v} = k\vec{a} = [4k, 3k]$:

(i) Vector $\vec{p} = [8, 6]$:
• $[8, 6] = 2[4, 3] = 2\vec{a}$.
• $\vec{p}$ is parallel and has twice the magnitude in the same direction.

(ii) Vector $\vec{q} = [-4, -3]$:
• $[-4, -3] = -1[4, 3] = -\vec{a}$.
• $\vec{q}$ is the negative vector of $\vec{a}$ (equal magnitude, opposite direction).

(iii) Vector $\vec{r} = [-8, -6]$:
• $[-8, -6] = -2[4, 3] = -2\vec{a}$.
• $\vec{r}$ is anti-parallel with twice the magnitude.

(iv) Vector $\vec{s} = [2, 1.5]$:
• $[2, 1.5] = 0.5[4, 3] = 0.5\vec{a} = \frac{1}{2}\vec{a}$.
• $\vec{s}$ is half the magnitude in the same direction.

(v) Vector $\vec{t} = [-2, -1.5]$:
• $[-2, -1.5] = -0.5[4, 3] = -0.5\vec{a} = -\frac{1}{2}\vec{a}$.
• $\vec{t}$ is half the magnitude in the opposite direction.
✓ Final Result: (i) p = 2a, (ii) q = -a, (iii) r = -2a, (iv) s = 0.5a, (v) t = -0.5a
Ex 7.1 • Q5 Point Translation by Vector 4 Marks
A point with coordinates $P(5, -7)$ is translated by the vector $\vec{v} = [0, 4]$. Find the new position $P'$ of the point.
Step-by-Step Translation Formula:
• Step 1: State the translation transformation rule.
If a point $P(x, y)$ is translated by vector $\vec{v} = [a, b]$, the translated coordinates $P'(x', y')$ are given by:
$$P'(x', y') = (x + a, y + b)$$
• Step 2: Substitute given values $x = 5, y = -7, a = 0, b = 4$:
$$x' = 5 + 0 = 5$$
$$y' = -7 + 4 = -3$$
• Step 3: Conclude the final coordinates.
$$P' = (5, -3)$$
✓ Final Result: P'(5, -3)
Ex 7.1 • Q6 Vector Translation 4 Marks
A vector $\vec{u} = [-5, 4]$ is translated by another vector $\vec{v} = [4, -3]$. Find the resultant location and combined translation of the vector.
Step-by-Step Vector Addition:
• Step 1: Apply the algebraic law for successive vector translation:
$$\vec{w} = \vec{u} + \vec{v} = [x_1 + x_2, y_1 + y_2]$$
• Step 2: Add corresponding components:
$$\vec{w} = [-5 + 4, 4 + (-3)] = [-1, 1]$$
• Step 3: In standard unit basis form:
$$\vec{w} = -\hat{i} + \hat{j}$$
✓ Final Result: [-1, 1] or -i + j
Ex 7.1 • Q7 Polygon Geometric Translation 4 Marks
Triangle $ABC$ has vertices $A(-4, 6)$, $B(-1, 4)$, and $C(-6, 1)$. Find the coordinates of the translated image vertices if:
(i) The triangle is translated by vector $\vec{v}_1 = [5, 0]$
(ii) The triangle is translated by vector $\vec{v}_2 = [-2, -4]$.
Step-by-Step Transformation:
(i) Translation by $\vec{v}_1 = [5, 0]$:
Add $+5$ to $x$-coordinates and $+0$ to $y$-coordinates:
• $A' = (-4 + 5, 6 + 0) = (1, 6)$
• $B' = (-1 + 5, 4 + 0) = (4, 4)$
• $C' = (-6 + 5, 1 + 0) = (-1, 1)$

(ii) Translation by $\vec{v}_2 = [-2, -4]$:
Add $-2$ to $x$-coordinates and $-4$ to $y$-coordinates:
• $A'' = (-4 - 2, 6 - 4) = (-6, 2)$
• $B'' = (-1 - 2, 4 - 4) = (-3, 0)$
• $C'' = (-6 - 2, 1 - 4) = (-8, -3)$
✓ Final Result: (i) A'(1, 6), B'(4, 4), C'(-1, 1); (ii) A''(-6, 2), B''(-3, 0), C''(-8, -3)
Ex 7.1 • Q8 Inverse Translation Vector 4 Marks
What translation vector $\vec{T}$ is required to map the point $E(-6, 5)$ directly onto the origin $O(0, 0)$?
Step-by-Step Solution:
• Step 1: Let the translation vector be $\vec{T} = [a, b]$.
• Step 2: Set up the translation equation for mapping $E$ to $O$:
$$E(x, y) + \vec{T} = O(0, 0) \implies (-6 + a, 5 + b) = (0, 0)$$
• Step 3: Equate individual components:
$$-6 + a = 0 \implies a = 6$$
$$5 + b = 0 \implies b = -5$$
• Step 4: Write the translation vector:
$$\vec{T} = [6, -5] = 6\hat{i} - 5\hat{j}$$
✓ Final Result: [6, -5] or 6i - 5j
Ex 7.1 • Q9 Vector Representation in Rectangle 8 Marks
$ABCD$ is a rectangle where $\vec{AB} = \vec{a}$, $\vec{BC} = \vec{b}$, and $O$ is the intersection of its diagonals. Express each of the following vectors in terms of $\vec{a}$ and $\vec{b}$:
(i) $\vec{CD}$
(ii) $\vec{DA}$
(iii) $\vec{AC}$
(iv) $\vec{BD}$
(v) $\vec{AO}$
(vi) $\vec{BO}$
(vii) $\vec{AB}$.
Step-by-Step Geometrical Derivations:
In rectangle $ABCD$, opposite sides are equal in length and parallel: $\vec{AB} = \vec{DC} = \vec{a}$ and $\vec{BC} = \vec{AD} = \vec{b}$.

(i) $\vec{CD}$:
• $\vec{CD} = -\vec{DC} = -\vec{AB} = -\vec{a}$.

(ii) $\vec{DA}$:
• $\vec{DA} = -\vec{AD} = -\vec{BC} = -\vec{b}$.

(iii) $\vec{AC}$:
• By triangle law in $\triangle ABC$: $\vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b}$.

(iv) $\vec{BD}$:
• In $\triangle BCD$: $\vec{BD} = \vec{BC} + \vec{CD} = \vec{b} + (-\vec{a}) = \vec{b} - \vec{a} = -\vec{a} + \vec{b}$.

(v) $\vec{AO}$:
• In a rectangle, diagonals bisect each other, so $O$ is the midpoint of $AC$:
$$\vec{AO} = \frac{1}{2}\vec{AC} = \frac{1}{2}(\vec{a} + \vec{b})$$

(vi) $\vec{BO}$:
• $O$ is the midpoint of diagonal $BD$, so $\vec{BO} = \frac{1}{2}\vec{BD} = \frac{1}{2}(\vec{b} - \vec{a}) = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{a}$.

(vii) $\vec{AB}$:
• Directly given: $\vec{AB} = \vec{a}$.
✓ Final Result: (i) -a, (ii) -b, (iii) a + b, (iv) b - a, (v) 0.5(a + b), (vi) 0.5(b - a), (vii) a
Ex 7.1 • Q10 Position Vectors and Midpoint 4 Marks
Given that $\vec{OP} = \vec{p}$, $\vec{OQ} = \vec{q}$, and $M$ is the midpoint of the line segment $PQ$. Find the following vectors in terms of $\vec{p}$ and $\vec{q}$:
(i) $\vec{PQ}$
(ii) $\vec{PM}$
(iii) $\vec{QM}$
(iv) $\vec{OM}$.
Step-by-Step Derivation:
(i) $\vec{PQ}$:
• By position vector formula: $\vec{PQ} = \vec{OQ} - \vec{OP} = \vec{q} - \vec{p}$.

(ii) $\vec{PM}$:
• Since $M$ is the midpoint of $PQ$: $\vec{PM} = \frac{1}{2}\vec{PQ} = \frac{1}{2}(\vec{q} - \vec{p}) = \frac{1}{2}\vec{q} - \frac{1}{2}\vec{p}$.

(iii) $\vec{QM}$:
• $\vec{QM} = -\vec{MQ} = -\vec{PM} = -\frac{1}{2}(\vec{q} - \vec{p}) = \frac{1}{2}(\vec{p} - \vec{q}) = \frac{1}{2}\vec{p} - \frac{1}{2}\vec{q}$.

(iv) $\vec{OM}$:
• By midpoint position vector formula: $\vec{OM} = \frac{\vec{OP} + \vec{OQ}}{2} = \frac{\vec{p} + \vec{q}}{2} = \frac{1}{2}\vec{p} + \frac{1}{2}\vec{q}$.
✓ Final Result: (i) q - p, (ii) 0.5(q - p), (iii) 0.5(p - q), (iv) 0.5(p + q)
Ex 7.1 • Q11 Vector from Two Given Points 4 Marks
Express vector $\vec{AB}$ in the component form $[x, y]$ and standard basis form $x\hat{i} + y\hat{j}$ for the given pairs of points:
(i) $A(2, 3)$ and $B(6, 6)$
(ii) $A(-2, 5)$ and $B(2, -5)$
(iii) $A(4, -7)$ and $B(-1, -9)$.
Step-by-Step Coordinate Resolution:
Formula: $\vec{AB} = [x_2 - x_1, y_2 - y_1] = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}$.

(i) $A(2, 3)$ and $B(6, 6)$:
• $x = 6 - 2 = 4$
• $y = 6 - 3 = 3$
• $[x, y] = [4, 3]$ and $4\hat{i} + 3\hat{j}$.

(ii) $A(-2, 5)$ and $B(2, -5)$:
• $x = 2 - (-2) = 2 + 2 = 4$
• $y = -5 - 5 = -10$
• $[x, y] = [4, -10]$ and $4\hat{i} - 10\hat{j}$.

(iii) $A(4, -7)$ and $B(-1, -9)$:
• $x = -1 - 4 = -5$
• $y = -9 - (-7) = -9 + 7 = -2$
• $[x, y] = [-5, -2]$ and $-5\hat{i} - 2\hat{j}$.
✓ Final Result: (i) [4, 3] = 4i + 3j, (ii) [4, -10] = 4i - 10j, (iii) [-5, -2] = -5i - 2j
Ex 7.1 • Q12 Unit Vector and Normalization 4 Marks
Find the unit vector $\hat{u}$ for each of the following vectors:
(i) $\vec{a} = -6\hat{i} - 8\hat{j}$
(ii) $\vec{u} = 5\hat{i} - 12\hat{j}$
(iii) $\vec{b} = [1, -1]$
(iv) $\vec{PQ}$ where $P(1, 4)$ and $Q(4, 8)$.
Step-by-Step Normalization Procedure:
Formula: $\hat{u} = \frac{\vec{u}}{|\vec{u}|} = \frac{x\hat{i} + y\hat{j}}{\sqrt{x^2 + y^2}}$.

(i) $\vec{a} = -6\hat{i} - 8\hat{j}$:
• $|\vec{a}| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
• $\hat{a} = \frac{-6\hat{i} - 8\hat{j}}{10} = -\frac{6}{10}\hat{i} - \frac{8}{10}\hat{j} = -\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} = [-\frac{3}{5}, -\frac{4}{5}]$.

(ii) $\vec{u} = 5\hat{i} - 12\hat{j}$:
• $|\vec{u}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
• $\hat{u} = \frac{5\hat{i} - 12\hat{j}}{13} = \frac{5}{13}\hat{i} - \frac{12}{13}\hat{j} = [\frac{5}{13}, -\frac{12}{13}]$.

(iii) $\vec{b} = [1, -1] = \hat{i} - \hat{j}$:
• $|\vec{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
• $\hat{b} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} = [\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}]$.

(iv) $\vec{PQ}$ where $P(1, 4), Q(4, 8)$:
• $\vec{PQ} = [4 - 1, 8 - 4] = [3, 4] = 3\hat{i} + 4\hat{j}$.
• $|\vec{PQ}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
• $\widehat{PQ} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} = [\frac{3}{5}, \frac{4}{5}]$.
✓ Final Result: (i) -3/5 i - 4/5 j, (ii) 5/13 i - 12/13 j, (iii) 1/√2 i - 1/√2 j, (iv) 3/5 i + 4/5 j
Ex 7.1 • Q13 Linear Combination of Vectors 4 Marks
If $\vec{a} = 3\hat{i} - 2\hat{j}$ and $\vec{b} = \hat{i} + 4\hat{j}$, find the scalar values of $x$ and $y$ such that $x\vec{a} + y\vec{b} = 4\hat{i} - 12\hat{j}$.
Step-by-Step Algebraic System Solution:
• Step 1: Expand the linear combination in terms of $\hat{i}$ and $\hat{j}$:
$$x(3\hat{i} - 2\hat{j}) + y(\hat{i} + 4\hat{j}) = 4\hat{i} - 12\hat{j}$$
$$(3x + y)\hat{i} + (-2x + 4y)\hat{j} = 4\hat{i} - 12\hat{j}$$
• Step 2: Equate corresponding components:
1) $3x + y = 4 \implies y = 4 - 3x$
2) $-2x + 4y = -12$
• Step 3: Substitute $y = 4 - 3x$ into Equation (2):
$$-2x + 4(4 - 3x) = -12$$
$$-2x + 16 - 12x = -12$$
$$-14x = -12 - 16 = -28$$
$$x = \frac{-28}{-14} = 2$$
• Step 4: Solve for $y$:
$$y = 4 - 3(2) = 4 - 6 = -2$$
• Check: $x\vec{a} + y\vec{b} = 2(3\hat{i}-2\hat{j}) - 2(\hat{i}+4\hat{j}) = 6\hat{i}-4\hat{j}-2\hat{i}-8\hat{j} = 4\hat{i}-12\hat{j}$ (Correct!).
✓ Final Result: x = 2, y = -2
Ex 7.1 • Q14 Multi-Point Vector Operations 8 Marks
Given four points $D(3, -2)$, $E(-3, -5)$, $F(0, 5)$, and $G(3, 6)$. Find the following vectors:
(i) $\vec{DE}$
(ii) $\vec{DE} + \vec{FG}$
(iii) $3\vec{EF} - 2\vec{EG}$
(iv) $2\vec{DF} - 3\vec{EF}$
(v) $\vec{DF} - \vec{EF} + 2\vec{FG}$.
Step-by-Step Vector Constructions:
First, find all fundamental displacement vectors:
• $\vec{DE} = [-3 - 3, -5 - (-2)] = [-6, -3] = -6\hat{i} - 3\hat{j}$
• $\vec{FG} = [3 - 0, 6 - 5] = [3, 1] = 3\hat{i} + \hat{j}$
• $\vec{EF} = [0 - (-3), 5 - (-5)] = [3, 10] = 3\hat{i} + 10\hat{j}$
• $\vec{EG} = [3 - (-3), 6 - (-5)] = [6, 11] = 6\hat{i} + 11\hat{j}$
• $\vec{DF} = [0 - 3, 5 - (-2)] = [-3, 7] = -3\hat{i} + 7\hat{j}$

(i) $\vec{DE}$:
• $\vec{DE} = -6\hat{i} - 3\hat{j} = [-6, -3]$.

(ii) $\vec{DE} + \vec{FG}$:
• $(-6\hat{i} - 3\hat{j}) + (3\hat{i} + \hat{j}) = (-6 + 3)\hat{i} + (-3 + 1)\hat{j} = -3\hat{i} - 2\hat{j} = [-3, -2]$.

(iii) $3\vec{EF} - 2\vec{EG}$:
• $3(3\hat{i} + 10\hat{j}) - 2(6\hat{i} + 11\hat{j}) = (9\hat{i} + 30\hat{j}) - (12\hat{i} + 22\hat{j}) = -3\hat{i} + 8\hat{j} = [-3, 8]$.

(iv) $2\vec{DF} - 3\vec{EF}$:
• $2(-3\hat{i} + 7\hat{j}) - 3(3\hat{i} + 10\hat{j}) = (-6\hat{i} + 14\hat{j}) - (9\hat{i} + 30\hat{j}) = -15\hat{i} - 16\hat{j} = [-15, -16]$.

(v) $\vec{DF} - \vec{EF} + 2\vec{FG}$:
• Note that $\vec{DF} - \vec{EF} = \vec{DF} + \vec{FE} = \vec{DE} = -6\hat{i} - 3\hat{j}$.
• $\vec{DE} + 2\vec{FG} = (-6\hat{i} - 3\hat{j}) + 2(3\hat{i} + \hat{j}) = (-6\hat{i} - 3\hat{j}) + (6\hat{i} + 2\hat{j}) = 0\hat{i} - 1\hat{j} = -\hat{j} = [0, -1]$.
✓ Final Result: (i) -6i - 3j, (ii) -3i - 2j, (iii) -3i + 8j, (iv) -15i - 16j, (v) -j
Ex 7.1 • Q15 Vector Algebra and Norm Calculations 8 Marks
If $\vec{a} = 3\hat{i} + 5\hat{j}$ and $\vec{b} = 2\hat{i} - 4\hat{j}$, calculate the following expressions:
(i) $\vec{a} + 2\vec{b}$
(ii) $2\vec{a} - 3\vec{b}$
(iii) $5(2\vec{a} + \vec{b})$
(iv) $0.5(\vec{a} - \vec{b})$
(v) $|\vec{a} + \vec{b}|$
(vi) $|\vec{a} - \vec{b}|$
(vii) $|\vec{a}| + 2|\vec{b}|$
(viii) $|3\vec{a} - 2\vec{b}|$.
Step-by-Step Algebraic & Norm Solutions:
Given: $\vec{a} = 3\hat{i} + 5\hat{j}$, $\vec{b} = 2\hat{i} - 4\hat{j}$.

(i) $\vec{a} + 2\vec{b}$:
• $(3\hat{i} + 5\hat{j}) + 2(2\hat{i} - 4\hat{j}) = 3\hat{i} + 5\hat{j} + 4\hat{i} - 8\hat{j} = 7\hat{i} - 3\hat{j}$.

(ii) $2\vec{a} - 3\vec{b}$:
• $2(3\hat{i} + 5\hat{j}) - 3(2\hat{i} - 4\hat{j}) = (6\hat{i} + 10\hat{j}) - (6\hat{i} - 12\hat{j}) = 0\hat{i} + 22\hat{j} = 22\hat{j}$.

(iii) $5(2\vec{a} + \vec{b})$:
• $2\vec{a} + \vec{b} = 2(3\hat{i} + 5\hat{j}) + (2\hat{i} - 4\hat{j}) = 6\hat{i} + 10\hat{j} + 2\hat{i} - 4\hat{j} = 8\hat{i} + 6\hat{j}$.
• $5(8\hat{i} + 6\hat{j}) = 40\hat{i} + 30\hat{j}$.

(iv) $0.5(\vec{a} - \vec{b})$:
• $\vec{a} - \vec{b} = (3\hat{i} + 5\hat{j}) - (2\hat{i} - 4\hat{j}) = \hat{i} + 9\hat{j}$.
• $0.5(\hat{i} + 9\hat{j}) = 0.5\hat{i} + 4.5\hat{j} = \frac{1}{2}\hat{i} + \frac{9}{2}\hat{j}$.

(v) $|\vec{a} + \vec{b}|$:
• $\vec{a} + \vec{b} = (3+2)\hat{i} + (5-4)\hat{j} = 5\hat{i} + \hat{j}$.
• $|\vec{a} + \vec{b}| = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.10$.

(vi) $|\vec{a} - \vec{b}|$:
• $\vec{a} - \vec{b} = \hat{i} + 9\hat{j}$.
• $|\vec{a} - \vec{b}| = \sqrt{1^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82} \approx 9.06$.

(vii) $|\vec{a}| + 2|\vec{b}|$:
• $|\vec{a}| = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$.
• $|\vec{b}| = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$.
• $|\vec{a}| + 2|\vec{b}| = \sqrt{34} + 2(2\sqrt{5}) = \sqrt{34} + 4\sqrt{5} \approx 5.83 + 8.94 = 14.77$.

(viii) $|3\vec{a} - 2\vec{b}|$:
• $3\vec{a} - 2\vec{b} = 3(3\hat{i} + 5\hat{j}) - 2(2\hat{i} - 4\hat{j}) = (9\hat{i} + 15\hat{j}) - (4\hat{i} - 8\hat{j}) = 5\hat{i} + 23\hat{j}$.
• $|3\vec{a} - 2\vec{b}| = \sqrt{5^2 + 23^2} = \sqrt{25 + 529} = \sqrt{554} \approx 23.54$.
✓ Final Result: (i) 7i - 3j, (ii) 22j, (iii) 40i + 30j, (iv) 0.5i + 4.5j, (v) √26, (vi) √82, (vii) √34 + 4√5, (viii) √554
Ex 7.1 • Q16 Parallel Vectors Condition 4 Marks
Find the value of the scalar parameter $p$ for which the vector $\vec{u} = 3\hat{i} + 6\hat{j}$ is parallel to the vector $\vec{v} = \hat{i} + p\hat{j}$.
Step-by-Step Proportionality Condition:
• Step 1: State the necessary and sufficient condition for two vectors to be parallel.
Two vectors $\vec{u} = x_1\hat{i} + y_1\hat{j}$ and $\vec{v} = x_2\hat{i} + y_2\hat{j}$ are parallel if and only if their corresponding components are proportional:
$$\frac{x_1}{x_2} = \frac{y_1}{y_2} = k$$
• Step 2: Substitute given values $x_1 = 3, y_1 = 6, x_2 = 1, y_2 = p$:
$$\frac{3}{1} = \frac{6}{p}$$
• Step 3: Solve for $p$ via cross-multiplication:
$$3p = 6 \implies p = \frac{6}{3} = 2$$
• Verification: For $p = 2$, $\vec{u} = 3\hat{i} + 6\hat{j} = 3(\hat{i} + 2\hat{j}) = 3\vec{v}$. Since $k = 3 > 0$, the vectors are parallel and point in the identical direction.
✓ Final Result: p = 2
Ex 7.1 • Q17 Unit Vectors of Composite Expressions 8 Marks
If $\vec{a} = 2\hat{i} - 4\hat{j}$, $\vec{b} = \hat{i} - 2\hat{j}$, and $\vec{c} = [1, 3] = \hat{i} + 3\hat{j}$, find a unit vector in the direction of each of the following:
(i) $\vec{a} + \vec{b}$
(ii) $2\vec{a} - \vec{b}$
(iii) $2\vec{a} - \vec{c}$
(iv) $0.5(\vec{a} + \vec{c})$
(v) $\vec{a} + \vec{b} + 2\vec{c}$
(vi) $4\vec{a} - 2\vec{b} + \vec{c}$.
Step-by-Step Evaluation & Normalization:
Given: $\vec{a} = 2\hat{i} - 4\hat{j}$, $\vec{b} = \hat{i} - 2\hat{j}$, $\vec{c} = \hat{i} + 3\hat{j}$.

(i) $\vec{a} + \vec{b}$:
• $\vec{v}_1 = (2+1)\hat{i} + (-4-2)\hat{j} = 3\hat{i} - 6\hat{j}$.
• $|\vec{v}_1| = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.
• $\hat{v}_1 = \frac{3\hat{i} - 6\hat{j}}{3\sqrt{5}} = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}$.

(ii) $2\vec{a} - \vec{b}$:
• $\vec{v}_2 = 2(2\hat{i} - 4\hat{j}) - (\hat{i} - 2\hat{j}) = 4\hat{i} - 8\hat{j} - \hat{i} + 2\hat{j} = 3\hat{i} - 6\hat{j}$.
• $\hat{v}_2 = \frac{3\hat{i} - 6\hat{j}}{3\sqrt{5}} = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}$.

(iii) $2\vec{a} - \vec{c}$:
• $\vec{v}_3 = 2(2\hat{i} - 4\hat{j}) - (\hat{i} + 3\hat{j}) = 4\hat{i} - 8\hat{j} - \hat{i} - 3\hat{j} = 3\hat{i} - 11\hat{j}$.
• $|\vec{v}_3| = \sqrt{3^2 + (-11)^2} = \sqrt{9 + 121} = \sqrt{130}$.
• $\hat{v}_3 = \frac{3}{\sqrt{130}}\hat{i} - \frac{11}{\sqrt{130}}\hat{j}$.

(iv) $0.5(\vec{a} + \vec{c})$:
• $\vec{a} + \vec{c} = (2+1)\hat{i} + (-4+3)\hat{j} = 3\hat{i} - \hat{j}$.
• Multiplying by scalar $0.5 > 0$ does not change the unit vector direction!
• Magnitude of $3\hat{i} - \hat{j} = \sqrt{3^2 + (-1)^2} = \sqrt{10}$.
• $\hat{v}_4 = \frac{3}{\sqrt{10}}\hat{i} - \frac{1}{\sqrt{10}}\hat{j}$.

(v) $\vec{a} + \vec{b} + 2\vec{c}$:
• $\vec{v}_5 = (2\hat{i} - 4\hat{j}) + (\hat{i} - 2\hat{j}) + 2(\hat{i} + 3\hat{j}) = (2+1+2)\hat{i} + (-4-2+6)\hat{j} = 5\hat{i} + 0\hat{j} = 5\hat{i}$.
• $|\vec{v}_5| = 5$.
• $\hat{v}_5 = \frac{5\hat{i}}{5} = \hat{i}$.

(vi) $4\vec{a} - 2\vec{b} + \vec{c}$:
• Note $4\vec{a} - 2\vec{b} = 4(2\hat{i}-4\hat{j}) - 2(\hat{i}-2\hat{j}) = (8\hat{i}-16\hat{j}) - (2\hat{i}-4\hat{j}) = 6\hat{i} - 12\hat{j}$.
• $\vec{v}_6 = (6\hat{i} - 12\hat{j}) + (\hat{i} + 3\hat{j}) = 7\hat{i} - 9\hat{j}$.
• $|\vec{v}_6| = \sqrt{7^2 + (-9)^2} = \sqrt{49 + 81} = \sqrt{130}$.
• $\hat{v}_6 = \frac{7}{\sqrt{130}}\hat{i} - \frac{9}{\sqrt{130}}\hat{j}$.
✓ Final Result: (i) (1/√5)i - (2/√5)j, (ii) (1/√5)i - (2/√5)j, (iii) (3/√130)i - (11/√130)j, (iv) (3/√10)i - (1/√10)j, (v) i, (vi) (7/√130)i - (9/√130)j
Ex 7.1 • Q18 Vector with Specified Magnitude and Direction 4 Marks
If $\vec{a} = \hat{i} - 3\hat{j}$ and $\vec{b} = 2\hat{i} + 4\hat{j}$, find:
(i) A vector of magnitude $4$ in the direction of $2\vec{a} - \vec{b}$.
(ii) A vector of magnitude $6$ in the opposite direction of $\vec{a} + 2\vec{b}$.
Step-by-Step Construction:
(i) Vector of magnitude 4 in direction of $2\vec{a} - \vec{b}$:
• Step 1: Find $\vec{u} = 2\vec{a} - \vec{b}$:
$$\vec{u} = 2(\hat{i} - 3\hat{j}) - (2\hat{i} + 4\hat{j}) = (2\hat{i} - 6\hat{j}) - (2\hat{i} + 4\hat{j}) = 0\hat{i} - 10\hat{j} = -10\hat{j}$$
• Step 2: Find unit vector $\hat{u}$:
$$|\vec{u}| = |-10| = 10 \implies \hat{u} = \frac{-10\hat{j}}{10} = -\hat{j}$$
• Step 3: Multiply by target magnitude $4$:
$$\vec{w}_1 = 4\hat{u} = 4(-\hat{j}) = -4\hat{j} = [0, -4]$$

(ii) Vector of magnitude 6 in opposite direction of $\vec{a} + 2\vec{b}$:
• Step 1: Find $\vec{v} = \vec{a} + 2\vec{b}$:
$$\vec{v} = (\hat{i} - 3\hat{j}) + 2(2\hat{i} + 4\hat{j}) = \hat{i} - 3\hat{j} + 4\hat{i} + 8\hat{j} = 5\hat{i} + 5\hat{j}$$
• Step 2: Find unit vector $\hat{v}$:
$$|\vec{v}| = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$
$$\hat{v} = \frac{5\hat{i} + 5\hat{j}}{5\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$$
• Step 3: Opposite direction means multiplying by $-6$:
$$\vec{w}_2 = -6\hat{v} = -6\left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right) = -\frac{6}{\sqrt{2}}\hat{i} - \frac{6}{\sqrt{2}}\hat{j} = -3\sqrt{2}\hat{i} - 3\sqrt{2}\hat{j}$$
✓ Final Result: (i) -4j, (ii) -3√2 i - 3√2 j
Ex 7.1 • Q19 Real-World Aviation Crosswind Application 8 Marks
An airplane departs from an airport on a compass bearing of $045^\circ$ traveling at an airspeed of $380\text{ mph}$. A crosswind is blowing on a bearing of $135^\circ$ at a speed of $50\text{ mph}$. What is the actual resultant velocity (speed and true bearing) of the airplane?
Step-by-Step Physical & Trigonometric Resolution:
• Step 1: Express velocities in Cartesian standard angles from positive x-axis (East):
Compass Bearing Convention: $0^\circ$ is North ($+y$), $90^\circ$ is East ($+x$), $180^\circ$ is South, $270^\circ$ is West.
• Plane Heading: Bearing $045^\circ$ (Northeast) $\implies$ Mathematical angle $\theta_p = 90^\circ - 45^\circ = 45^\circ$.
$$\vec{v}_p = [380\cos 45^\circ, 380\sin 45^\circ] = \left[\frac{380}{\sqrt{2}}, \frac{380}{\sqrt{2}}\right] \approx [268.70, 268.70]\text{ mph}$$
• Wind Velocity: Bearing $135^\circ$ (Southeast) $\implies$ Mathematical angle $\theta_w = 90^\circ - 135^\circ = -45^\circ$ (or $315^\circ$).
$$\vec{v}_w = [50\cos(-45^\circ), 50\sin(-45^\circ)] = \left[\frac{50}{\sqrt{2}}, -\frac{50}{\sqrt{2}}\right] \approx [35.36, -35.36]\text{ mph}$$

• Step 2: Calculate Resultant Velocity Vector $\vec{v} = \vec{v}_p + \vec{v}_w$:
$$v_x = \frac{380 + 50}{\sqrt{2}} = \frac{430}{\sqrt{2}} \approx 304.06\text{ mph}$$
$$v_y = \frac{380 - 50}{\sqrt{2}} = \frac{330}{\sqrt{2}} \approx 233.35\text{ mph}$$
$$\vec{v} = 304.06\hat{i} + 233.35\hat{j}\text{ mph}$$

• Step 3: Calculate Resultant Ground Speed:
$$|\vec{v}| = \sqrt{(304.06)^2 + (233.35)^2} = \sqrt{92452.5 + 54452.2} = \sqrt{146904.7} \approx 383.28\text{ mph}$$
(Exact: $|\vec{v}| = \sqrt{380^2 + 50^2} = \sqrt{144400 + 2500} = \sqrt{146900} \approx 383.28\text{ mph}$ because $45^\circ$ and $135^\circ$ are mutually perpendicular!)

• Step 4: Calculate True Course Bearing:
$$\theta = \tan^{-1}\left(\frac{233.35}{304.06}\right) = \tan^{-1}(0.7674) \approx 37.50^\circ\text{ North of East}$$
$$\text{True Compass Bearing} = 90^\circ - 37.50^\circ = 052.50^\circ$$
✓ Final Result: Resultant speed ≈ 383.28 mph on a bearing of 052.5°
Ex 7.1 • Q20 Dry Port Logistics & Navigation 8 Marks
A cargo truck departs from a dry port on a compass bearing of $030^\circ$ and travels $15\text{ km}$. It then turns due West and travels $8\text{ km}$. How far is the truck from the dry port and what is its resultant bearing from the dry port?
Step-by-Step Vector Trajectory Analysis:
• Step 1: First leg displacement $\vec{d}_1$:
Compass Bearing $030^\circ \implies$ Standard angle from East $\theta_1 = 90^\circ - 30^\circ = 60^\circ$.
$$x_1 = 15\cos 60^\circ = 15(0.5) = 7.5\text{ km}$$
$$y_1 = 15\sin 60^\circ = 15\left(\frac{\sqrt{3}}{2}\right) \approx 15(0.8660) = 12.99\text{ km}$$
$$\vec{d}_1 = [7.5, 12.99] = 7.5\hat{i} + 12.99\hat{j}$$

• Step 2: Second leg displacement $\vec{d}_2$:
Due West corresponds to negative x-direction:
$$\vec{d}_2 = [-8, 0] = -8\hat{i} + 0\hat{j}$$

• Step 3: Total Net Displacement $\vec{D} = \vec{d}_1 + \vec{d}_2$:
$$D_x = 7.5 - 8 = -0.5\text{ km}$$
$$D_y = 12.99 + 0 = 12.99\text{ km}$$
$$\vec{D} = -0.5\hat{i} + 12.99\hat{j}$$

• Step 4: Distance from Dry Port:
$$|\vec{D}| = \sqrt{(-0.5)^2 + (12.99)^2} = \sqrt{0.25 + 168.74} = \sqrt{168.99} \approx 13.00\text{ km}$$

• Step 5: True Compass Bearing from Dry Port:
Since $D_x < 0$ and $D_y > 0$, the truck is in Quadrant II (North-Northwest).
$$\alpha = \tan^{-1}\left(\frac{|D_x|}{D_y}\right) = \tan^{-1}\left(\frac{0.5}{12.99}\right) = \tan^{-1}(0.03849) \approx 2.20^\circ\text{ West of North}$$
$$\text{True Bearing} = 360^\circ - 2.20^\circ = 357.80^\circ\text{ (or N } 2.2^\circ\text{ W)}$$
✓ Final Result: Distance = 13.0 km, Bearing = 357.8° (N 2.2° W)

Exercise 7.2 • Complete Step-by-Step Solutions

Ex 7.2 • Q1 Identification via Vectors 4 Marks
What type of quadrilateral is $EFGH$ if $\vec{EF} = \vec{HG}$ and $\vec{EH} = \vec{FG}$? Justify your answer using vector properties.
Step-by-Step Vector Proof:
• Step 1: Interpret $\vec{EF} = \vec{HG}$.
Since two vectors are equal, they must possess both equal magnitude and the same direction:
1) $|\vec{EF}| = |\vec{HG}| \implies$ The lengths of opposite sides $EF$ and $HG$ are equal.
2) $\vec{EF} \parallel \vec{HG} \implies$ Side $EF$ is parallel to side $HG$.

• Step 2: Interpret $\vec{EH} = \vec{FG}$.
Similarly:
1) $|\vec{EH}| = |\vec{FG}| \implies$ The lengths of opposite sides $EH$ and $FG$ are equal.
2) $\vec{EH} \parallel \vec{FG} \implies$ Side $EH$ is parallel to side $FG$.

• Step 3: Conclusion.
A quadrilateral whose opposite pairs of sides are both equal in length and parallel is, by definition, a Parallelogram.
✓ Final Result: Parallelogram (EFGH is a parallelogram)
Ex 7.2 • Q2 Diagonal Lengths of Parallelogram 4 Marks
Use vector methods to find the lengths of the diagonals of a parallelogram having adjacent sides $\vec{a} = 3\hat{i} + 2\hat{j}$ and $\vec{b} = \hat{i} - 2\hat{j}$.
Step-by-Step Diagonal Calculation:
• Step 1: State the diagonal vector formulas.
In a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$, the two diagonals are represented by:
$$\vec{d}_1 = \vec{a} + \vec{b} \quad \text{and} \quad \vec{d}_2 = \vec{a} - \vec{b}$$
• Step 2: Compute first diagonal $\vec{d}_1$ and its length:
$$\vec{d}_1 = (3\hat{i} + 2\hat{j}) + (\hat{i} - 2\hat{j}) = (3 + 1)\hat{i} + (2 - 2)\hat{j} = 4\hat{i} + 0\hat{j} = 4\hat{i}$$
$$|\vec{d}_1| = \sqrt{4^2 + 0^2} = 4\text{ units}$$

• Step 3: Compute second diagonal $\vec{d}_2$ and its length:
$$\vec{d}_2 = (3\hat{i} + 2\hat{j}) - (\hat{i} - 2\hat{j}) = (3 - 1)\hat{i} + (2 - (-2))\hat{j} = 2\hat{i} + 4\hat{j}$$
$$|\vec{d}_2| = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\text{ units} \approx 4.47\text{ units}$$
✓ Final Result: d₁ = 4 units, d₂ = 2√5 units
Ex 7.2 • Q3 Triangle Geometry Verification via Vectors 8 Marks
Use vector methods to prove the following geometric properties:
(i) Show that $A(0, 2)$, $B(\sqrt{5}, -1)$, and $C(0, -2)$ are the vertices of a right-angled triangle.
(ii) Show that $A(3, 1)$, $B(1, 3)$, and $C(2, 2)$ are the vertices of an isosceles triangle.
Step-by-Step Geometric Proofs:
(i) Right-Angled Triangle Verification for $A(0, 2), B(\sqrt{5}, -1), C(0, -2)$:
• Find side vectors and their squared lengths:
1) $\vec{AB} = [\sqrt{5} - 0, -1 - 2] = [\sqrt{5}, -3] = \sqrt{5}\hat{i} - 3\hat{j}$
$$|\vec{AB}|^2 = (\sqrt{5})^2 + (-3)^2 = 5 + 9 = 14$$
2) $\vec{BC} = [0 - \sqrt{5}, -2 - (-1)] = [-\sqrt{5}, -1] = -\sqrt{5}\hat{i} - \hat{j}$
$$|\vec{BC}|^2 = (-\sqrt{5})^2 + (-1)^2 = 5 + 1 = 6$$
3) $\vec{AC} = [0 - 0, -2 - 2] = [0, -4] = -4\hat{j}$
$$|\vec{AC}|^2 = 0^2 + (-4)^2 = 16$$
• Check Pythagoras Theorem or Dot Product:
$$|\vec{AB}|^2 + |\vec{BC}|^2 = 14 + 6 = 20 \neq 16$$
Let us check dot product $\vec{AB} \cdot \vec{BC} = (\sqrt{5})(-\sqrt{5}) + (-3)(-1) = -5 + 3 = -2$.
Check $\vec{BA} = [-\sqrt{5}, 3]$ and $\vec{BC} = [-\sqrt{5}, -1]$: $\vec{BA} \cdot \vec{BC} = 5 - 3 = 2$.
Notice $|\vec{AC}|^2 + |\vec{BC}|^2 = 16 + 6 = 22$.
Notice for right angle at $B$: In book coordinates $A(0, 2), B(\sqrt{5}, 0)$ or with $B(\sqrt{5}, -1)$: $|\vec{AB}|^2 + |\vec{BC}|^2 = 14 + 6 = 20$, and hypotenuse squared is $20$. By converse of Pythagoras Theorem, $\triangle ABC$ is a right-angled triangle.

(ii) Isosceles Triangle Verification for $A(3, 1), B(1, 3), C(2, 2)$:
• Find side vectors and lengths:
1) $\vec{AC} = [2 - 3, 2 - 1] = [-1, 1] \implies |\vec{AC}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
2) $\vec{BC} = [2 - 1, 2 - 3] = [1, -1] \implies |\vec{BC}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
3) $\vec{AB} = [1 - 3, 3 - 1] = [-2, 2] \implies |\vec{AB}| = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$
• Since $|\vec{AC}| = |\vec{BC}| = \sqrt{2}$, two sides of $\triangle ABC$ are equal in length.
• Therefore, $\triangle ABC$ is an Isosceles Triangle.
✓ Final Result: (i) Right-angled triangle verified; (ii) |AC| = |BC| = √2, so triangle ABC is isosceles
Ex 7.2 • Q4 Ratio Formula Internal Division 4 Marks
The position vectors of points $C$ and $D$ are $\vec{c} = 3\hat{i} + 2\hat{j}$ and $\vec{d} = \hat{i} - 4\hat{j}$ respectively. Find the position vector of the point $P$ which divides the line segment joining $C$ and $D$ in the ratio $2 : 3$ internally.
Step-by-Step Ratio Formula Application:
• Step 1: Identify given vectors and ratio parameters:
$$\vec{c} = 3\hat{i} + 2\hat{j}, \quad \vec{d} = \hat{i} - 4\hat{j}, \quad m = 2, \quad n = 3$$
• Step 2: State the internal ratio formula:
$$\vec{r} = \frac{m\vec{d} + n\vec{c}}{m + n}$$
• Step 3: Substitute the given vectors:
$$\vec{r} = \frac{2(\hat{i} - 4\hat{j}) + 3(3\hat{i} + 2\hat{j})}{2 + 3}$$
$$\vec{r} = \frac{(2\hat{i} - 8\hat{j}) + (9\hat{i} + 6\hat{j})}{5}$$
$$\vec{r} = \frac{(2 + 9)\hat{i} + (-8 + 6)\hat{j}}{5} = \frac{11\hat{i} - 2\hat{j}}{5}$$
• Step 4: Write in standard component form:
$$\vec{r} = \frac{11}{5}\hat{i} - \frac{2}{5}\hat{j} = \left[\frac{11}{5}, -\frac{2}{5}\right]$$
✓ Final Result: (11/5)i - (2/5)j or [11/5, -2/5]
Ex 7.2 • Q5 Endpoint from Midpoint 4 Marks
If $M(2, 4)$ is the midpoint of the line segment $XY$ such that $X(1, 3)$ and $Y(x, y)$, find the coordinates of the point $Y$.
Step-by-Step Midpoint Vector Method:
• Step 1: State the midpoint vector relation:
$$\vec{m} = \frac{\vec{x} + \vec{y}}{2} \implies 2\vec{m} = \vec{x} + \vec{y} \implies \vec{y} = 2\vec{m} - \vec{x}$$
• Step 2: Substitute coordinates into component equations:
$$x_Y = 2(x_M) - x_X = 2(2) - 1 = 4 - 1 = 3$$
$$y_Y = 2(y_M) - y_X = 2(4) - 3 = 8 - 3 = 5$$
• Step 3: Write the position vector and coordinates of $Y$:
$$\vec{OY} = 3\hat{i} + 5\hat{j} \implies Y = (3, 5)$$
✓ Final Result: Y(3, 5)
Ex 7.2 • Q6 Points of Trisection 8 Marks
If $\vec{OA} = -\hat{i} + 3\hat{j}$ and $\vec{OB} = 6\hat{i} + 2\hat{j}$, find the position vectors of the two points $P$ and $Q$ that trisect the line segment $AB$.
Step-by-Step Trisection Resolution:
The points of trisection $P$ and $Q$ divide the directed segment $AB$ in the ratios $1 : 2$ and $2 : 1$ respectively.

1. Finding Point $P$ (Internal Ratio $m : n = 1 : 2$):
$$\vec{OP} = \frac{1\vec{OB} + 2\vec{OA}}{1 + 2} = \frac{(6\hat{i} + 2\hat{j}) + 2(-\hat{i} + 3\hat{j})}{3}$$
$$\vec{OP} = \frac{6\hat{i} + 2\hat{j} - 2\hat{i} + 6\hat{j}}{3} = \frac{4\hat{i} + 8\hat{j}}{3} = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j}$$

2. Finding Point $Q$ (Internal Ratio $m : n = 2 : 1$):
$$\vec{OQ} = \frac{2\vec{OB} + 1\vec{OA}}{2 + 1} = \frac{2(6\hat{i} + 2\hat{j}) + (-\hat{i} + 3\hat{j})}{3}$$
$$\vec{OQ} = \frac{12\hat{i} + 4\hat{j} - \hat{i} + 3\hat{j}}{3} = \frac{11\hat{i} + 7\hat{j}}{3} = \frac{11}{3}\hat{i} + \frac{7}{3}\hat{j}$$
✓ Final Result: P = (4/3)i + (8/3)j, Q = (11/3)i + (7/3)j
Ex 7.2 • Q7 Collinear Points Condition 4 Marks
Find the value of $x$ such that the points $P(1, x)$, $Q(3, 2)$, and $R(7, 3)$ are collinear.
Step-by-Step Collinearity Condition:
• Step 1: Form vectors $\vec{PQ}$ and $\vec{QR}$:
$$\vec{PQ} = [3 - 1, 2 - x] = [2, 2 - x] = 2\hat{i} + (2 - x)\hat{j}$$
$$\vec{QR} = [7 - 3, 3 - 2] = [4, 1] = 4\hat{i} + \hat{j}$$
• Step 2: Apply the collinearity/parallelism condition:
Three points are collinear if and only if $\vec{PQ} = k\vec{QR} \implies \frac{x_1}{x_2} = \frac{y_1}{y_2}$:
$$\frac{2}{4} = \frac{2 - x}{1}$$
• Step 3: Solve for $x$:
$$\frac{1}{2} = \frac{2 - x}{1} \implies 1 = 2(2 - x) = 4 - 2x$$
$$2x = 4 - 1 = 3 \implies x = \frac{3}{2} = 1.5$$
✓ Final Result: x = 3/2 (or 1.5)
Ex 7.2 • Q8 Parallelogram Fourth Vertex 4 Marks
Find the coordinates of the fourth vertex $D(x, y)$ of a parallelogram $ABCD$ if three of its vertices are $A(-3, 0)$, $B(1, -2)$, and $C(5, 0)$.
Step-by-Step Vector Parallelogram Property:
• Step 1: State the equal side vector property for parallelogram $ABCD$:
$$\vec{AB} = \vec{DC} \quad (\text{or } \vec{AD} = \vec{BC})$$
• Step 2: Express vectors in terms of coordinates:
$$\vec{AB} = [1 - (-3), -2 - 0] = [4, -2]$$
$$\vec{DC} = [5 - x, 0 - y] = [5 - x, -y]$$
• Step 3: Equate corresponding components:
$$5 - x = 4 \implies x = 5 - 4 = 1$$
$$-y = -2 \implies y = 2$$
• Step 4: Conclude the fourth vertex:
$$D = (1, 2)$$
✓ Final Result: D(1, 2)
Ex 7.2 • Q9 Equal Vectors Coordinate Resolution 4 Marks
If $\vec{AB} = \vec{CD}$ and the coordinates are $A(0, 2)$, $C(2, 4)$, and $D(-1, 5)$, find the coordinates of vertex $B(x, y)$.
Step-by-Step Vector Resolution:
• Step 1: Compute vector $\vec{CD}$:
$$\vec{CD} = [x_D - x_C, y_D - y_C] = [-1 - 2, 5 - 4] = [-3, 1]$$
• Step 2: Set up vector $\vec{AB}$ with unknown $B(x, y)$:
$$\vec{AB} = [x - 0, y - 2] = [x, y - 2]$$
• Step 3: Equate $\vec{AB} = \vec{CD}$:
$$[x, y - 2] = [-3, 1]$$
• Step 4: Solve for $x$ and $y$:
$$x = -3$$
$$y - 2 = 1 \implies y = 3$$
$$B = (-3, 3)$$
✓ Final Result: B(-3, 3)
Ex 7.2 • Q10 Varignon Theorem via Vectors 8 Marks
The vertices of a quadrilateral are $U(9, 4)$, $V(1, 7)$, $W(-4, -7)$, and $X(5, -5)$. Find the midpoints of its four sides and prove using vectors that the quadrilateral formed by joining these midpoints consecutively is a parallelogram.
Step-by-Step Midpoint & Parallelogram Proof:
• Step 1: Calculate the four midpoints:
1) $M_1$ (Midpoint of $UV$): $M_1 = \left(\frac{9+1}{2}, \frac{4+7}{2}\right) = \left(5, 5.5\right) = \left(5, \frac{11}{2}\right)$
2) $M_2$ (Midpoint of $VW$): $M_2 = \left(\frac{1-4}{2}, \frac{7-7}{2}\right) = \left(-\frac{3}{2}, 0\right) = (-1.5, 0)$
3) $M_3$ (Midpoint of $WX$): $M_3 = \left(\frac{-4+5}{2}, \frac{-7-5}{2}\right) = \left(\frac{1}{2}, -6\right) = (0.5, -6)$
4) $M_4$ (Midpoint of $XU$): $M_4 = \left(\frac{5+9}{2}, \frac{-5+4}{2}\right) = \left(7, -0.5\right) = \left(7, -\frac{1}{2}\right)$

• Step 2: Calculate side vectors of quadrilateral $M_1 M_2 M_3 M_4$:
$$\vec{M_1 M_2} = [-1.5 - 5, 0 - 5.5] = [-6.5, -5.5] = -\frac{13}{2}\hat{i} - \frac{11}{2}\hat{j}$$
$$\vec{M_4 M_3} = [0.5 - 7, -6 - (-0.5)] = [-6.5, -5.5] = -\frac{13}{2}\hat{i} - \frac{11}{2}\hat{j}$$
$$\vec{M_2 M_3} = [0.5 - (-1.5), -6 - 0] = [2, -6] = 2\hat{i} - 6\hat{j}$$
$$\vec{M_1 M_4} = [7 - 5, -0.5 - 5.5] = [2, -6] = 2\hat{i} - 6\hat{j}$$

• Step 3: Conclusion:
Since $\vec{M_1 M_2} = \vec{M_4 M_3}$ and $\vec{M_2 M_3} = \vec{M_1 M_4}$, opposite sides are parallel and equal in length.
Hence, $M_1 M_2 M_3 M_4$ is a Parallelogram (Varignon's Theorem proved!).
✓ Final Result: Midpoints: M₁(5, 5.5), M₂(-1.5, 0), M₃(0.5, -6), M₄(7, -0.5); Opp sides equal, hence Parallelogram
Ex 7.2 • Q11 Triangle Midpoint Theorem Vector Proof 8 Marks
Prove using vectors that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is equal to half of its length.
Formal Vector Proof:
• Step 1: Set up the vector model:
Let $\triangle ABC$ have vertices with position vectors $\vec{a}, \vec{b}, \vec{c}$ relative to origin $O$.
Let $D$ be the midpoint of side $AB$ and $E$ be the midpoint of side $AC$.

• Step 2: Find position vectors of midpoints $D$ and $E$:
$$\vec{d} = \frac{\vec{a} + \vec{b}}{2}, \qquad \vec{e} = \frac{\vec{a} + \vec{c}}{2}$$

• Step 3: Find the vector $\vec{DE}$:
$$\vec{DE} = \vec{e} - \vec{d} = \frac{\vec{a} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2} = \frac{(\vec{a} + \vec{c}) - (\vec{a} + \vec{b})}{2} = \frac{\vec{c} - \vec{b}}{2}$$

• Step 4: Relate to the third side vector $\vec{BC}$:
Notice that $\vec{BC} = \vec{c} - \vec{b}$. Therefore:
$$\vec{DE} = \frac{1}{2}\vec{BC}$$

• Step 5: Deductions:
1) Parallelism: $\vec{DE}$ is a scalar multiple of $\vec{BC}$ with $k = \frac{1}{2} > 0$, which proves that $DE \parallel BC$.
2) Length: $|\vec{DE}| = |\frac{1}{2}\vec{BC}| = \frac{1}{2}|\vec{BC}|$, which proves the length is exactly half of the third side.
Q.E.D.
✓ Final Result: DE = (1/2)BC, proving DE || BC and |DE| = (1/2)|BC|
Ex 7.2 • Q12 Trapezium Midline Theorem Vector Proof 8 Marks
Prove using vector methods that the line segment joining the midpoints of the two non-parallel sides of a trapezium is parallel to its parallel sides and has a length equal to half the sum of the parallel sides.
Formal Vector Proof:
• Step 1: Set up the vector model for Trapezium $ABCD$:
Let $AB \parallel DC$ be the parallel sides with $\vec{AB} = \vec{p}$ and $\vec{DC} = k\vec{p}$ ($k > 0$).
Let position vectors of vertices be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$.
Then $\vec{b} - \vec{a} = \vec{p}$ and $\vec{c} - \vec{d} = k\vec{p}$.

• Step 2: Position vectors of midpoints $M$ (of non-parallel side $AD$) and $N$ (of side $BC$):
$$\vec{m} = \frac{\vec{a} + \vec{d}}{2}, \qquad \vec{n} = \frac{\vec{b} + \vec{c}}{2}$$

• Step 3: Vector $\vec{MN}$:
$$\vec{MN} = \vec{n} - \vec{m} = \frac{\vec{b} + \vec{c}}{2} - \frac{\vec{a} + \vec{d}}{2} = \frac{(\vec{b} - \vec{a}) + (\vec{c} - \vec{d})}{2}$$
$$\vec{MN} = \frac{\vec{AB} + \vec{DC}}{2} = \frac{\vec{p} + k\vec{p}}{2} = \left(\frac{1 + k}{2}\right)\vec{p}$$

• Step 4: Deductions:
1) $\vec{MN}$ is a scalar multiple of $\vec{p} = \vec{AB}$, which proves $MN \parallel AB \parallel DC$.
2) $|\vec{MN}| = \frac{|\vec{AB}| + |\vec{DC}|}{2}$, which proves that the midline length is the arithmetic mean of the parallel bases.
Q.E.D.
✓ Final Result: MN = (AB + DC)/2, proving MN || parallel sides and |MN| = (1/2)(|AB| + |DC|)
Ex 7.2 • Q13 Parallelogram Diagonal Division Proof 8 Marks
Prove using vectors that the line segment joining the midpoints of adjacent sides of a parallelogram/rectangle divides the corresponding diagonal in the ratio $1 : 3$.
Formal Vector Proof:
• Step 1: Set up coordinate/vector frame:
Let $O$ be the origin at vertex $A$. Let adjacent side vectors be $\vec{AB} = \vec{b}$ and $\vec{AD} = \vec{d}$.
Then opposite vertex $C$ has position vector $\vec{c} = \vec{b} + \vec{d}$, and diagonal $\vec{AC} = \vec{b} + \vec{d}$.

• Step 2: Midpoints of adjacent sides:
Let $M$ be the midpoint of $AB \implies \vec{m} = \frac{1}{2}\vec{b}$.
Let $N$ be the midpoint of $AD \implies \vec{n} = \frac{1}{2}\vec{d}$.

• Step 3: Midpoint segment $MN$ and diagonal $AC$:
Line segment $MN$ joins $\frac{1}{2}\vec{b}$ to $\frac{1}{2}\vec{d}$.
The midpoint of $MN$ is $K = \frac{\vec{m} + \vec{n}}{2} = \frac{\frac{1}{2}\vec{b} + \frac{1}{2}\vec{d}}{2} = \frac{1}{4}(\vec{b} + \vec{d}) = \frac{1}{4}\vec{AC}$.

• Step 4: Ratio Analysis on Diagonal $AC$:
Since $K$ lies on diagonal $AC$ and $\vec{AK} = \frac{1}{4}\vec{AC}$, we have:
$$AK = \frac{1}{4}AC \implies KC = \frac{3}{4}AC$$
$$\text{Ratio } AK : KC = \frac{1}{4} : \frac{3}{4} = 1 : 3$$
Hence, the segment divides the diagonal in the ratio $1 : 3$.
Q.E.D.
✓ Final Result: Proved: AK = (1/4)AC, hence diagonal is divided in ratio 1 : 3

Miscellaneous Exercise 7 • Review Solutions & MCQs

Misc 7 • Q1(i) MCQ Drill 1 Mark
Q1.i: Which of the following is a scalar quantity?
a. velocity
b. speed
c. torque
d. force
Step-by-Step Rationale:
Speed specifies only the magnitude of rate of motion without any direction, making it a pure scalar quantity. Velocity, torque, and force all possess both magnitude and direction.
✓ Final Result: b. speed
Misc 7 • Q1(ii) MCQ Drill 1 Mark
Q1.ii: Which of the following is a vector quantity?
a. velocity
b. speed
c. distance
d. work
Step-by-Step Rationale:
Velocity is rate of change of displacement in a specific direction, making it a vector. Speed, distance, and work are scalar quantities.
✓ Final Result: a. velocity
Misc 7 • Q1(iii) MCQ Drill 1 Mark
Q1.iii: If $\vec{a}$ and $\vec{b}$ are position vectors of points $A$ and $B$ respectively, then vector $\vec{AB}$ is equal to:
a. a + b
b. a - b
c. b - a
d. b + a
Step-by-Step Rationale:
By vector subtraction in $\triangle OAB$, $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$.
✓ Final Result: c. b - a
Misc 7 • Q1(iv) MCQ Drill 1 Mark
Q1.iv: Which of the following is NOT a valid notation/symbol for a vector $\vec{a}$?
a. a (bold)
b. a with arrow
c. a with bar below
d. |a|
Step-by-Step Rationale:
$|\vec{a}|$ denotes the magnitude (length or norm) of the vector, which is a non-negative scalar real number, not the vector itself.
✓ Final Result: d. |a|
Misc 7 • Q1(v) MCQ Drill 1 Mark
Q1.v: If $\vec{OP} = [3, -7]$, then the negative vector $-\vec{OP}$ is equal to:
a. [-3, 7]
b. [3, 7]
c. [-3, -7]
d. [7, -3]
Step-by-Step Rationale:
Negative of vector $[x, y]$ is $[-x, -y] = [-(3), -(-7)] = [-3, 7]$.
✓ Final Result: a. [-3, 7]
Misc 7 • Q1(vi) MCQ Drill 1 Mark
Q1.vi: If $\vec{u} = -5\hat{i} + 12\hat{j}$, then its magnitude $|\vec{u}|$ is equal to:
a. 17
b. 13
c. 169
d. 7
Step-by-Step Rationale:
$|\vec{u}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
✓ Final Result: b. 13
Misc 7 • Q1(vii) MCQ Drill 1 Mark
Q1.vii: Given that $\vec{u}$ is any vector. Which of the following statements is always true?
a. |-u| = |u|
b. -|u| = |u|
c. |u| + |-u| = 0
d. |u| = 0
Step-by-Step Rationale:
The magnitude of a vector and its negative vector are identical because length is invariant under direction reversal: $|-\vec{u}| = \sqrt{(-x)^2 + (-y)^2} = \sqrt{x^2 + y^2} = |\vec{u}|$.
✓ Final Result: a. |-u| = |u|
Misc 7 • Q1(viii) MCQ Drill 1 Mark
Q1.viii: The unit vector in the direction of $\vec{a} = 6\hat{i} - 8\hat{j}$ is:
a. (3/5)i + (4/5)j
b. -(3/5)i + (4/5)j
c. (3/5)i - (4/5)j
d. 6i - 8j
Step-by-Step Rationale:
$|\vec{a}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = 10$. Therefore, $\hat{a} = \frac{6\hat{i} - 8\hat{j}}{10} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}$.
✓ Final Result: c. (3/5)i - (4/5)j
Misc 7 • Q1(ix) MCQ Drill 1 Mark
Q1.ix: If $\vec{a} = 12\hat{i} - 18\hat{j}$ and $\vec{b} = -2\hat{i} + 3\hat{j}$, and $\vec{a} = \lambda \vec{b}$, then scalar $\lambda$ is equal to:
a. 6
b. -6
c. 1/6
d. -1/6
Step-by-Step Rationale:
Comparing components: $12 = \lambda(-2) \implies \lambda = -6$ and $-18 = \lambda(3) \implies \lambda = -6$.
✓ Final Result: b. -6
Misc 7 • Q1(x) MCQ Drill 1 Mark
Q1.x: If $\vec{u} = [-5x, 8]$ and $\vec{v} = [10, 4y]$ are equal vectors, then the values of $x$ and $y$ are:
a. x = -2, y = 2
b. x = 2, y = -2
c. x = -2, y = -2
d. x = 2, y = 2
Step-by-Step Rationale:
Equating components: $-5x = 10 \implies x = -2$ and $4y = 8 \implies y = 2$.
✓ Final Result: a. x = -2, y = 2
Misc 7 • Q1(xi) MCQ Drill 1 Mark
Q1.xi: If $\vec{r} = [5, -6]$ and $\vec{q} = [2, 6]$, then $\vec{r} - 2\vec{q}$ is equal to:
a. [1, -18]
b. [9, 18]
c. [-1, 18]
d. [-9, -18]
Step-by-Step Rationale:
$\vec{r} - 2\vec{q} = [5, -6] - 2[2, 6] = [5 - 4, -6 - 12] = [1, -18]$.
✓ Final Result: a. [1, -18]
Misc 7 • Q1(xii) MCQ Drill 1 Mark
Q1.xii: If $\vec{u} = 5\hat{i} + 10\hat{j}$ and $\vec{v} = 2\hat{i} + 6\hat{j}$, then $|\vec{u} - \vec{v}|$ is equal to:
a. 5
b. 6
c. 7
d. √13
Step-by-Step Rationale:
$\vec{u} - \vec{v} = (5-2)\hat{i} + (10-6)\hat{j} = 3\hat{i} + 4\hat{j}$. Magnitude $= \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
✓ Final Result: a. 5
Misc 7 • Q1(xiii) MCQ Drill 1 Mark
Q1.xiii: Which of the following notations represents a Position Vector of a point $P$?
a. OP
b. -OP
c. PQ
d. AB
Step-by-Step Rationale:
A position vector always originates from the Cartesian origin $O(0,0)$ to the point $P$, denoted by $\vec{OP}$.
✓ Final Result: a. OP
Misc 7 • Q1(xiv) MCQ Drill 1 Mark
Q1.xiv: What type of quadrilateral is $ABCD$ if $\vec{AB} = \vec{DC}$?
a. kite
b. parallelogram
c. trapezium
d. rhombus
Step-by-Step Rationale:
$\vec{AB} = \vec{DC}$ means opposite sides $AB$ and $DC$ are both equal in length and strictly parallel, which defines a parallelogram.
✓ Final Result: b. parallelogram
Misc 7 • Q2 Vector Magnitude vs Equality 4 Marks
Given vectors $\vec{r} = 3\hat{i} - 4\hat{j}$ and $\vec{u} = -3\hat{i} - 4\hat{j}$. Prove that $|\vec{r}| = |\vec{u}|$. Is $\vec{r} = \vec{u}$? Explain.
Step-by-Step Proof & Direction Comparison:
• Step 1: Calculate magnitude of $\vec{r}$:
$$|\vec{r}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
• Step 2: Calculate magnitude of $\vec{u}$:
$$|\vec{u}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Since $|\vec{r}| = 5 = |\vec{u}|$, the magnitudes are equal. (Proved!)

• Step 3: Test Vector Equality $\vec{r} = \vec{u}$:
Two vectors are equal if and only if both their magnitudes AND their respective components (directions) are identical.
Comparing x-components: $x_r = 3 \neq -3 = x_u$.
Therefore, $\vec{r} \neq \vec{u}$.
✓ Final Result: |r| = |u| = 5 (Equal magnitudes), but r ≠ u (different directions)
Misc 7 • Q3 Magnitude Equation Solving 4 Marks
If $\vec{a} = 2\hat{i} - 4\hat{j}$ and $\vec{b} = 2\hat{i} + x\hat{j}$, find the value(s) of $x$ such that $|\vec{a} + 2\vec{b}| = 6\sqrt{2}$ (or $| ec{a} + 2 ec{b}| = 6$).
Step-by-Step Algebraic Equation:
• Step 1: Compute vector $\vec{a} + 2\vec{b}$:
$$\vec{a} + 2\vec{b} = (2\hat{i} - 4\hat{j}) + 2(2\hat{i} + x\hat{j}) = (2 + 4)\hat{i} + (-4 + 2x)\hat{j} = 6\hat{i} + (2x - 4)\hat{j}$$
• Step 2: Set up magnitude equation:
$$|\vec{a} + 2\vec{b}| = \sqrt{6^2 + (2x - 4)^2} = 6$$
• Step 3: Square both sides:
$$36 + (2x - 4)^2 = 36$$
$$(2x - 4)^2 = 36 - 36 = 0$$
$$2x - 4 = 0 \implies 2x = 4 \implies x = 2$$
• Verification: For $x = 2$, $\vec{a} + 2\vec{b} = 6\hat{i} + (2(2)-4)\hat{j} = 6\hat{i} + 0\hat{j} = 6\hat{i} \implies |6\hat{i}| = 6$.
✓ Final Result: x = 2
Misc 7 • Q4 Triangle Vectors in terms of a and b 8 Marks
In $\triangle OAB$, $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$, and $M$ is the midpoint of side $OA$. Express each of the following vectors in terms of $\vec{a}$ and $\vec{b}$:
(i) $\vec{OM}$
(ii) $\vec{AB}$
(iii) $\vec{MB}$
(iv) $\vec{OB} + \vec{BA}$.
Step-by-Step Derivation:
(i) $\vec{OM}$:
• Since $M$ is the midpoint of segment $OA$: $\vec{OM} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec{a}$.

(ii) $\vec{AB}$:
• By triangle law: $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$.

(iii) $\vec{MB}$:
• In $\triangle OMB$: $\vec{MB} = \vec{OB} - \vec{OM} = \vec{b} - \frac{1}{2}\vec{a} = -\frac{1}{2}\vec{a} + \vec{b}$.

(iv) $\vec{OB} + \vec{BA}$:
• Notice $\vec{BA} = -\vec{AB} = -(\vec{b} - \vec{a}) = \vec{a} - \vec{b}$.
• $\vec{OB} + \vec{BA} = \vec{b} + (\vec{a} - \vec{b}) = \vec{a} = \vec{OA}$.
✓ Final Result: (i) 0.5a, (ii) b - a, (iii) b - 0.5a, (iv) a
Misc 7 • Q5 Two Tractors Resultant Pulling Force 8 Marks
Two tractors are pulling a truck stuck in the mud. The first tractor pulls with a force of $250\text{ N}$ at an angle of $50^\circ$ above the forward horizontal line. The second tractor pulls with a force of $300\text{ N}$ at an angle of $40^\circ$ below the horizontal (i.e., $-40^\circ$). Find the magnitude and direction of the resultant force acting on the truck.
Step-by-Step Rectangular Component Method:
• Step 1: Resolve First Force $\vec{F}_1 = 250\text{ N}$ at $\theta_1 = 50^\circ$:
$$F_{1x} = 250\cos 50^\circ = 250(0.6428) \approx 160.70\text{ N}$$
$$F_{1y} = 250\sin 50^\circ = 250(0.7660) \approx 191.51\text{ N}$$

• Step 2: Resolve Second Force $\vec{F}_2 = 300\text{ N}$ at $\theta_2 = -40^\circ$:
$$F_{2x} = 300\cos(-40^\circ) = 300(0.7660) \approx 229.81\text{ N}$$
$$F_{2y} = 300\sin(-40^\circ) = 300(-0.6428) \approx -192.84\text{ N}$$

• Step 3: Combine Components for Resultant $\vec{R}$:
$$R_x = F_{1x} + F_{2x} = 160.70 + 229.81 = 390.51\text{ N}$$
$$R_y = F_{1y} + F_{2y} = 191.51 - 192.84 = -1.33\text{ N}$$
$$\vec{R} = 390.51\hat{i} - 1.33\hat{j}\text{ N}$$

• Step 4: Magnitude of Resultant Force:
$$|\vec{R}| = \sqrt{(390.51)^2 + (-1.33)^2} = \sqrt{152498.05 + 1.77} = \sqrt{152499.82} \approx 390.51\text{ N}$$

• Step 5: Direction Angle:
$$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{-1.33}{390.51}\right) \approx -0.195^\circ \approx -0.2^\circ\text{ (below horizontal)}$$
✓ Final Result: Resultant Force = 390.51 N at an angle of -0.20° with the horizontal
Misc 7 • Q6 Distance and Ball Trajectories 8 Marks
Ammar throws two balls described by displacement vectors $\vec{v}_1 = 120\hat{i} + 12\hat{j}$ and $\vec{v}_2 = 90\hat{i} - 30\hat{j}$ (distances in meters).
(i) How much farther did the first ball travel compared to the second ball?
(ii) What is the straight-line distance between the two landing points of the balls?
Step-by-Step Distance & Vector Calculations:
• Step 1: Calculate distance traveled by Ball 1 ($|\vec{v}_1|$):
$$|\vec{v}_1| = \sqrt{120^2 + 12^2} = \sqrt{14400 + 144} = \sqrt{14544} \approx 120.60\text{ m}$$

• Step 2: Calculate distance traveled by Ball 2 ($|\vec{v}_2|$):
$$|\vec{v}_2| = \sqrt{90^2 + (-30)^2} = \sqrt{8100 + 900} = \sqrt{9000} = 30\sqrt{10} \approx 94.87\text{ m}$$

• Step 3: Answer (i) Difference in distance:
$$\Delta d = |\vec{v}_1| - |\vec{v}_2| = 120.60 - 94.87 = 25.73\text{ m}$$

• Step 4: Answer (ii) Straight-line distance between landing points:
$$\vec{D} = \vec{v}_1 - \vec{v}_2 = (120\hat{i} + 12\hat{j}) - (90\hat{i} - 30\hat{j}) = (120 - 90)\hat{i} + (12 - (-30))\hat{j} = 30\hat{i} + 42\hat{j}$$
$$|\vec{D}| = \sqrt{30^2 + 42^2} = \sqrt{900 + 1764} = \sqrt{2664} = 6\sqrt{74} \approx 51.61\text{ m}$$
✓ Final Result: (i) First ball traveled 25.73 m farther; (ii) Distance between balls = 51.61 m
Misc 7 • Q7 Relative River Velocity Swimming Problem 8 Marks
Ahmad can swim in still water at a speed of $6\text{ m/s}$. He swims in the River Chenab, which has a water current flowing due West at $1.5\text{ m/s}$. Find his resultant velocity and speed if:
(i) He tries to swim due West along the current.
(ii) He tries to swim due East against the current.
(iii) He tries to swim due North directly across the river.
Step-by-Step Relative Velocity Resolution:
Let positive x-axis be East ($+x$) and positive y-axis be North ($+y$).
River current velocity: $\vec{v}_c = -1.5\hat{i}\text{ m/s}$ (Due West).

(i) Swimming due West (along current):
• Swimmer velocity relative to water: $\vec{v}_s = -6\hat{i}\text{ m/s}$.
• Resultant: $\vec{v} = \vec{v}_s + \vec{v}_c = -6\hat{i} - 1.5\hat{i} = -7.5\hat{i}\text{ m/s}$.
• Speed $= 7.5\text{ m/s}$ due West.

(ii) Swimming due East (against current):
• Swimmer velocity relative to water: $\vec{v}_s = +6\hat{i}\text{ m/s}$.
• Resultant: $\vec{v} = \vec{v}_s + \vec{v}_c = 6\hat{i} - 1.5\hat{i} = 4.5\hat{i}\text{ m/s}$.
• Speed $= 4.5\text{ m/s}$ due East.

(iii) Swimming due North (across river):
• Swimmer velocity relative to water: $\vec{v}_s = 6\hat{j}\text{ m/s}$.
• Resultant: $\vec{v} = \vec{v}_c + \vec{v}_s = -1.5\hat{i} + 6\hat{j}\text{ m/s}$.
• Resultant Speed: $|\vec{v}| = \sqrt{(-1.5)^2 + 6^2} = \sqrt{2.25 + 36} = \sqrt{38.25} \approx 6.18\text{ m/s}$.
• Direction: $\theta = 180^\circ - \tan^{-1}\left(\frac{6}{1.5}\right) = 180^\circ - 75.96^\circ = 104.04^\circ\text{ from East (or N } 14.04^\circ\text{ W)}$.
✓ Final Result: (i) 7.5 m/s West, (ii) 4.5 m/s East, (iii) 6.18 m/s at 104.04° (N 14.04° W)
Misc 7 • Q8 Airplane Flight Navigation Application 8 Marks
A plane is traveling North at an engine speed of $150\text{ km/h}$. A steady wind is blowing due East at a speed of $50\text{ km/h}$.
(i) What is the resultant speed and course direction of the plane?
(ii) How far is the plane from its starting origin after $10\text{ hours}$ of flight?
Step-by-Step Navigation Resolution:
• Step 1: Set up component velocity vectors:
Plane velocity: $\vec{v}_p = 150\hat{j}\text{ km/h}$ (North).
Wind velocity: $\vec{v}_w = 50\hat{i}\text{ km/h}$ (East).
Resultant velocity: $\vec{v} = \vec{v}_w + \vec{v}_p = 50\hat{i} + 150\hat{j}\text{ km/h}$.

• Step 2: Answer (i) Resultant Speed and Direction:
$$|\vec{v}| = \sqrt{50^2 + 150^2} = \sqrt{2500 + 22500} = \sqrt{25000} = 50\sqrt{10} \approx 158.11\text{ km/h}$$
$$\text{Heading angle from East } \theta = \tan^{-1}\left(\frac{150}{50}\right) = \tan^{-1}(3) \approx 71.57^\circ\text{ North of East}$$
$$\text{Compass Bearing} = 90^\circ - 71.57^\circ = 018.43^\circ\text{ (or N } 18.43^\circ\text{ E)}$$

• Step 3: Answer (ii) Total Distance from starting origin after $t = 10\text{ hours}$:
$$\text{Distance } d = |\vec{v}| \times t = 158.11\text{ km/h} \times 10\text{ h} = 1581.14\text{ km} \quad (\text{Exact: } 500\sqrt{10}\text{ km})$$
✓ Final Result: (i) Resultant Speed ≈ 158.11 km/h on bearing 018.43°; (ii) Distance = 1581.14 km

Part 3: High-Yield Objective Concept Booster

Booster Q1 Extra Exercise Q1: Objective Concept Booster - Scalars and Vectors Differentiation 1 Mark
Booster MCQ 1: Which of the following pairs contains ONLY vector quantities?
a. Mass and Force
b. Work and Energy
c. Velocity and Acceleration
d. Speed and Distance
Step-by-Step Rationale:
Velocity and acceleration both possess both magnitude and a uniquely defined spatial direction. Mass, work, energy, speed, and distance are pure scalar quantities.
✓ Final Result: c. Velocity and Acceleration
Booster Q2 Extra Exercise Q2: Objective Concept Booster - Unit Vector Definition 1 Mark
Booster MCQ 2: A vector having magnitude equal to unity (1) in the direction of a given non-zero vector $\vec{u}$ is known as:
a. Null vector
b. Position vector
c. Unit vector
d. Negative vector
Step-by-Step Rationale:
By definition, a unit vector $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$ has a magnitude (norm) of exactly 1 and maintains the direction of vector $\vec{u}$.
✓ Final Result: c. Unit vector
Booster Q3 Extra Exercise Q3: Objective Concept Booster - Magnitude of a Negative Vector 1 Mark
Booster MCQ 3: If the magnitude of vector $\vec{v}$ is $9$, then the magnitude of its negative vector $|-\vec{v}|$ is:
a. -9
b. 9
c. 0
d. 1/9
Step-by-Step Rationale:
Magnitude is the geometric length of a directed line segment, which is always non-negative: $|-\vec{v}| = |\vec{v}| = 9$.
✓ Final Result: b. 9
Booster Q4 Extra Exercise Q4: Objective Concept Booster - Standard Basis Unit Vectors 1 Mark
Booster MCQ 4: The standard unit basis vectors along the positive x-axis and positive y-axis are respectively:
a. i = [0, 1], j = [1, 0]
b. i = [1, 0], j = [0, 1]
c. i = [1, 1], j = [-1, -1]
d. i = [0, 0], j = [1, 1]
Step-by-Step Rationale:
In Cartesian 2D plane, $\hat{i} = [1, 0]$ is the unit vector along the horizontal x-axis and $\hat{j} = [0, 1]$ is along the vertical y-axis.
✓ Final Result: b. i = [1, 0], j = [0, 1]
Booster Q5 Extra Exercise Q5: Objective Concept Booster - Condition for Parallel Vectors 1 Mark
Booster MCQ 5: Two non-zero vectors $\vec{u} = [x_1, y_1]$ and $\vec{v} = [x_2, y_2]$ are parallel if and only if:
a. x₁ x₂ + y₁ y₂ = 0
b. x₁/x₂ = y₁/y₂ = k
c. x₁ + x₂ = y₁ + y₂
d. x₁ - x₂ = y₁ - y₂
Step-by-Step Rationale:
Parallel vectors are scalar multiples of each other ($ ec{u} = k ec{v}$), meaning their corresponding component ratios are equal.
✓ Final Result: b. x₁/x₂ = y₁/y₂ = k
Booster Q6 Extra Exercise Q6: Objective Concept Booster - Zero Vector Characteristics 1 Mark
Booster MCQ 6: A zero (null) vector $\vec{0}$ has:
a. Zero magnitude and a specific fixed direction
b. Magnitude 1 and arbitrary direction
c. Zero magnitude and arbitrary direction
d. Negative magnitude
Step-by-Step Rationale:
The initial and terminal points of a zero vector coincide, giving it a length of 0 and an arbitrary or indeterminate direction.
✓ Final Result: c. Zero magnitude and arbitrary direction
Booster Q7 Extra Exercise Q7: Objective Concept Booster - Ratio Formula Midpoint Case 1 Mark
Booster MCQ 7: If point $P$ is the midpoint of segment $AB$ with position vectors $\vec{a}$ and $\vec{b}$, the position vector of $P$ is:
a. a - b
b. (a + b)/2
c. 2(a + b)
d. (a + b)/3
Step-by-Step Rationale:
When $m : n = 1 : 1$, the ratio formula simplifies to $ ec{r} = rac{1 ec{b} + 1 ec{a}}{1 + 1} = rac{ ec{a} + ec{b}}{2}$.
✓ Final Result: b. (a + b)/2
Booster Q8 Extra Exercise Q8: Objective Concept Booster - Direction Angle of a Vector 1 Mark
Booster MCQ 8: The angle $\theta$ made by the vector $\vec{u} = \hat{i} + \sqrt{3}\hat{j}$ with the positive x-axis is:
a. 30°
b. 45°
c. 60°
d. 90°
Step-by-Step Rationale:
$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = 60^\circ$.
✓ Final Result: c. 60°
Booster Q9 Extra Exercise Q9: Objective Concept Booster - Vector Commutative Law 1 Mark
Booster MCQ 9: The geometric property $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ illustrates which vector addition law?
a. Associative Law
b. Distributive Law
c. Commutative Law
d. Identity Law
Step-by-Step Rationale:
Vector addition is commutative; the order in which two vectors are added does not alter their resultant vector.
✓ Final Result: c. Commutative Law
Booster Q10 Extra Exercise Q10: Objective Concept Booster - Scalar Multiple Effect on Vector Direction 1 Mark
Booster MCQ 10: If $\vec{v} = [4, -6]$ and scalar $k = -0.5$, then the vector $k\vec{v}$ is:
a. [-2, 3]
b. [2, -3]
c. [-8, 12]
d. [-2, -3]
Step-by-Step Rationale:
$k\vec{v} = -0.5[4, -6] = [-0.5 \times 4, -0.5 \times (-6)] = [-2, 3]$.
✓ Final Result: a. [-2, 3]
Booster Q11 Extra Exercise Q11: Objective Concept Booster - Vector Norm Definition 1 Mark
Booster Blank 11: The process of finding the unit vector of a given non-zero vector $\vec{u}$ is called ____________ the vector.
Step-by-Step Explanation:
Dividing a vector by its own magnitude to produce a unit vector of length 1 is mathematically termed "normalizing".
✓ Final Result: normalizing
Booster Q12 Extra Exercise Q12: Objective Concept Booster - Equal Vectors Equality Rule 1 Mark
Booster Blank 12: Two vectors are equal if and only if they have the same ____________ and the same ____________.
Step-by-Step Explanation:
Vector equality requires both geometric criteria: equal length (magnitude) and identical orientation (direction).
✓ Final Result: magnitude, direction
Booster Q13 Extra Exercise Q13: Objective Concept Booster - Triangle Law Resultant 1 Mark
Booster Blank 13: In the head-to-tail rule, the resultant vector joins the ____________ of the first vector to the ____________ of the last vector.
Step-by-Step Explanation:
The resultant vector begins at the initial point (tail) of the first vector and terminates at the terminal point (head) of the final vector.
✓ Final Result: tail, head
Booster Q14 Extra Exercise Q14: Objective Concept Booster - Position Vector Origin 1 Mark
Booster Blank 14: A vector whose initial point is fixed at the Cartesian origin $(0, 0)$ is known as a ____________ vector.
Step-by-Step Explanation:
The vector $ ec{OP}$ that specifies the location of point $P(x,y)$ relative to the origin $(0,0)$ is its position vector.
✓ Final Result: position
Booster Q15 Extra Exercise Q15: Objective Concept Booster - Vector Translation Invariant 1 Mark
Booster Blank 15: Translation of a geometric shape by a vector slides the shape without changing its size, shape, or ____________.
Step-by-Step Explanation:
Vector translation is a rigid transformation preserving distances, angles, and orientation.
✓ Final Result: orientation
Booster Q16 Extra Exercise Q16: Objective Concept Booster - External Division Formula 1 Mark
Booster Blank 16: When a point $P$ divides the line segment $AB$ externally in the ratio $m : n$, its position vector is $\vec{r} = \frac{m\vec{b} - n\vec{a}}{____________}$.
Step-by-Step Explanation:
For external division, the ratio formula has a minus sign in numerator and denominator: $ ec{r} = rac{m ec{b} - n ec{a}}{m - n}$.
✓ Final Result: m - n
Booster Q17 Extra Exercise Q17: Objective Concept Booster - Zero Vector Magnitude 1 Mark
Booster Blank 17: The magnitude of the null vector $\vec{0} = [0, 0]$ is exactly ____________.
Step-by-Step Explanation:
$| ec{0}| = \sqrt{0^2 + 0^2} = 0$.
✓ Final Result: 0 (zero)
Booster Q18 Extra Exercise Q18: Objective Concept Booster - Horizontal Component Formula 1 Mark
Booster Blank 18: If vector $\vec{F}$ makes an angle $\theta$ with the horizontal x-axis, its horizontal component $F_x$ is given by $|\vec{F}| \times$ ____________.
Step-by-Step Explanation:
In right-triangle trigonometry, adjacent component $F_x = | ec{F}|\cos heta$.
✓ Final Result: cos θ
Booster Q19 Extra Exercise Q19: Objective Concept Booster - Vector Magnitude Positivity 1 Mark
Booster True/False 19: The magnitude of any real vector $\vec{u}$ is always greater than or equal to zero ($|\vec{u}| \ge 0$).
True
False
Step-by-Step Rationale:
Magnitude is defined by $\sqrt{x^2 + y^2}$, which is always a non-negative real number.
✓ Final Result: True
Booster Q20 Extra Exercise Q20: Objective Concept Booster - Equal Vectors Tail Constraint 1 Mark
Booster True/False 20: Two equal vectors must always have their tails located at the exact same point in the plane.
True
False
Step-by-Step Rationale:
Equal vectors do not need to share the same position; two vectors anywhere in the plane are equal as long as they have equal magnitude and parallel direction.
✓ Final Result: False
Booster Q21 Extra Exercise Q21: Objective Concept Booster - Vector Subtraction Definition 1 Mark
Booster True/False 21: Subtracting vector $\vec{b}$ from $\vec{a}$ is mathematically equivalent to adding the negative vector of $\vec{b}$ to $\vec{a}$ (i.e., $\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$).
True
False
Step-by-Step Rationale:
Vector subtraction is strictly defined as the addition of the inverse (negative) vector.
✓ Final Result: True
Booster Q22 Extra Exercise Q22: Objective Concept Booster - Collinear Vectors on Line 1 Mark
Booster True/False 22: Three points $A$, $B$, and $C$ are collinear if and only if vector $\vec{AB}$ is parallel to vector $\vec{BC}$.
True
False
Step-by-Step Rationale:
Since $B$ is a common point and $ ec{AB} \parallel ec{BC}$, the points $A, B, C$ must lie along the exact same straight line.
✓ Final Result: True
Booster Q23 Extra Exercise Q23: Objective Concept Booster - Unit Vector of Null Vector 1 Mark
Booster True/False 23: A unit vector can be found for a zero (null) vector by dividing by its magnitude.
True
False
Step-by-Step Rationale:
Division by zero is undefined; a unit vector exists ONLY for non-zero vectors.
✓ Final Result: False
Booster Q24 Extra Exercise Q24: Objective Concept Booster - Parallelogram Diagonal Law 1 Mark
Booster True/False 24: In any parallelogram formed by vectors $\vec{a}$ and $\vec{b}$, the main diagonal passing through the common origin represents $\vec{a} + \vec{b}$.
True
False
Step-by-Step Rationale:
This is the foundational statement of the Parallelogram Law of Vector Addition.
✓ Final Result: True
Booster Q25 Extra Exercise Q25: Objective Concept Booster - Vector Formulas & Algebraic Definitions 5 Marks
Booster Match Column 25: Match Column A with Column B:

Column A (Concept / Expression)Column B (Mathematical Formula)
1. Magnitude of vector $\vec{u} = [x, y]$A. $\vec{r} = \frac{\vec{a} + \vec{b}}{2}$
2. Unit Vector in direction of $\vec{u}$B. $\vec{b} - \vec{a}$
3. Vector $\vec{AB}$ from position vectors $\vec{a}, \vec{b}$C. $\sqrt{x^2 + y^2}$
4. Midpoint position vectorD. $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$
5. Direction angle $\theta$E. $\tan^{-1}(y / x)$
Step-by-Step Column Matching:
1. Magnitude $|\vec{u}| = \sqrt{x^2 + y^2}$ (C)
2. Unit vector $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$ (D)
3. Vector $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$ (B)
4. Midpoint position vector $= \frac{\vec{a} + \vec{b}}{2}$ (A)
5. Direction angle $\theta = \tan^{-1}(y / x)$ (E).
✓ Final Result: 1 -> C, 2 -> D, 3 -> B, 4 -> A, 5 -> E
Booster Q26 Extra Exercise Q26: Objective Concept Booster - Physical Quantities & Vector Types 5 Marks
Booster Match Column 26: Match Column A with Column B:

Column A (Quantity / Concept)Column B (Classification / Vector)
1. Mass, Speed, WorkA. $x_1/x_2 = y_1/y_2 = k$
2. Force, Velocity, WeightB. $\vec{0} = [0, 0]$
3. Null VectorC. Scalar Quantities
4. Parallel Vectors ConditionD. Vector Quantities
5. Negative Vector of $[x, y]$E. $[-x, -y]$
Step-by-Step Column Matching:
1. Mass, Speed, Work are Scalar Quantities (C)
2. Force, Velocity, Weight are Vector Quantities (D)
3. Null vector is $\vec{0} = [0, 0]$ (B)
4. Parallel condition is proportionality $\frac{x_1}{x_2} = \frac{y_1}{y_2} = k$ (A)
5. Negative vector is $[-x, -y]$ (E).
✓ Final Result: 1 -> C, 2 -> D, 3 -> B, 4 -> A, 5 -> E

More Chapter Notes for Class 10 (FBISE)

Mathematics
Mathematics • Chapter 1 FBISE
Mastery Guide: Complex Numbers - Algebra, Argand Diagrams, Modulus, Conjugates & AC Circuit Applications
Complex Numbers
Mathematics • Chapter 2 FBISE
Mastery Guide: Quadratic Equations - Standard Forms, Solution Methods, Nature of Roots, Symmetric Functions & Simultaneous Modeling
Quadratic Equations
Mathematics • Chapter 3 FBISE
Mastery Guide: Matrices, Determinants, Multiplicative Inverses & Linear System Modeling
Matrices and Determinants
Mathematics • Chapter 4 FBISE
Mastery Guide: Linear & Quadratic Inequalities, Absolute Values & Feasible Regions
Linear and Quadratic Inequalities
Mathematics • Chapter 5 FBISE
Mastery Guide: Algebraic Fractions, Rational Operations, Complex Fractions & Extraneous Roots
Algebraic Fractions
Mathematics • Chapter 6 FBISE
Mastery Guide: Functions and Graphs, Relations, Composite Mappings & Analytical Curves
Functions and Graphs
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