Class 10 Mathematics Chapter 7: Vectors in Plane Online Practice Test (2026) - Apex Rankers Sample Questions & Solved Answers

Question 1

Draw and represent the following vectors geometrically with appropriate scale and direction:
(i) $10 ext{ N}$ force along the positive x-axis
(ii) $50 ext{ m/s}$ velocity at $150^\circ$ with the positive x-axis
(iii) $220 ext{ m}$ displacement towards North
(iv) $24 ext{ m/s}^2$ acceleration at $45^\circ$ with the positive x-axis.

Accepted Answer

Correct Answer: (i) 10i N, (ii) -25√3 i + 25j m/s, (iii) 220j m, (iv) 12√2 i + 12√2 j m/s². Solution: Step-by-Step Geometrical Construction & Resolution:
(i) $10 ext{ N}$ force along positive x-axis:
• Scale: Let $1 ext{ cm} = 2 ext{ N} \implies ext{Length} = \frac{10}{2} = 5 ext{ cm}$.
• Direction: Directed along positive x-axis (Angle $ heta = 0^\circ$).
• Component Form: $\vec{F} = [10\cos 0^\circ, 10\sin 0^\circ] = [10, 0] = 10\hat{i} ext{ N}$.

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

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

(iv) $24 ext{ m/s}^2$ acceleration at $45^\circ$ with positive x-axis:
• Scale: Let $1 ext{ cm} = 6 ext{ m/s}^2 \implies ext{Length} = \frac{24}{6} = 4 ext{ cm}$.
• Direction: In Quadrant I making an angle of $45^\circ$ with the horizontal.
• Component Form: $\vec{a} = [24\cos 45^\circ, 24\sin 45^\circ] = [24(\frac{1}{\sqrt{2}}), 24(\frac{1}{\sqrt{2}})] = [12\sqrt{2}, 12\sqrt{2}] \approx [16.97, 16.97] ext{ m/s}^2 = 12\sqrt{2}\hat{i} + 12\sqrt{2}\hat{j} ext{ m/s}^2$.

Question 2

Given vector $\vec{F}$ is $4 ext{ cm}$ 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) $-0.5\vec{F}$.

Accepted Answer

Correct Answer: (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°. Solution: Step-by-Step Geometrical & Scalar Properties:
Given: $|\vec{F}| = 4 ext{ cm}$ and direction $ heta = 45^\circ$.
• If scalar $k > 0$, the vector maintains direction $ heta = 45^\circ$ and length becomes $k|\vec{F}|$.
• If scalar $k < 0$, the direction reverses by $180^\circ$ (new direction $ heta = 45^\circ + 180^\circ = 225^\circ$) and length is $|k||\vec{F}|$.

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

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

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

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

(v) $-0.5\vec{F}$:
• Length $= |-0.5| imes 4 ext{ cm} = 2 ext{ cm}$.
• Direction $= 45^\circ + 180^\circ = 225^\circ$ (Opposite direction).

Question 3

Given vectors $\vec{a} = 3 ext{ units West}$ ($[-3, 0]$) and $\vec{b} = 3 ext{ units North}$ ($[0, 3]$). 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) $-(\vec{a} - 2\vec{b})$.

Accepted Answer

Correct Answer: (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]. Solution: 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 $= \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} ext{ units} \approx 6.71$.

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

(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 $= \sqrt{(-9)^2 + (4.5)^2} = \sqrt{81 + 20.25} = \sqrt{101.25} = 4.5\sqrt{5} ext{ units} \approx 10.06$.

(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 $= \sqrt{(-9)^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} ext{ units} \approx 9.49$.

(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 $= \sqrt{(-1.5)^2 + (1.5)^2} = \sqrt{2.25 + 2.25} = \sqrt{4.5} = 1.5\sqrt{2} ext{ units} \approx 2.12$.

(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 $= \sqrt{(-9)^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117} = 3\sqrt{13} ext{ units} \approx 10.82$.

(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 $= \sqrt{(-6)^2 + (-7.5)^2} = \sqrt{36 + 56.25} = \sqrt{92.25} \approx 9.60 ext{ units}$.

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

Question 4

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 $\vec{t} = [-2, -1.5]$.

Accepted Answer

Correct Answer: (i) p = 2a, (ii) q = -a, (iii) r = -2a, (iv) s = 0.5a, (v) t = -0.5a. Solution: 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.

Question 5

A point with coordinates $P(5, -7)$ is translated by the vector $\vec{v} = [0, 4]$. Find the new position $P'$ of the point.

Accepted Answer

Correct Answer: P'(5, -3). Solution: 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.
$$P' = (5, -3)$$

Question 6

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.

Accepted Answer

Correct Answer: [-1, 1] or -i + j. Solution: 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:
$$\vec{w} = -\hat{i} + \hat{j}$$

Question 7

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 $\vec{v}_2 = [-2, -4]$.

Accepted Answer

Correct Answer: (i) A'(1, 6), B'(4, 4), C'(-1, 1); (ii) A''(-6, 2), B''(-3, 0), C''(-8, -3). Solution: 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:
• $A'' = (-4 - 2, 6 - 4) = (-6, 2)$
• $B'' = (-1 - 2, 4 - 4) = (-3, 0)$
• $C'' = (-6 - 2, 1 - 4) = (-8, -3)$

Question 8

What translation vector $\vec{T}$ is required to map the point $E(-6, 5)$ directly onto the origin $O(0, 0)$?

Accepted Answer

Correct Answer: [6, -5] or 6i - 5j. Solution: 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:
$$\vec{T} = [6, -5] = 6\hat{i} - 5\hat{j}$$

Question 9

$ABCD$ 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) $\vec{AB}$.

Accepted Answer

Correct Answer: (i) -a, (ii) -b, (iii) a + b, (iv) b - a, (v) 0.5(a + b), (vi) 0.5(b - a), (vii) a. Solution: 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 $ riangle ABC$: $\vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b}$.

(iv) $\vec{BD}$:
• In $ riangle 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: $\vec{AB} = \vec{a}$.

Question 10

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) $\vec{OM}$.

Accepted Answer

Correct Answer: (i) q - p, (ii) 0.5(q - p), (iii) 0.5(p - q), (iv) 0.5(p + q). Solution: 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: $\vec{OM} = \frac{\vec{OP} + \vec{OQ}}{2} = \frac{\vec{p} + \vec{q}}{2} = \frac{1}{2}\vec{p} + \frac{1}{2}\vec{q}$.

Class 10 Mathematics - Ch 7: Vectors in Plane

Class 10 Mathematics - Ch 7: Vectors in Plane

🎯 Question Types & Curriculum Breakdown

💡 Strategic Preparation & Exam Hall Guidelines

📝 Pre-Rendered Solved Sample Questions & Detailed Solutions

Question statement goes here?

Correct Answer: A

Practice Completed!

Class 10 Mathematics - Ch 7: Vectors in Plane

Class 10 Mathematics - Ch 7: Vectors in Plane

🎯 Question Types & Curriculum Breakdown

💡 Strategic Preparation & Exam Hall Guidelines

📝 Pre-Rendered Solved Sample Questions & Detailed Solutions

Question statement goes here?

Correct Answer: A

Practice Completed!

✏️ Handwriting Recognition Board