Vector Subtraction — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Vector Subtraction MCQs with step-by-step solutions (35 questions). Part of Vectors. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vector Subtraction · medium · theory
If $\vec A = 5\hat{i} + 3\hat{j}$ and $\vec B = 2\hat{i} + \hat{j}$, then $|\vec A - \vec B|$ is
A. $\sqrt{13}$ ✓ Correct
B. $\sqrt{5}$
C. $5$
D. $13$
Solution: $\vec A - \vec B = 3\hat{i} + 2\hat{j}$, so $|\vec A - \vec B| = \sqrt{9+4} = \sqrt{13}$.
Q2 — Vector Subtraction · medium · theory
Two vectors each of magnitude $a$ are inclined at an angle $\theta$. The magnitude of their difference is
A. $2a\sin(\theta/2)$ ✓ Correct
B. $2a\cos(\theta/2)$
C. $2a\tan(\theta/2)$
D. $a\sin\theta$
Solution: $|\vec A - \vec B|^2 = a^2 + a^2 - 2a^2\cos\theta = 2a^2(1-\cos\theta) = 4a^2\sin^2(\theta/2)$, so the magnitude is $2a\sin(\theta/2)$.
Q3 — Vector Subtraction · medium · numerical
Two vectors of magnitudes $4$ units and $3$ units are perpendicular. The magnitude of their difference is
A. $\sqrt{7}$
B. $5$ ✓ Correct
C. $1$
D. $7$
Solution: For perpendicular vectors $|\vec A - \vec B| = \sqrt{4^2 + 3^2} = 5$ units (the $\cos\theta$ term vanishes).
Q4 — Vector Subtraction · easy · numerical
Two vectors each of magnitude $5$ units are inclined at $60^\circ$. The magnitude of their difference is
A. $10$
B. $5\sqrt{3}$
C. $5$ ✓ Correct
D. $5\sqrt{2}$
Solution: $|\vec A - \vec B| = 2(5)\sin(60^\circ/2) = 10\sin 30^\circ = 5$ units.
Q5 — Vector Subtraction · medium · theory
If $|\vec A - \vec B| = |\vec A| = |\vec B|$, the angle between $\vec A$ and $\vec B$ is
A. $60^\circ$ ✓ Correct
B. $90^\circ$
C. $30^\circ$
D. $120^\circ$
Solution: With $|A|=|B|=a$: $2a\sin(\theta/2) = a \Rightarrow \sin(\theta/2) = \tfrac{1}{2} \Rightarrow \theta = 60^\circ$.
Q6 — Vector Subtraction · medium · numerical
A particle moves with velocity $10$ m/s towards east. Later its velocity is $10$ m/s towards north. The magnitude of the change in velocity is
A. Zero
B. $10\sqrt{2}$ m/s ✓ Correct
C. $20$ m/s
D. $10$ m/s
Solution: $\Delta\vec v = \vec v_2 - \vec v_1 = 10\hat{j} - 10\hat{i}$, so $|\Delta\vec v| = \sqrt{100+100} = 10\sqrt{2}$ m/s (directed north-west).
Q7 — Vector Subtraction · medium · theory
A ball moving with speed $v$ hits a wall normally and rebounds with the same speed. The magnitude of the change in its velocity is
A. $v$
B. $2v$ ✓ Correct
C. $v\sqrt{2}$
D. Zero
Solution: Velocity reverses: $\Delta\vec v = (-v) - (+v) = -2v$ along the normal, so the magnitude is $2v$.
Q8 — Vector Subtraction · medium · theory
A particle moves on a circle with constant speed $v$. The magnitude of the change in velocity after half a revolution is
A. $v\sqrt{2}$
B. Zero
C. $2v$ ✓ Correct
D. $v$
Solution: After half a revolution the velocity direction reverses, so $|\Delta\vec v| = |{-\vec v} - \vec v| = 2v$.
Q9 — Vector Subtraction · medium · theory
A particle moves on a circle with constant speed $v$. The magnitude of the change in velocity after a quarter revolution is
A. $v\sqrt{2}$ ✓ Correct
B. Zero
C. $2v$
D. $v$
Solution: The two velocities are perpendicular with equal magnitudes: $|\Delta\vec v| = \sqrt{v^2+v^2} = v\sqrt{2}$.
Q10 — Vector Subtraction · medium · numerical
Two vectors have magnitudes $7$ and $4$ units. The minimum possible magnitude of their difference is
A. $0$
B. $11$
C. $7$
D. $3$ ✓ Correct
Solution: $|\vec A - \vec B|$ ranges from $|7-4| = 3$ (parallel) to $7+4 = 11$ (antiparallel). Minimum is $3$.
Q11 — Vector Subtraction · easy · numerical
Two vectors have magnitudes $7$ and $4$ units. Which of the following CANNOT be the magnitude of $\vec A - \vec B$?
A. $3$
B. $11$
C. $2$ ✓ Correct
D. $7$
Solution: The difference must lie between $3$ and $11$; a value of $2$ is impossible.
Q12 — Vector Subtraction · medium · theory
If $\vec A = 3\hat{i} + 4\hat{j}$ and $\vec B = \hat{i} + 2\hat{j}$, the unit vector along $\vec A - \vec B$ is
A. $\hat{i}$
B. $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ ✓ Correct
C. $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$
D. $\frac{2\hat{i}+2\hat{j}}{4}$
Solution: $\vec A - \vec B = 2\hat{i} + 2\hat{j}$ with magnitude $2\sqrt{2}$, so the unit vector is $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
Q13 — Vector Subtraction · medium · numerical
If $|\vec A + \vec B| = 10$ and $|\vec A - \vec B| = 6$, then $\vec A \cdot \vec B$ equals
A. $32$
B. $8$
C. $64$
D. $16$ ✓ Correct
Solution: $|\vec A+\vec B|^2 - |\vec A-\vec B|^2 = 4\,\vec A\cdot\vec B$. So $100 - 36 = 64 = 4\,\vec A\cdot\vec B \Rightarrow \vec A\cdot\vec B = 16$.
Q14 — Vector Subtraction · medium · numerical
Rain falls vertically at $3$ m/s. A man walks horizontally at $4$ m/s. The velocity of the rain relative to the man has magnitude
A. $3.5$ m/s
B. $7$ m/s
C. $1$ m/s
D. $5$ m/s ✓ Correct
Solution: $\vec v_{rain,man} = \vec v_{rain} - \vec v_{man}$; the two are perpendicular, so the magnitude is $\sqrt{3^2+4^2} = 5$ m/s.
Q15 — Vector Subtraction · medium · numerical
Car A moves at $40$ m/s towards east and car B at $30$ m/s towards north. The magnitude of the velocity of A relative to B is
A. $50$ m/s ✓ Correct
B. $70$ m/s
C. $35$ m/s
D. $10$ m/s
Solution: $\vec v_{A,B} = \vec v_A - \vec v_B = 40\hat{i} - 30\hat{j}$, magnitude $\sqrt{1600+900} = 50$ m/s.
Q16 — Vector Subtraction · medium · theory
If $\vec A = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec B = \hat{i} + \hat{j} - \hat{k}$, then $|\vec A - \vec B|$ is
A. $\sqrt{6}$
B. $\sqrt{3}$
C. $9$
D. $3$ ✓ Correct
Solution: $\vec A - \vec B = \hat{i} - 2\hat{j} + 2\hat{k}$, so $|\vec A - \vec B| = \sqrt{1+4+4} = 3$.
Q17 — Vector Subtraction · medium · theory
Two vectors each of magnitude $a$ are inclined at $120^\circ$. The magnitude of their difference is
A. $a\sqrt{3}$ ✓ Correct
B. $2a$
C. $a$
D. $a\sqrt{2}$
Solution: $|\vec A - \vec B| = 2a\sin(120^\circ/2) = 2a\sin 60^\circ = a\sqrt{3}$.
Q18 — Vector Subtraction · medium · theory
If $|\vec A - \vec B| > |\vec A + \vec B|$, the angle between $\vec A$ and $\vec B$ is
A. Acute (less than $90^\circ$)
B. Obtuse (greater than $90^\circ$) ✓ Correct
C. Exactly $90^\circ$
D. Zero
Solution: $|\vec A-\vec B|^2 - |\vec A+\vec B|^2 = -4AB\cos\theta > 0$ requires $\cos\theta < 0$, i.e. an obtuse angle.
Q19 — Vector Subtraction · medium · theory
$|\vec A - \vec B| = |\vec A| + |\vec B|$ is possible only when
A. Never
B. The vectors are parallel ($\theta = 0^\circ$)
C. The vectors are antiparallel ($\theta = 180^\circ$) ✓ Correct
D. The vectors are perpendicular
Solution: The difference is largest when the vectors point opposite ways: then subtracting adds the magnitudes.
Q20 — Vector Subtraction · medium · theory
For two non-zero vectors with $|\vec A| > |\vec B|$, $|\vec A - \vec B| = |\vec A| - |\vec B|$ happens when
A. The angle is $60^\circ$
B. They are perpendicular
C. They are antiparallel
D. They are parallel, in the same direction ✓ Correct
Solution: The difference is smallest (equal to the difference of magnitudes) when both vectors point the same way.
Q21 — Vector Subtraction · medium · theory
A particle moves from position $(2, 3)$ m to $(5, 7)$ m. The magnitude of its displacement is
A. $\sqrt{7}$ m
B. $7$ m
C. $25$ m
D. $5$ m ✓ Correct
Solution: $\Delta\vec r = (5-2)\hat{i} + (7-3)\hat{j} = 3\hat{i}+4\hat{j}$, magnitude $\sqrt{9+16} = 5$ m.
Q22 — Vector Subtraction · medium · theory
A projectile is fired with speed $u$ at angle $\theta$ to the horizontal. The magnitude of the change in its velocity between launch and landing (same level) is
A. Zero
B. $2u\sin\theta$ ✓ Correct
C. $u\sin\theta$
D. $2u\cos\theta$
Solution: The horizontal component is unchanged; the vertical component reverses from $+u\sin\theta$ to $-u\sin\theta$. Hence $|\Delta\vec v| = 2u\sin\theta$ (directed vertically downward).
Q23 — Vector Subtraction · medium · numerical
If $\vec A + \vec B = 6\hat{i} + 2\hat{j}$ and $\vec A - \vec B = 2\hat{i} + 4\hat{j}$, then $|\vec A|$ is
A. $\sqrt{5}$
B. $3$
C. $5$ ✓ Correct
D. $4$
Solution: Adding the equations: $2\vec A = 8\hat{i}+6\hat{j} \Rightarrow \vec A = 4\hat{i}+3\hat{j}$, so $|\vec A| = 5$.
Q24 — Vector Subtraction · easy · theory
A cyclist rides along a circular track at constant speed $v$. After turning through $60^\circ$ of arc, the magnitude of the change in velocity is
A. $v$ ✓ Correct
B. $v/2$
C. $v\sqrt{3}$
D. $2v$
Solution: $|\Delta\vec v| = 2v\sin(\theta/2) = 2v\sin 30^\circ = v$.
Q25 — Vector Subtraction · medium · numerical
Two vectors of magnitudes $5$ and $3$ are inclined at $120^\circ$. The magnitude of their difference is
A. $\sqrt{19}$
B. $7$ ✓ Correct
C. $4$
D. $8$
Solution: $|\vec A-\vec B|^2 = 25 + 9 - 2(5)(3)\cos 120^\circ = 34 + 15 = 49$, so $|\vec A-\vec B| = 7$.
Q26 — Vector Subtraction · medium · theory
If $(\vec A - \vec B)$ is perpendicular to $(\vec A + \vec B)$, then
A. $\vec A = \vec B$
B. $|\vec A| = |\vec B|$ ✓ Correct
C. $\vec A \parallel \vec B$
D. $\vec A \perp \vec B$
Solution: $(\vec A-\vec B)\cdot(\vec A+\vec B) = A^2 - B^2 = 0 \Rightarrow |\vec A| = |\vec B|$ (the diagonals of a rhombus are perpendicular).
Q27 — Vector Subtraction · medium · numerical
Particle P has velocity $3\hat{i}$ m/s and particle Q has velocity $3\hat{j}$ m/s. The velocity of Q relative to P has magnitude
A. Zero
B. $3\sqrt{2}$ m/s ✓ Correct
C. $3$ m/s
D. $6$ m/s
Solution: $\vec v_{Q,P} = 3\hat{j} - 3\hat{i}$, magnitude $\sqrt{9+9} = 3\sqrt{2}$ m/s.
Q28 — Vector Subtraction · medium · theory
$|\vec A - \vec B|^2$ always equals
A. $(A - B)^2$
B. $A^2 + B^2 - 2\vec A\cdot\vec B$ ✓ Correct
C. $A^2 + B^2$
D. $A^2 - B^2$
Solution: $(\vec A-\vec B)\cdot(\vec A-\vec B) = A^2 + B^2 - 2\vec A\cdot\vec B$. It equals $(A-B)^2$ only for parallel vectors.
Q29 — Vector Subtraction · medium · theory
Two vectors have equal magnitudes $a$, and $|\vec A - \vec B| = a\sqrt{3}$. The angle between them is
A. $150^\circ$
B. $90^\circ$
C. $60^\circ$
D. $120^\circ$ ✓ Correct
Solution: $2a\sin(\theta/2) = a\sqrt{3} \Rightarrow \sin(\theta/2) = \tfrac{\sqrt{3}}{2} \Rightarrow \theta/2 = 60^\circ \Rightarrow \theta = 120^\circ$.
Q30 — Vector Subtraction · medium · numerical
For two perpendicular vectors of magnitude 3 and 4, |A − B| is:
A. 7
B. 12
C. 5 ✓ Correct
D. 1
Solution: |A − B| = √(3²+4²) = 5 (perpendicular, so same as |A+B|).