Vector Addition — NEET Physics MCQs with Solutions

Solution: For perpendicular vectors, R = √(A² + B²) = √(3² + 4²) = √25 = 5 N.

Solution: The triangle law states that if two vectors are represented by two sides of a triangle taken in order, the third side (in reverse order) gives their resultant. It applies to any two vectors.

Solution: R = √(A² + B² + 2AB·cosθ) = √(100 + 100 + 2·100·cos60°) = √(200 + 100) = √300 = 10√3 units.

Solution: Resultant is maximum when the vectors are parallel (7 + 3 = 10) and minimum when antiparallel (7 − 3 = 4).

Solution: Vector addition is commutative: the order of addition does not change the resultant, so A + B = B + A.

Solution: With |A| = |B| = a and R = a: a² = a² + a² + 2a²cosθ ⇒ 1 = 2(1 + cosθ) ⇒ cosθ = −1/2 ⇒ θ = 120°.

Solution: R = √(A² + B² + 2AB·cosθ) is largest when cosθ is largest, i.e. cosθ = 1 at θ = 0° (vectors parallel).

Solution: (5√2)² = 5² + 5² + 2·5·5·cosθ ⇒ 50 = 50 + 50cosθ ⇒ cosθ = 0 ⇒ θ = 90°.

Solution: The resultant must lie between |2 − 1| = 1 and 2 + 1 = 3 units. A value of 4 units is impossible.

Solution: This equality holds only when θ = 0°, i.e. the vectors point in the same direction, so the resultant adds up arithmetically.

Solution: 5² = 3² + 4² + 2·3·4·cosθ ⇒ 25 = 25 + 24cosθ ⇒ cosθ = 0 ⇒ θ = 90°.

Solution: From F² = F² + F² + 2F²cosθ we get cosθ = −1/2. New resultant: √(F² + (2F)² + 2·F·2F·(−1/2)) = √(F² + 4F² − 2F²) = √(3F²) = F√3.

Solution: Ax = 10·cos30° = 10·(√3/2) = 5√3; Ay = 10·sin30° = 10·(1/2) = 5.

Solution: The parallelogram law states the resultant is the diagonal of the parallelogram drawn from the common tail of the two vectors.

Solution: Three equal vectors symmetrically separated by 120° cancel out — their components sum to zero in every direction, so the resultant is zero.

Solution: Check with $R^2 = A^2 + B^2 + 2AB\cos\theta$: here $10^2 = 6^2 + 8^2$, i.e. $100 = 36 + 64$, so $2AB\cos\theta = 0 \Rightarrow \theta = 90^\circ$. (A 6-8-10 triangle is right-angled.)

Solution: $R = \sqrt{5^2 + 5^2 + 2\cdot5\cdot5\cos120^\circ} = \sqrt{25 + 25 - 25} = \sqrt{25} = 5$ units.

Solution: With $|A|=|B|=a$ and $R=a$: $a^2 = 2a^2(1+\cos\theta) \Rightarrow \cos\theta = -\tfrac{1}{2} \Rightarrow \theta = 120^\circ$.

Solution: The resultant must lie between $|4-3| = 1$ and $4+3 = 7$. A value of $8$ is outside this range, so it is impossible.

Solution: Squaring both sides: $A^2+B^2+2\vec A\cdot\vec B = A^2+B^2-2\vec A\cdot\vec B \Rightarrow \vec A\cdot\vec B = 0$, so the vectors are perpendicular.

Solution: $\vec A + \vec B = 3\hat{i} + 2\hat{j}$, so $|\vec A + \vec B| = \sqrt{3^2 + 2^2} = \sqrt{13}$.

Solution: $\vec A + \vec B = 2\hat{i}$. The unit vector is $\frac{2\hat{i}}{2} = \hat{i}$.

Solution: With $|A|=|B|=a$: $|\vec A+\vec B|^2 = 2a^2(1+\cos\theta)$ and $|\vec A-\vec B|^2 = 2a^2(1-\cos\theta)$. Setting $2a^2(1+\cos\theta) = n^2 \cdot 2a^2(1-\cos\theta)$ gives $\cos\theta = \frac{n^2-1}{n^2+1}$. (A well-known JEE Main pattern.)

Solution: $R^2 = 2F^2(1+\cos\theta) = 3F^2 \Rightarrow \cos\theta = \tfrac{1}{2} \Rightarrow \theta = 60^\circ$.

Solution: Let the forces be $P$ (smaller) and $Q$ with $P+Q=16$. Resultant $\perp P$ means $Q^2 = P^2 + 8^2$. So $(Q-P)(Q+P) = 64 \Rightarrow Q-P = 4$. Solving: $P = 6$ N, $Q = 10$ N. (Classic AIEEE/JEE problem.)

Solution: If $\vec R = \vec P + \vec Q$ is $\perp \vec P$, then $\vec R \cdot \vec P = 0 \Rightarrow P^2 + PQ\cos\theta = 0 \Rightarrow \cos\theta = -\frac{P}{Q}$.

Solution: The resultant of the two forces is $10\sqrt{2}$ N towards north-east. Equilibrium needs an equal and opposite force: $10\sqrt{2}$ N towards south-west.

Solution: $\vec C = -(\vec A + \vec B)$, so $49 = 9 + 25 + 2(3)(5)\cos\theta \Rightarrow \cos\theta = \tfrac{1}{2} \Rightarrow \theta = 60^\circ$.

Solution: Two vectors summing to zero must be equal and opposite (equal magnitudes), so unequal magnitudes need at least three vectors forming a closed triangle.

Solution: $\tan\theta = \frac{A_y}{A_x} = \frac{4}{3}$, so $\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ$.