Vector Algebra — JEE MCQs with answers

Answer: B — Observation 4 of the vector product (p.364) gives $\hat{i}\times\hat{i} = \hat{j}\times\hat{j} = \hat{k}\times\hat{k} = \vec{0}$: a vector is parallel to itself, the angle is $0$ and $\sin 0 = 0$. For the scalar product the angle $0$ gives $\cos 0 = 1$, so $\hat{k}\cdot\hat{k} = |\hat{k}|^2 = 1$. The pair $\vec{0}$ and $0$ would be right for the vector product but wrong for the scalar product; $\hat{k}$ is not the value of $\hat{k}\times\hat{k}$. The answer is option B.

Answer: C — Example 2 (p.342) classifies 30 km/hr as 'Speed-scalar' and 10 Newton as 'Force-vector'. The printed answers to EXERCISE 10.1 Q2 give (ii) 2 meters north-west as a vector and (vi) $20\ \text{m/s}^2$ as a vector. So of the four, only 30 km/hr is classified as a scalar, which is option C.

Answer: B — Let $\vec{a}$, $\vec{b}$ and $\vec{r}$ be the position vectors of $P$, $Q$ and $R$. The internal formula with $m : n = 2 : 3$ gives $\vec{r} = \dfrac{2\vec{b} + 3\vec{a}}{5}$, so $\vec{b} = \dfrac{5\vec{r} - 3\vec{a}}{2} = \dfrac{(15 - 3)\hat{i} + (0 + 6)\hat{j} + (10 - 12)\hat{k}}{2} = 6\hat{i} + 3\hat{j} - \hat{k}$. So $Q$ is $(6, 3, -1)$, option B. Check: $\overrightarrow{PR} = 2\hat{i} + 2\hat{j} - 2\hat{k}$ and $\overrightarrow{RQ} = 3\hat{i} + 3\hat{j} - 3\hat{k}$, so $PR : RQ = 2 : 3$. Option A interchanges the multipliers, solving $\vec{r} = \dfrac{3\vec{b} + 2\vec{a}}{5}$, which gives $\vec{b} = \dfrac{5\vec{r} - 2\vec{a}}{3}$. Option C uses the external formula, solving $\vec{r} = \dfrac{2\vec{b} - 3\vec{a}}{2 - 3}$, which gives $\vec{b} = \dfrac{3\vec{a} - \vec{r}}{2}$. Option D makes both slips, solving $\vec{r} = \dfrac{3\vec{b} - 2\vec{a}}{3 - 2}$, which gives $\vec{b} = \dfrac{\vec{r} + 2\vec{a}}{3}$.

Answer: A — $|\vec{a}| = \sqrt{2 + 1 + 1} = 2$, and $\cos\alpha = \dfrac{a_{1}}{|\vec{a}|} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$, so $\alpha = \dfrac{\pi}{4}$, which is option A. Option B uses the $\hat{j}$ component, $\cos^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$; that is the angle with the $y$-axis. Option C divides by $|\vec{a}|^{2} = 4$ instead of $|\vec{a}| = 2$. Option D makes both slips.

Answer: C — The triangle law in triangle ABD gives $\overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AD}$, so $\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} = \vec{b} - \vec{a}$, which is option C. Option D is the same diagonal with its arrow reversed: $\overrightarrow{DB} = \vec{a} - \vec{b}$. Option A is the diagonal through the common point A of the two sides, which by the parallelogram law is the sum $\overrightarrow{AC} = \vec{a} + \vec{b}$, and option B is its negative, $\overrightarrow{CA} = -\vec{a} - \vec{b}$.

Answer: D — The sentence in the Introduction reads: the answer “involves muscular strength (magnitude) and direction (in which another player is positioned)”. So the magnitude is the muscular strength and the direction is the one in which another player is positioned, which is option D. Option A changes the magnitude to the speed of the ball, option B changes the direction to that of the goal, and option C makes both changes; neither the speed of the ball nor the goal is mentioned in this sentence.

Answer: A — Let $|\vec{a}| = |\vec{b}| = m$. From $\vec{a}\cdot\vec{b} = m^2\cos 60^\circ = \dfrac{m^2}{2} = \dfrac{9}{2}$ we get $m^2 = 9$ (the method of EXERCISE 10.3, Question 8). Then, as in Example 17 with a plus sign, $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 9 + 9 + 9 = 27$, so $|\vec{a} + \vec{b}| = 3\sqrt{3}$, which is option A. Option C finds $m^2$ by multiplying by $\cos 60^\circ$, $m^2 = \dfrac{9}{4}$, giving $\dfrac{9}{4} + 9 + \dfrac{9}{4} = \dfrac{27}{2}$. Option B has $m^2 = 9$ but drops the term $2\,\vec{a}\cdot\vec{b}$: $9 + 9 = 18$. Option D makes both slips: $\dfrac{9}{4} + \dfrac{9}{4} = \dfrac{9}{2}$.

Answer: D — For the vector $\overrightarrow{AB}$ the textbook calls the point A, from where the vector starts, its initial point, and the point B, where it ends, its terminal point; the starting point is written first. A vector that starts at M and ends at N is therefore $\overrightarrow{MN}$, with initial point M and terminal point N, which is option D. Option B swaps the names of the two points. Option A writes the letters in the reverse order, which would be the vector that starts at N. Option C makes both slips.

Answer: B — $\overrightarrow{AC} = \hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{CB} = 2\hat{i} - 2\hat{j} + 4\hat{k} = 2\,\overrightarrow{AC}$. The two vectors have the same direction, so $C$ lies between $A$ and $B$, and $AC : CB = 1 : 2$, which is option B. Check with the section formula: $\dfrac{1(4\hat{i} - 2\hat{j} + 7\hat{k}) + 2(\hat{i} + \hat{j} + \hat{k})}{1 + 2} = 2\hat{i} + 3\hat{k}$, the position vector of $C$. Option D inverts the ratio. For $B$ to lie between $A$ and $C$, $\overrightarrow{AB}$ and $\overrightarrow{BC}$ would have to point the same way, but $\overrightarrow{AB} = 3\hat{i} - 3\hat{j} + 6\hat{k}$ and $\overrightarrow{BC} = -2\hat{i} + 2\hat{j} - 4\hat{k}$ are opposite in direction. Option C follows the order in which the points are named and compares only the lengths $|\overrightarrow{AB}| = 3\sqrt{6}$ and $|\overrightarrow{BC}| = 2\sqrt{6}$; option A also inverts that ratio.

Answer: D — In a regular hexagon the sides AB and ED are opposite and parallel, and the diagonal FC through the centre is parallel to both and twice as long as a side. (With the vertices on a circle of radius 1: $A(1,0)$, $B(\frac12,\frac{\sqrt3}{2})$, $C(-\frac12,\frac{\sqrt3}{2})$, $D(-1,0)$, $E(-\frac12,-\frac{\sqrt3}{2})$, $F(\frac12,-\frac{\sqrt3}{2})$, the displacements are $\overrightarrow{AB}$: $(-\frac12,\frac{\sqrt3}{2})$, $\overrightarrow{FC}$: $(-1,\sqrt3)$, $\overrightarrow{DE}$: $(\frac12,-\frac{\sqrt3}{2})$.) All three are parallel to the same line, so they are collinear. $\overrightarrow{DE}$ has the magnitude of $\overrightarrow{AB}$ but the opposite direction, so it is its negative, not its equal; $\overrightarrow{FC}$ has twice the magnitude of either, so it equals neither. So option D is correct.