Q.If $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}+5\hat{j}$, $3\hat{i}+2\hat{j}-3\hat{k}$ and $\hat{i}-6\hat{j}-\hat{k}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

Showing the 12 most recent of 25 on this concept.

1. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.Let $\vec{AB}=2\hat{i}+10\hat{j}+11\hat{k}$ and $\vec{AC}=-\hat{i}+2\hat{j}+2\hat{k}$. If $\theta$ is the angle between $\vec{AB}$ and $\vec{AC}$ then $\sin\theta =$ (A) $\dfrac{\sqrt{13}}{9}$ (B) $\dfrac{\sqrt{15}}{9}$ (C) $\dfrac{\sqrt{14}}{9}$ (D) $\dfrac{\sqrt{17}}{9}$ (E) $\dfrac{4}{9}$

   ›Reveal solutionSolution

   Using the cross product, $|\vec{AB}\times\vec{AC}|=5\sqrt{17}$, and dividing by $|\vec{AB}||\vec{AC}|=45$ gives $\sin\theta=\dfrac{\sqrt{17}}{9}$.

   $\vec{AB}=(2,10,11)$, $|\vec{AB}|=\sqrt{4+100+121}=\sqrt{225}=15$.

   $\vec{AC}=(-1,2,2)$, $|\vec{AC}|=\sqrt{1+4+4}=3$.

   Cross product:

   $\vec{AB}\times\vec{AC}=\big(10\cdot2-11\cdot2,\;-(2\cdot2-11\cdot(-1)),\;2\cdot2-10\cdot(-1)\big)=(-2,-15,14).$

   $|\vec{AB}\times\vec{AC}|=\sqrt{4+225+196}=\sqrt{425}=5\sqrt{17}.$

   Therefore $\sin\theta=\dfrac{5\sqrt{17}}{15\cdot3}=\dfrac{\sqrt{17}}{9}$.

   ✓Final answer

   The correct option is (D).

2. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.If $\vec{a}=2\vec{i}+4\vec{j}+7\vec{k}$ and $\vec{b}=4\vec{i}+7\vec{j}+2\vec{k}$, then the angle between $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is equal to (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{2}$ (D) $\frac{2\pi}{3}$ (E) $\frac{2\pi}{5}$

   ›Reveal solutionSolution

   $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2$. Because $|\vec a|^2=|\vec b|^2=69$, the dot product is $0$, so the angle is $\frac{\pi}{2}$.

   We have

   $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2.$

   Computing the magnitudes:

   $|\vec a|^2=2^2+4^2+7^2=4+16+49=69,$

   $|\vec b|^2=4^2+7^2+2^2=16+49+4=69.$

   Since $|\vec a|^2=|\vec b|^2$, the dot product is $0$, so $\vec a+\vec b$ and $\vec a-\vec b$ are perpendicular and the angle between them is $\dfrac{\pi}{2}$.

   ✓Final answer

   The correct option is (C).

3. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.Let $|\vec{a}|=2$, $|\vec{b}|=\sqrt{13+6\sqrt{3}}$ and $|\vec{c}|=3$. If $\vec{a}-\vec{b}+\vec{c}=0$, then the angle between $\vec{a}$ and $\vec{c}$ is (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{12}$ (D) $\frac{\pi}{2}$ (E) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   Rearrange to $\vec{b}=\vec{a}+\vec{c}$, square both sides, and read off $\vec{a}\cdot\vec{c}$.

   Since $\vec{a}-\vec{b}+\vec{c}=0$, we have $\vec{b}=\vec{a}+\vec{c}$.

   Then $|\vec{b}|^2=|\vec{a}|^2+|\vec{c}|^2+2\vec{a}\cdot\vec{c}=4+9+2\vec{a}\cdot\vec{c}=13+2\vec{a}\cdot\vec{c}$.

   Given $|\vec{b}|^2=13+6\sqrt{3}$, so $2\vec{a}\cdot\vec{c}=6\sqrt{3}$, i.e. $\vec{a}\cdot\vec{c}=3\sqrt{3}$.

   $\cos\theta=\dfrac{\vec{a}\cdot\vec{c}}{|\vec{a}||\vec{c}|}=\dfrac{3\sqrt{3}}{2\cdot 3}=\dfrac{\sqrt{3}}{2}$, hence $\theta=\dfrac{\pi}{6}$.

   ✓Final answer

   The correct option is (E).

4. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.Let $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{a} + \vec{b} = 4\hat{i} - 2\hat{j} + \lambda\hat{k}$. If $\vec{a} \cdot \vec{b} = 4$, then the value of $\lambda$ is equal to (A) $3$ (B) $-3$ (C) $-6$ (D) $6$ (E) $0$

   ›Reveal solutionSolution

   $\lambda=-6$.

   Concept and Intuition

   Find $\vec b$ by subtracting $\vec a$ from $\vec a+\vec b$, then apply the dot-product condition.

   Step-by-Step Solution

   1. $\vec b=(\vec a+\vec b)-\vec a=(4-1)\hat i+(-2-2)\hat j+(\lambda-(-3))\hat k=3\hat i-4\hat j+(\lambda+3)\hat k$.
   2. $\vec a\cdot\vec b=(1)(3)+(2)(-4)+(-3)(\lambda+3)=3-8-3\lambda-9=-14-3\lambda$.
   3. Set $-14-3\lambda=4\Rightarrow -3\lambda=18\Rightarrow \lambda=-6$.

   Common Mistakes

   • Sign error in the $\hat k$ component: $\lambda-(-3)=\lambda+3$.
   • Miscomputing $(-3)(\lambda+3)$.

   ✓Final answer

   The correct option is (C) — $-6$.

   ANSWER: C

5. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.If $\vec{a}$ and $\vec{b}$ are two nonzero vectors and if $\left|\vec{a}\times\vec{b}\right| = \left|\vec{a}\cdot\vec{b}\right|$, then the angle between $\vec{a}$ and $\vec{b}$ is equal to (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{6}$ (E) $\dfrac{2\pi}{3}$

   ›Reveal solutionSolution

   Equating magnitudes gives $\tan\theta=1\Rightarrow\theta=\frac{\pi}{4}$.

   $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$ and $|\vec a\cdot\vec b|=|\vec a||\vec b||\cos\theta|$.

   Setting them equal: $\sin\theta=|\cos\theta|\Rightarrow\tan\theta=1\Rightarrow\theta=\frac{\pi}{4}$.

   ✓Final answer

   The correct option is (B).

6. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.Let $\vec{a}$ and $\vec{b}$ be two unit vectors, and $\theta$ be the angle between them. If $\vec{a} - \vec{b}$ is a unit vector, then $\theta$ is equal to (A) $\dfrac{\pi}{3}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{2\pi}{3}$ (E) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   From $|\vec a-\vec b|=1$ with unit vectors, $\cos\theta=\tfrac12$, so $\theta=\tfrac{\pi}{3}$.

   Since $\vec a$ and $\vec b$ are unit vectors,

   $|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\,\vec a\cdot\vec b=2-2\cos\theta.$

   Given $|\vec a-\vec b|=1$, so $2-2\cos\theta=1$, giving $\cos\theta=\dfrac12$ and $\theta=\dfrac{\pi}{3}$.

   ✓Final answer

   The correct option is (A).

7. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.If $|\vec{a}| = 8$, $|\vec{b}| = 5$, and $|\vec{a} - \vec{b}| = 7$ then the angle between $\vec{a}$ and $\vec{b}$ is equal to (A) $\frac{3\pi}{4}$ (B) $\frac{2\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{6}$ (E) $\frac{\pi}{3}$

   ›Reveal solutionSolution

   Expand $|\vec a-\vec b|^2$ to get $\vec a\cdot\vec b=20$, then $\cos\theta = \tfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\tfrac12$.

   $|\vec a - \vec b|^2 = |\vec a|^2 + |\vec b|^2 - 2\,\vec a\cdot\vec b.$

   $7^2 = 8^2 + 5^2 - 2\,\vec a\cdot\vec b \;\Rightarrow\; 49 = 89 - 2\,\vec a\cdot\vec b \;\Rightarrow\; \vec a\cdot\vec b = 20.$

   $\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|} = \frac{20}{8\cdot 5} = \frac{20}{40} = \frac12 \;\Rightarrow\; \theta = \frac{\pi}{3}.$

   ✓Final answer

   The correct option is (E).

8. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.If $\vec{a}$ and $\vec{b}$ are position vectors of the points $(\alpha, 3, 0)$ and $(1, 0, 0)$ respectively and if the angle between the vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$, then the value of $\alpha$ is equal to (A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $5$

   ›Reveal solutionSolution

   $\alpha=3$.

   Concept and Intuition

   The angle between position vectors satisfies $\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$. Here $\vec a=(\alpha,3,0)$ and $\vec b=(1,0,0)$.

   Step-by-Step Solution

   1. $\vec a\cdot\vec b=\alpha$; $|\vec a|=\sqrt{\alpha^2+9}$, $|\vec b|=1$.
   2. $\cos\frac{\pi}{4}=\frac{\alpha}{\sqrt{\alpha^2+9}}=\frac{1}{\sqrt2}$.
   3. Square: $\frac{\alpha^2}{\alpha^2+9}=\frac12\Rightarrow 2\alpha^2=\alpha^2+9\Rightarrow\alpha^2=9$.
   4. Take $\alpha=3$ (positive, so the angle is acute).

   Common Mistakes

   • Forgetting the acute-angle requirement rules out $\alpha=-3$.
   • Errors squaring the equation.

   ✓Final answer

   The correct option is (C) — $3$.

   ANSWER: C

9. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The angle between the line $\vec{r} = \hat{i} + 2\hat{j} + t(3\hat{i} + 2\hat{j} - \hat{k})$ and the plane $2x - 3y - z = 1$ is (A) $\sin^{-1}\left(\frac{1}{196}\right)$ (B) $\sin^{-1}\left(\frac{1}{14}\right)$ (C) $\cos^{-1}\left(\frac{1}{14}\right)$ (D) $\cos^{-1}\left(\frac{13}{14}\right)$ (E) $\sin^{-1}\left(\frac{13}{14}\right)$

   ›Reveal solutionSolution

   The angle is $\sin^{-1}(1/14)$.

   Concept and Intuition

   The angle between a line and a plane is the complement of the angle between the line's direction and the plane's normal, so it uses the sine: $\sin\theta=|\vec d\cdot\vec n|/(|\vec d||\vec n|)$.

   Step-by-Step Solution

   1. Direction $\vec d=(3,2,-1)$, normal $\vec n=(2,-3,-1)$.
   2. $\vec d\cdot\vec n=6-6+1=1$; $|\vec d|=\sqrt{14}$, $|\vec n|=\sqrt{14}$.
   3. $\sin\theta=\dfrac{1}{\sqrt{14}\cdot\sqrt{14}}=\dfrac{1}{14}$, so $\theta=\sin^{-1}(1/14)$.

   Common Mistakes

   • Using $\cos$ instead of $\sin$ (that gives the line-normal angle, not the line-plane angle).

   ✓Final answer

   The correct option is (B) — $\sin^{-1}\left(\frac{1}{14}\right)$.

   ANSWER: B

10. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.The angle subtended by the vector $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ with the $y$-axis is (A) $\cos^{-1}\left(\frac{2}{\sqrt{3}}\right)$ (B) $\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$ (C) $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$ (D) $\sin^{-1}\left(\frac{2}{\sqrt{3}}\right)$ (E) $\frac{\pi}{2}$

    ›Reveal solutionSolution

    The angle with the $y$-axis satisfies $\cos\theta=\tfrac{A_y}{|\vec{A}|}=\tfrac{1}{\sqrt{3}}$.

    Component along $y$-axis. For $\vec{A}=\hat{i}+\hat{j}+\hat{k}$, the $y$-component is $A_y=1$ and the magnitude is

    $|\vec{A}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}.$

    Direction cosine with the $y$-axis.

    $\cos\theta = \frac{\vec{A}\cdot\hat{j}}{|\vec{A}|} = \frac{1}{\sqrt{3}} \\;\Rightarrow\\; \theta = \cos^{-1}\\!\left(\frac{1}{\sqrt{3}}\right).$

    ✓Final answer

    The correct option is (C).

11. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.Let $\vec{a}, \vec{b}$ and $\vec{c}$ be the sides of a triangle ABC such that $\vec{BC} = \vec{a}$, $\vec{CA} = \vec{b}$ and $\vec{AB} = \vec{c}$. If $BC = AC = 3$ and $\vec{b}\cdot\vec{c} = -9$ then $\vec{a}\cdot\vec{b}$ is equal to (A) $27$ (B) $9$ (C) $3\sqrt{3}$ (D) $0$ (E) $-3\sqrt{3}$

    ›Reveal solutionSolution

    Use the closure $\vec a+\vec b+\vec c=0$ and the given $\vec b\cdot\vec c=-9$, $|\vec b|=3$ to get $\vec a\cdot\vec b=0$.

    For a triangle with $\vec{BC}=\vec a$, $\vec{CA}=\vec b$, $\vec{AB}=\vec c$, we have $\vec a+\vec b+\vec c=\vec 0$, so $\vec c=-(\vec a+\vec b)$.

    Here $|\vec b|=CA=3$, so $|\vec b|^2=9$. Then

    $\vec b\cdot\vec c=\vec b\cdot\big(-(\vec a+\vec b)\big)=-(\vec a\cdot\vec b+|\vec b|^2)=-(\vec a\cdot\vec b+9).$

    Given $\vec b\cdot\vec c=-9$:

    $-(\vec a\cdot\vec b+9)=-9\implies \vec a\cdot\vec b=0.$

    ✓Final answer

    The correct option is (D).

12. KEAM 2026Set eng-2026-04194 marksMCQ
    Q.If $|\vec{a}| = \sqrt{26}$, $|\vec{b}| = \sqrt{3}$ and $\vec{a} \times \vec{b} = 5\hat{i} + \hat{j} - 4\hat{k}$, then $\vec{a} \cdot \vec{b} =$ (A) $\pm 12$ (B) $\pm 4$ (C) $\pm 10$ (D) $\pm 8$ (E) $\pm 6$

    ›Reveal solutionSolution

    Use $|\vec a|^2|\vec b|^2=(\vec a\cdot\vec b)^2+|\vec a\times\vec b|^2$.

    $|\vec a\times\vec b|^2=5^2+1^2+(-4)^2=42$ and $|\vec a|^2|\vec b|^2=26\cdot3=78$.

    By the identity,

    $(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2-|\vec a\times\vec b|^2=78-42=36\Rightarrow \vec a\cdot\vec b=\pm6.$

    ✓Final answer

    The correct option is (E).