Q.Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$.

Showing the 12 most recent of 25 on this concept.

1. 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).

2. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.If the angle between the vectors $\vec{a} = x\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = x\hat{i} - x\hat{j} + 2\hat{k}$ is acute, then $x$ lies is (A) $(-\infty, -1) \cup (1, \infty)$ (B) $(-\infty, -1) \cup (2, \infty)$ (C) $(-\infty, 1) \cup (2, \infty)$ (D) $(-\infty, 0) \cup (1, \infty)$ (E) $(-\infty, 0) \cup (2, \infty)$

   ›Reveal solutionSolution

   The angle is acute iff the dot product is positive.

   $\vec a\cdot\vec b=x\cdot x+3(-x)+1\cdot2=x^2-3x+2$.

   Acute $\Rightarrow x^2-3x+2>0\Rightarrow(x-1)(x-2)>0$.

   Solution: $x<1$ or $x>2$, i.e. $(-\infty,1)\cup(2,\infty)$.

   ✓Final answer

   The correct option is (C).

3. 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).

4. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.Let $\theta$ be the angle between the unit vectors $\hat{a}$ and $\hat{b}$ . If $|\hat{a} - \hat{b}| = \dfrac{\sqrt{3}}{2}$ , then the value of $\cos\theta$ is (A) $\dfrac{3}{8}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{5}{8}$ (D) $\dfrac{3}{4}$ (E) $\dfrac{7}{8}$

   ›Reveal solutionSolution

   Use $|\hat a-\hat b|^2=2-2\cos\theta$ for unit vectors.

   $|\hat a-\hat b|^2=|\hat a|^2+|\hat b|^2-2\hat a\cdot\hat b=1+1-2\cos\theta=2-2\cos\theta$.

   Given $|\hat a-\hat b|=\dfrac{\sqrt3}{2}$, so $|\hat a-\hat b|^2=\dfrac34$. Then $2-2\cos\theta=\dfrac34\Rightarrow 2\cos\theta=\dfrac54\Rightarrow\cos\theta=\dfrac58$.

   ✓Final answer

   The correct option is (C).

5. 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).

6. 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).

7. 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

8. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The acute angle between the planes $2x - y - 3z = 7$ and $x + 2y + 2z = 0$ is (A) $\cos^{-1}\left(\frac{-\sqrt{14}}{14}\right)$ (B) $\pi - \cos^{-1}\left(\frac{-\sqrt{14}}{7}\right)$ (C) $\cos^{-1}\left(\frac{\sqrt{14}}{11}\right)$ (D) $\pi - \cos^{-1}\left(\frac{-\sqrt{14}}{21}\right)$ (E) $\pi - \cos^{-1}\left(\frac{\sqrt{14}}{7}\right)$

   ›Reveal solutionSolution

   The acute angle equals $\pi-\cos^{-1}(-\sqrt{14}/7)\approx57.7^\circ$.

   Concept and Intuition

   The angle between two planes is the angle between their normals. If the signed cosine is negative, that formula returns the obtuse angle, and the acute angle is its supplement $\pi$ minus it.

   Step-by-Step Solution

   1. Normals $\vec n_1=(2,-1,-3)$, $\vec n_2=(1,2,2)$; $\vec n_1\cdot\vec n_2=2-2-6=-6$.
   2. $|\vec n_1|=\sqrt{14}$, $|\vec n_2|=3$, so $\cos\theta=\dfrac{-6}{3\sqrt{14}}=\dfrac{-2}{\sqrt{14}}=\dfrac{-\sqrt{14}}{7}$.
   3. This $\theta=\cos^{-1}(-\sqrt{14}/7)\approx122.3^\circ$ is obtuse; the acute angle is $\pi-\cos^{-1}(-\sqrt{14}/7)\approx57.7^\circ$.

   Common Mistakes

   • Taking the raw obtuse value as the answer, or dropping the sign and matching $\pi-\cos^{-1}(\sqrt{14}/7)$ (which is itself obtuse).

   ✓Final answer

   The correct option is (B) — $\pi - \cos^{-1}\left(\frac{-\sqrt{14}}{7}\right)$.

   ANSWER: B

9. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.If $|\vec{a}| = 2$, $\vec{b} = 2\hat{i} - \hat{j} - 3\hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$, then $\vec{a} \cdot \vec{b}$ is equal to (A) $14\sqrt{2}$ (B) $2\sqrt{7}$ (C) $\sqrt{30}$ (D) $\sqrt{7}$ (E) $\sqrt{14}$

   ›Reveal solutionSolution

   $\vec a\cdot\vec b=2\sqrt7$.

   Concept and Intuition

   The dot product equals the product of the magnitudes times the cosine of the angle between the vectors: $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$.

   Step-by-Step Solution

   1. $|\vec b|=\sqrt{2^2+(-1)^2+(-3)^2}=\sqrt{14}$.
   2. $\vec a\cdot\vec b=2\cdot\sqrt{14}\cdot\cos\tfrac{\pi}{4}=2\sqrt{14}\cdot\tfrac{\sqrt2}{2}=\sqrt{28}=2\sqrt7$.

   Common Mistakes

   • Forgetting to compute $|\vec b|$, or mishandling $\sqrt{14}\cdot\sqrt2=\sqrt{28}=2\sqrt7$.

   ✓Final answer

   The correct option is (B) — $2\sqrt{7}$.

   ANSWER: B

10. 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).

11. 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).

12. 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).