Q.Find the angle between the vectors $2\hat{i}-\hat{j}+\hat{k}$ and $3\hat{i}+4\hat{j}-\hat{k}$.

Showing the 12 most recent of 25 on this concept.

1. 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:

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

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

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2. 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. …

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3. 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. $\vec d\cdot\vec n=6-6+1=1$; $|\vec d|=\sqrt{14}$, $|\vec n|=\sqrt{14}$. …

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4. 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. $\vec a\cdot\vec b=2\cdot\sqrt{14}\cdot\cos\tfrac{\pi}{4}=2\sqrt{14}\cdot\tfrac{\sqrt2}{2}=\sqrt{28}=2\sqrt7$. …

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5. 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 $\cos\theta = \tfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\tfrac12$.

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

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

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6. 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:

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

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7. 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 $\cos\theta=\dfrac{-6}{3\sqrt{14}}=\dfrac{-2}{\sqrt{14}}=\dfrac{-\sqrt{14}}{7}$. …

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8. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.If $\vec{a}$ and $\vec{b}$ are two unit vectors and if $\dfrac{\pi}{4}$ is the angle between $\vec{a}$ and $\vec{b}$, then $\left(\vec{a} + \left(\vec{a}\cdot\vec{b}\right)\vec{b}\right)\cdot\left(\vec{a} - \left(\vec{a}\cdot\vec{b}\right)\vec{b}\right)$ is equal to (A) $\dfrac{1}{4}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{3}{2}$ (D) $\dfrac{1}{2}$ (E) $\dfrac{5}{4}$

   ›Reveal solutionSolution

   $(\vec a+k\vec b)\cdot(\vec a-k\vec b)=|\vec a|^2-k^2|\vec b|^2$ with $k=\vec a\cdot\vec b=\cos\frac\pi4$.

   Here $\vec a\cdot\vec b=\cos\frac\pi4=\frac{1}{\sqrt2}$, and $|\vec a|=|\vec b|=1$. …

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9. 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,

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

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10. 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. $\vec{a}\cdot\vec{c}=3\sqrt{3}$. …

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11. 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 $|\vec a\cdot\vec b|=|\vec a||\vec b||\cos\theta|$. …

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12. 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: $\frac{\alpha^2}{\alpha^2+9}=\frac12\Rightarrow 2\alpha^2=\alpha^2+9\Rightarrow\alpha^2=9$. …

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