Q.If $\theta$ is the angle between two vectors $\hat{i} - 2\hat{j} + \hat{k}$ and $3\hat{i} - 2\hat{j} + \hat{k}$, find $\sin\theta$.

Showing the 12 most recent of 25 on this concept.

1. 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.

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

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2. 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, …

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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. $\vec{a}\cdot\vec{c}=3\sqrt{3}$. …

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4. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}|=7,|\vec{b}|=1$ and $|\vec{a}\times\vec{b}|^{2}=k^{2}-(\vec{a}\cdot\vec{b})^{2}$ then the value(s) of $k$ is/are (A) $5$ (B) $-5$ (C) $3$ (D) $-3$ (E) $\pm 7$

   ›Reveal solutionSolution

   The Lagrange identity gives $k^2=|\vec a|^2|\vec b|^2=49$, so $k=\pm7$.

   By the identity $|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2$. Comparing with the given $|\vec a\times\vec b|^2=k^2-(\vec a\cdot\vec b)^2$: …

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

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

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

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

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7. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.If $|\vec{a}| = 3$, $|\vec{b}| = 2$, then the value of $(2\vec{a} + 3\vec{b}) \cdot (2\vec{a} - 3\vec{b})$ is equal to (A) $6$ (B) $-12$ (C) $12$ (D) $-6$ (E) $0$

   ›Reveal solutionSolution

   This is a difference-of-squares dot product: $4|\vec a|^2 - 9|\vec b|^2$, which equals $0$.

   $(2\vec a + 3\vec b)\cdot(2\vec a - 3\vec b) = 4\,\vec a\cdot\vec a - 9\,\vec b\cdot\vec b = 4|\vec a|^2 - 9|\vec b|^2.$ …

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8. 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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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 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 …

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11. 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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12. KEAM 2025Set pha-2025-0424F4 marksMCQ
    Q.The angle between two unit vectors $\hat{A}$ and $\hat{B}$ is $60^{\circ}$. The value of $|\hat{A} - \hat{B}|$ is (A) $\frac{1}{2}$ (B) $\frac{3}{4}$ (C) $\frac{1}{4}$ (D) $1$ (E) $\frac{1}{8}$

    ›Reveal solutionSolution

    For unit vectors at $60^\circ$, $|\hat A-\hat B|=\sqrt{2-2\cos60^\circ}=\sqrt{1}=1$.

    Derivation: The magnitude of a vector difference is

    $|\hat A-\hat B|^2 = |\hat A|^2 + |\hat B|^2 - 2|\hat A||\hat B|\cos\theta.$

    With $|\hat A|=|\hat B|=1$ and $\theta = 60^\circ$ ($\cos60^\circ = \tfrac12$): …

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