Q.Find angle $\theta$ between the vectors $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$.

Showing the 12 most recent of 25 on this concept.

1. 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).

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.

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

   ✓Final answer

   The correct option is (C).

3. 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).

4. 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).

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

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

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

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

   ✓Final answer

   The correct option is (E).

6. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.If $|\vec{a}| = \sqrt{14}$, $|\vec{b}| = \sqrt{10}$, $|\vec{a} - \vec{b}| = \sqrt{24}$ and $\theta$ is angle between $\vec{a}$ and $\vec{b}$, then $\cos\theta =$ (A) $\frac{\sqrt{35}}{70}$ (B) $\frac{\sqrt{6}}{12}$ (C) $\frac{\sqrt{15}}{60}$ (D) $\frac{\sqrt{210}}{35}$ (E) $0$

   ›Reveal solutionSolution

   $\cos\theta=0$.

   Concept and Intuition

   The law of cosines for vectors gives $\vec a\cdot\vec b$, and $\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$.

   Step-by-Step Solution

   1. $|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b$.
   2. $24=14+10-2\vec a\cdot\vec b=24-2\vec a\cdot\vec b$.
   3. So $\vec a\cdot\vec b=0$, and $\cos\theta=\frac{0}{\sqrt{14}\sqrt{10}}=0$.

   Common Mistakes

   • Using $|\vec a+\vec b|$ formula (wrong sign) instead of $|\vec a-\vec b|$.
   • Forgetting to divide by the magnitudes when finding $\cos\theta$.

   ✓Final answer

   The correct option is (E) — $0$.

   ANSWER: E

7. 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).

8. 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).

9. 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).

10. 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

11. 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 $\theta = 60^\circ$ ($\cos60^\circ = \tfrac12$):

    $|\hat A-\hat B|^2 = 1 + 1 - 2(1)(1)\left(\tfrac12\right) = 2 - 1 = 1.$

    Hence $|\hat A-\hat B| = 1$.

    ✓Final answer

    The correct option is (D).

12. 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 $|\vec a\times\vec b|^2=k^2-(\vec a\cdot\vec b)^2$:

    $k^2=|\vec a|^2|\vec b|^2=7^2\cdot1^2=49\;\Rightarrow\;k=\pm7.$

    ✓Final answer

    The correct option is (E).