Q.Find the angle between the following pair of lines:

Showing the 12 most recent of 15 on this concept.

1. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The angle between the lines $\frac{x-3}{-4} = \frac{y+2}{3} = \frac{z-1}{5}$ and $\frac{x-2}{2} = \frac{y-4}{1} = \frac{z+3}{3}$ is (A) $cos^{-1}\left(\frac{1}{3\sqrt{7}}\right)$ (B) $cos^{-1}\left(\frac{\sqrt{2}}{1}\right)$ (C) $cos^{-1}\left(\frac{2}{7}\right)$ (D) $cos^{-1}\left(\frac{1}{2\sqrt{7}}\right)$ (E) $cos^{-1}\left(\frac{1}{\sqrt{7}}\right)$

   ›Reveal solutionSolution

   Use direction ratios and $\cos\theta = \tfrac{|\vec d_1\cdot\vec d_2|}{|\vec d_1||\vec d_2|}$; it simplifies to $\tfrac{1}{\sqrt7}$.

   Directions: $\vec d_1 = (-4, 3, 5)$, $\vec d_2 = (2, 1, 3)$.

   $\vec d_1 \cdot \vec d_2 = -8 + 3 + 15 = 10.$

   $|\vec d_1| = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt2,\qquad |\vec d_2| = \sqrt{4 + 1 + 9} = \sqrt{14}.$ …

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2. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.The angle between the lines $\dfrac{x-1}{2}=\dfrac{2y+3}{4}=\dfrac{z+5}{-2}$ and $\dfrac{x-3}{4}=\dfrac{y+1}{-4}=\dfrac{z+3}{-4}$ is equal to (A) $\cos^{-1}\left(\dfrac{1}{8}\right)$ (B) $\cos^{-1}\left(\dfrac{1}{3}\right)$ (C) $\cos^{-1}\left(\dfrac{1}{4}\right)$ (D) $\cos^{-1}\left(\dfrac{1}{12}\right)$ (E) $\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$

   ›Reveal solutionSolution

   Rewriting line 1 as $\dfrac{x-1}{2}=\dfrac{y+3/2}{2}=\dfrac{z+5}{-2}$ gives direction $(2,2,-2)$; line 2 has direction $(4,-4,-4)$. Then $\cos\theta=\dfrac{|(2)(4)+(2)(-4)+(-2)(-4)|}{\sqrt{12}\,\sqrt{48}}=\dfrac{8}{24}=\dfrac13$.

   For line 1, $\dfrac{2y+3}{4}=\dfrac{y+3/2}{2}$, so the direction ratios are $(2,2,-2)$.

   For line 2, the direction ratios are $(4,-4,-4)$.

   Dot product: $(2)(4)+(2)(-4)+(-2)(-4)=8-8+8=8$. …

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3. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.The angle between the lines $\dfrac{x-1}{6}=\dfrac{y-5}{8}=\dfrac{z-3}{10}$ and $\dfrac{x+1}{2}=\dfrac{2y+3}{2}=\dfrac{z+3}{2}$ is (A) $\cos^{-1}\left(\dfrac{\sqrt{2}}{6}\right)$ (B) $\cos^{-1}\left(\dfrac{2\sqrt{2}}{3}\right)$ (C) $\cos^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$ (D) $\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$ (E) $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$

   ›Reveal solutionSolution

   Extract direction ratios, use the dot-product angle formula.

   Line 1 direction $(6,8,10)\parallel(3,4,5)$. For line 2, $\dfrac{x+1}{2}=\dfrac{2y+3}{2}=\dfrac{z+3}{2}$ rewrites as $\dfrac{x+1}{2}=\dfrac{y+3/2}{1}=\dfrac{z+3}{2}$, direction $(2,1,2)$. …

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4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.The angle between the lines $\dfrac{x-3}{1} = \dfrac{y+1}{-1} = \dfrac{z-2}{-1}$ and $\dfrac{x+1}{2} = \dfrac{y-2}{2} = \dfrac{z+3}{-2}$ is (A) $\cos^{-1}\left(\dfrac{\sqrt{2}}{6}\right)$ (B) $\cos^{-1}\left(\dfrac{\sqrt{6}}{6}\right)$ (C) $\cos^{-1}\left(\dfrac{\sqrt{2}}{2}\right)$ (D) $\cos^{-1}\left(\dfrac{1}{3}\right)$ (E) $\cos^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$

   ›Reveal solutionSolution

   The angle is $\cos^{-1}\!\left(\dfrac13\right)$.

   Concept and Intuition

   The angle between two lines uses their direction vectors: $\cos\theta=\dfrac{|\vec d_1\cdot\vec d_2|}{|\vec d_1||\vec d_2|}$.

   Step-by-Step Solution

   1. $\vec d_1=(1,-1,-1)$, $\vec d_2=(2,2,-2)$.
   2. $\vec d_1\cdot\vec d_2=2-2+2=2$; $|\vec d_1|=\sqrt3$, $|\vec d_2|=2\sqrt3$.
   3. $\cos\theta=\dfrac{2}{\sqrt3\cdot2\sqrt3}=\dfrac{2}{6}=\dfrac13\Rightarrow \theta=\cos^{-1}\!\left(\tfrac13\right)$.

   Common Mistakes …

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5. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.The angle between the two straight lines $\vec{r} = \left(4\hat{i} - \hat{k}\right) + t\left(2\hat{i} + \hat{j} - 2\hat{k}\right),\ t\in\mathbb{R}$ and $\vec{r} = \left(\hat{i} - \hat{j} + 2\hat{k}\right) + s\left(2\hat{i} - 2\hat{j} + \hat{k}\right),\ s\in\mathbb{R}$ is (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{6}$ (D) 0 (E) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Zero dot product of direction vectors $\Rightarrow$ angle $=\frac{\pi}{2}$.

   Direction vectors: $\vec d_1=(2,1,-2)$, $\vec d_2=(2,-2,1)$.

   $\vec d_1\cdot\vec d_2=4-2-2=0.$ …

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6. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.The angle between the lines $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{1}$ and $\dfrac{x}{0} = \dfrac{y}{1} = \dfrac{z}{-1}$ is (A) $\dfrac{\pi}{2}$ (B) $0$ (C) $\pi$ (D) $\dfrac{\pi}{4}$ (E) $\sin^{-1}\left(\sqrt{2}\right)$

   ›Reveal solutionSolution

   Dot product of direction vectors determines the angle.

   Direction vectors $\vec d_1=(1,1,1)$ and $\vec d_2=(0,1,-1)$:

   $\vec d_1\cdot\vec d_2=(1)(0)+(1)(1)+(1)(-1)=0.$ …

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7. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.If $\theta$ is angle between the lines $\frac{x}{1} = \frac{y+1}{2} = \frac{z-1}{3}$ and $\frac{x+1}{3} = \frac{y}{2} = \frac{z}{1}$, then $\cos\theta =$ (A) $\frac{5}{9}$ (B) $\frac{5}{8}$ (C) $\frac{5}{6}$ (D) $\frac{5}{7}$ (E) $\frac{6}{7}$

   ›Reveal solutionSolution

   $\cos\theta=\frac{5}{7}$.

   Concept and Intuition

   The angle between two lines equals the angle between their direction vectors: $\cos\theta=\frac{\vec d_1\cdot\vec d_2}{|\vec d_1||\vec d_2|}$.

   Step-by-Step Solution

   1. Direction vectors: $(1,2,3)$ and $(3,2,1)$.
   2. Dot product $=1\cdot3+2\cdot2+3\cdot1=3+4+3=10$.
   3. Magnitudes: $\sqrt{1+4+9}=\sqrt{14}$ each.
   4. $\cos\theta=\frac{10}{14}=\frac57$. …

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8. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The angle between the lines $\vec{r} = \hat{i} + 4\hat{k} + \lambda(2\hat{i} + \hat{j} - \hat{k})$ and $\vec{r} = 2\hat{i} - \hat{j} + 3\hat{k} + \mu(3\hat{i} + \hat{k})$ is (A) $\cos^{-1}\left(\frac{\sqrt{5}}{6}\right)$ (B) $\cos^{-1}\left(\frac{\sqrt{15}}{6}\right)$ (C) $\cos^{-1}\left(\frac{1}{12}\right)$ (D) $\cos^{-1}\left(\frac{\sqrt{15}}{15}\right)$ (E) $\cos^{-1}\left(\frac{\sqrt{3}}{30}\right)$

   ›Reveal solutionSolution

   The angle is $\cos^{-1}(\sqrt{15}/6)$.

   Concept and Intuition

   The angle between two lines equals the angle between their direction vectors, found via $\cos\theta=|\vec d_1\cdot\vec d_2|/(|\vec d_1||\vec d_2|)$.

   Step-by-Step Solution

   1. $\vec d_1=(2,1,-1)$, $\vec d_2=(3,0,1)$.
   2. $\vec d_1\cdot\vec d_2=6+0-1=5$; $|\vec d_1|=\sqrt6$, $|\vec d_2|=\sqrt{10}$. …

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9. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The angle between the lines $\vec{r} = (3 + \alpha)\hat{i} + 2(1 + \alpha)\hat{j} + 2(-2 + \alpha)\hat{k}$ and $\vec{r} = (5 + 3\beta)\hat{i} + 2(1 + \beta)\hat{j} + 6\beta\hat{k}$, where $\alpha$ and $\beta$ are parameters, is (A) $\cos^{-1}\left(\frac{17}{21}\right)$ (B) $\cos^{-1}\left(\frac{19}{12}\right)$ (C) $\cos^{-1}\left(\frac{9}{21}\right)$ (D) $2\cos^{-1}\left(\frac{19}{21}\right)$ (E) $\cos^{-1}\left(\frac{19}{21}\right)$

   ›Reveal solutionSolution

   Read each line's direction as the coefficient of its parameter, then use the dot-product angle formula.

   Line 1 direction (coefficient of $\alpha$): $\hat i+2\hat j+2\hat k=(1,2,2)$, magnitude $3$.

   Line 2 direction (coefficient of $\beta$): $3\hat i+2\hat j+6\hat k=(3,2,6)$, magnitude $7$. …

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10. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.The angle between the lines $\vec{r} = (3\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = (5\hat{i} - 2\hat{j}) + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is (A) $\cos^{-1}\left(\dfrac{9}{13}\right)$ (B) $\cos^{-1}\left(\dfrac{3}{19}\right)$ (C) $\cos^{-1}\left(\dfrac{19}{21}\right)$ (D) $\cos^{-1}\left(\dfrac{13}{17}\right)$ (E) $\cos^{-1}\left(\dfrac{3}{17}\right)$

    ›Reveal solutionSolution

    Direction vectors $(1,2,2)$ and $(3,2,6)$: $d_1\cdot d_2=19$, $|d_1|=3,|d_2|=7$, so $\cos\theta=\frac{19}{21}$. …

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11. KEAM 2023Set eng-2023-P2-B24 marksMCQ
    Q.The angle between the lines, whose direction cosines are proportional to $4,\sqrt{3}-1,-\sqrt{3}-1$ and $4,-\sqrt{3}-1,\sqrt{3}-1$, is (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$ (E) $\pi$

    ›Reveal solutionSolution

    The angle between the lines is $\tfrac{\pi}{3}$.

    Concept and Intuition

    Use $\cos\phi=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ with the given direction ratios.

    Step-by-Step Solution

    1. $\vec a\cdot\vec b=16+(\sqrt3-1)(-\sqrt3-1)+(-\sqrt3-1)(\sqrt3-1)=16-2-2=12$.
    2. $|\vec a|^2=16+(\sqrt3-1)^2+(\sqrt3+1)^2=16+8=24$; similarly $|\vec b|^2=24$. …

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12. KEAM 2026Set eng-2026-04184 marksMCQ
    Q.The direction ratios of a straight line $L_{1}$ are $2,-1,2$ and that of another straight line $L_{2}$ are $3,6,-2$. Then the angle between $L_{1}$ and $L_{2}$ is (A) $\cos^{-1}\left(\dfrac{-4}{21}\right)$ (B) $\cos^{-1}\left(\dfrac{8}{21}\right)$ (C) $\cos^{-1}\left(\dfrac{5}{21}\right)$ (D) $\cos^{-1}\left(\dfrac{-4}{7}\right)$ (E) $\cos^{-1}\left(\dfrac{10}{21}\right)$

    ›Reveal solutionSolution

    The dot product of the direction ratios over the magnitude product gives $\cos\theta=-\tfrac{4}{21}$.

    For direction ratios $(2,-1,2)$ and $(3,6,-2)$: …

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