Q.Show that the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.

Showing the 12 most recent of 15 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The straight line passing through the points $(3, 2, 3)$ and $(5, -1, -2)$ is perpendicular to the straight line passing through the points $(1, 3, 1)$ and $(\alpha, \alpha, \alpha)$ . Then the value of $\alpha$ is equal to (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

   ›Reveal solutionSolution

   Set the dot product of the two direction vectors to zero.

   First line direction: $(5-3,\,-1-2,\,-2-3)=(2,-3,-5)$.

   Second line direction: $(\alpha-1,\,\alpha-3,\,\alpha-1)$.

   Perpendicular: $2(\alpha-1)-3(\alpha-3)-5(\alpha-1)=0$. …

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2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.Let $\vec{a}=(\sin^{2}\alpha)\hat{i}+(\cos 2\alpha)\hat{j}+(\cos^{2}\alpha)\hat{k},\ 0\le\alpha\le\dfrac{\pi}{2}$ and $\vec{b}=\hat{i}-2\hat{j}+\hat{k}$. If $\vec{a}$ and $\vec{b}$ are perpendicular to each other, then the value of $\alpha$ is equal to (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$ (E) $0$

   ›Reveal solutionSolution

   The dot product simplifies to $1-2\cos2\alpha=0$, giving $\alpha=\tfrac{\pi}{6}$.

   $\vec a\cdot\vec b=0$:

   $(\sin^{2}\alpha)(1)+(\cos2\alpha)(-2)+(\cos^{2}\alpha)(1)=0.$

   Since $\sin^{2}\alpha+\cos^{2}\alpha=1$: …

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3. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $\vec{a} = 2\hat{i} - 2\hat{j} + 4\hat{k}$, $\vec{b} = -5\hat{i} - \hat{j} + 8\hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j} - \lambda\hat{k}$. If $\vec{a} + \vec{b} + \vec{c}$ and $\vec{a} - \vec{b} + \vec{c}$ are perpendicular, then the values of $\lambda$ are (A) 4 and -12 (B) -2 and 12 (C) -6 and 14 (D) -3 and 12 (E) -4 and 12

   ›Reveal solutionSolution

   Compute the two vectors, set their dot product to zero; only the $k$-terms survive.

   $\vec a+\vec b+\vec c=(2-5+3,\,-2-1+1,\,4+8-\lambda)=(0,-2,12-\lambda)$.

   $\vec a-\vec b+\vec c=(2+5+3,\,-2+1+1,\,4-8-\lambda)=(10,0,-4-\lambda)$. …

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4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.Let $\vec{a} + \vec{b} = \lambda\hat{i} + 16\hat{j} - 18\hat{k}$ and $\vec{a} - \vec{b} = 2\hat{i} + 8\hat{j} + \lambda\hat{k}$. If $\vec{a} + \vec{b}$ is perpendicular to $\vec{a} - \vec{b}$, then $|\vec{a}| =$ (A) $5\sqrt{13}$ (B) $\sqrt{174}$ (C) $\sqrt{184}$ (D) $13\sqrt{5}$ (E) $\sqrt{194}$

   ›Reveal solutionSolution

   $|\vec a|=\sqrt{194}$.

   Concept and Intuition

   Perpendicularity gives a dot-product equation that fixes $\lambda$; then $\vec a=\tfrac12[(\vec a+\vec b)+(\vec a-\vec b)]$.

   Step-by-Step Solution

   1. $(\vec a+\vec b)\cdot(\vec a-\vec b)=2\lambda+16\cdot8+(-18)\lambda=2\lambda+128-18\lambda=0\Rightarrow \lambda=8$.
   2. $\vec a+\vec b=8\hat i+16\hat j-18\hat k$, $\vec a-\vec b=2\hat i+8\hat j+8\hat k$. …

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5. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.A straight line passing through $(6, 1, 3)$ meets the line $\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z-2}{3}$ at $Q$. If the lines are perpendicular to each other, then the coordinates of $Q$ are (A) $(2, 1, 3)$ (B) $(1, 2, 3)$ (C) $(3, 1, 5)$ (D) $(2, -1, 3)$ (E) $(-1, 2, 3)$

   ›Reveal solutionSolution

   $Q=(3,1,5)$.

   Concept and Intuition

   Parametrize $Q$ on the given line, form vector $\vec{PQ}$ from $P(6,1,3)$, and impose that $\vec{PQ}$ is perpendicular to the line's direction.

   Step-by-Step Solution

   1. $Q=(1+2t,\,t,\,2+3t)$; direction $(2,1,3)$.
   2. $\vec{PQ}=(2t-5,\,t-1,\,3t-1)$; set $\vec{PQ}\cdot(2,1,3)=0$: $2(2t-5)+(t-1)+3(3t-1)=14t-14=0$.
   3. $t=1\Rightarrow Q=(3,1,5)$. …

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6. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.A straight line through the point $(1,-1,0)$ meets the line $\dfrac{x-1}{1}=\dfrac{y+1}{1}=\dfrac{z-1}{-1}$ at right angle. It's equation is (A) $\dfrac{x-1}{1}=\dfrac{y+1}{1}=\dfrac{z}{2}$ (B) $\dfrac{x-1}{1}=\dfrac{y-1}{1}=\dfrac{z}{4}$ (C) $\dfrac{x-1}{-1}=\dfrac{y-1}{1}=\dfrac{z}{6}$ (D) $\dfrac{x-1}{-1}=\dfrac{y-1}{-1}=\dfrac{z}{3}$ (E) $\dfrac{x-1}{1}=\dfrac{y+1}{1}=\dfrac{z}{-2}$

   ›Reveal solutionSolution

   Find where the perpendicular from $(1,-1,0)$ meets the given line: the connecting vector must be $\perp$ to $(1,1,-1)$, yielding direction $(1,1,2)$ and line $\dfrac{x-1}{1}=\dfrac{y+1}{1}=\dfrac{z}{2}$.

   The given line has direction $\vec{d}=(1,1,-1)$ and points $(1+t,\,-1+t,\,1-t)$.

   The vector from $(1,-1,0)$ to a general point on it is $(t,\,t,\,1-t)$. For a right angle it must be perpendicular to $\vec{d}$:

   $(t,t,1-t)\cdot(1,1,-1)=t+t-(1-t)=3t-1=0\Rightarrow t=\tfrac13.$ …

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7. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.If $\vec{a} = \alpha\hat{i} + \beta\hat{j}$ and $\vec{b} = \alpha\hat{i} - \beta\hat{j}$ are perpendicular, where $\alpha \ne \beta$, then $\alpha + \beta$ is equal to (A) $\alpha\beta$ (B) $\alpha - \beta$ (C) $\dfrac{1}{\alpha - \beta}$ (D) $\dfrac{1}{2\alpha\beta}$ (E) 0

   ›Reveal solutionSolution

   $\vec a\cdot\vec b=0\Rightarrow\alpha^2-\beta^2=0$, and $\alpha\ne\beta$ forces $\alpha+\beta=0$.

   $\vec a\cdot\vec b=(\alpha)(\alpha)+(\beta)(-\beta)=\alpha^2-\beta^2=0$. …

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8. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.The vectors $\vec{a}=4\hat{i}-3\hat{j}-\hat{k}$ and $\vec{b}=3\hat{i}+2\hat{j}+\lambda\hat{k}$ are perpendicular to each other. Then the value of $\lambda$ is equal to (A) $3$ (B) $4$ (C) $-3$ (D) $-4$ (E) $6$

   ›Reveal solutionSolution

   Perpendicular vectors have zero dot product.

   $\vec a\cdot\vec b=(4)(3)+(-3)(2)+(-1)(\lambda)=12-6-\lambda=6-\lambda.$ …

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9. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.If the lines $\frac{x-1}{2}=\frac{y-2}{2}=\frac{z-3}{\alpha}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ are perpendicular, then the value of $\alpha$ is (A) $6$ (B) $4$ (C) $3$ (D) $-3$ (E) $-2$

   ›Reveal solutionSolution

   The dot product of the direction ratios $(2,2,\alpha)$ and $(2,1,-2)$ must be zero: $4+2-2\alpha=0\Rightarrow\alpha=3$.

   Two lines are perpendicular when the dot product of their direction ratios is zero. Here the direction ratios are $(2,2,\alpha)$ and $(2,1,-2)$: …

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10. KEAM 2024Set eng-2024-06084 marksMCQ
    Q.A vector of magnitude 6 and perpendicular to $\vec{a}=2\vec{i}+2\vec{j}+\vec{k}$ and $\vec{b}=\vec{i}-2\vec{j}+2\vec{k}$, is (A) $\pm(2\vec{i}-\vec{j}-2\vec{k})$ (B) $\pm2(2\vec{i}-\vec{j}+2\vec{k})$ (C) $\pm3(2\vec{i}-\vec{j}-2\vec{k})$ (D) $\pm2(2\vec{i}+\vec{j}-2\vec{k})$ (E) $\pm2(2\vec{i}-\vec{j}-2\vec{k})$

    ›Reveal solutionSolution

    Compute $\vec a\times\vec b=(6,-3,-6)$, magnitude $9$; scale to length $6$ to get $\pm 2(2\vec i-\vec j-2\vec k)$.

    A vector perpendicular to both $\vec a=(2,2,1)$ and $\vec b=(1,-2,2)$ is their cross product:

    $\vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\2&2&1\\1&-2&2\end{vmatrix}=\vec i(4+2)-\vec j(4-1)+\vec k(-4-2)=(6,-3,-6).$

    Its magnitude is

    $|\vec a\times\vec b|=\sqrt{6^2+3^2+6^2}=\sqrt{81}=9.$ …

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11. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.If two vectors $\vec{a} = \cos\alpha\,\hat{i} + \sin\alpha\,\hat{j} + \sin\dfrac{\alpha}{2}\,\hat{k}$ and $\vec{b} = \sin\alpha\,\hat{i} - \cos\alpha\,\hat{j} + \cos\dfrac{\alpha}{2}\,\hat{k}$ are perpendicular, then the values of $\alpha$ are (A) $0$ and $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$ (C) $0$ and $\pi$ (D) $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$ (E) $0$ and $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    Set the dot product to zero and simplify.

    Compute $\vec a\cdot\vec b$:

    $\vec a\cdot\vec b=\cos\alpha\sin\alpha+\sin\alpha(-\cos\alpha)+\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}.$

    The first two terms cancel, and $\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}=\dfrac{1}{2}\sin\alpha$, so …

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12. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.The lines $\dfrac{x+3}{-2} = \dfrac{y}{1} = \dfrac{z-4}{3}$ and $\dfrac{x}{\mu} = \dfrac{y-1}{\mu+1} = \dfrac{z}{\mu+2}$ are perpendicular to each other. Then the value of $\mu$ is (A) $\dfrac{-5}{3}$ (B) $3$ (C) $4$ (D) $\dfrac{-1}{4}$ (E) $\dfrac{-7}{2}$

    ›Reveal solutionSolution

    Set the dot product of the two direction vectors to zero.

    Direction vectors: $\vec d_1=(-2,1,3)$ and $\vec d_2=(\mu,\mu+1,\mu+2)$. Perpendicular $\Rightarrow \vec d_1\cdot\vec d_2=0$: …

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