Q.Find the foot of perpendicular from the point $(2, 3, -8)$ to the line $\dfrac{4-x}{2} = \dfrac{y}{6} = \dfrac{1-z}{3}$. Also, find the perpendicular distance from the given point to the line.

Showing the 12 most recent of 18 on this concept.

1. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.The shortest distance between the point $(2, 3, 4)$ and the line $\dfrac{x - 4}{-2} = \dfrac{y - 4}{2} = \dfrac{z - 6}{1}$ is (A) $12$ (B) $9$ (C) $3$ (D) $\sqrt{5}$ (E) $\sqrt{3}$

   ›Reveal solutionSolution

   Use the point-to-line formula $\dfrac{|\vec{AP}\times\vec d|}{|\vec d|}$ with $A=(4,4,6)$, $\vec d=(-2,2,1)$.

   The line passes through $A=(4,4,6)$ with direction $\vec d=(-2,2,1)$, $|\vec d|=3$. For $P=(2,3,4)$:

   $\vec{AP}=P-A=(-2,-1,-2).$

   Compute the cross product: …

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2. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The perpendicular distance between the lines $3x + 4y - 6 = 0$ and $6x + 8y + 18 = 0$ is (A) 15 (B) 12 (C) 9 (D) 3 (E) 0

   ›Reveal solutionSolution

   Make the two lines have identical $x,y$ coefficients, then apply the parallel-line distance formula.

   Divide $6x+8y+18=0$ by $2$: $3x+4y+9=0$. This is parallel to $3x+4y-6=0$. …

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3. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The distance of the point $P(1,-3)$ from the line $2y-3x=4$ is (A) $\sqrt{13}$ units (B) $13$ units (C) $7$ units (D) $\sqrt{7}$ units (E) $2\sqrt{13}$ units

   ›Reveal solutionSolution

   Applying the point–line distance formula gives $\frac{13}{\sqrt{13}}=\sqrt{13}$.

   Write the line $2y-3x=4$ as $-3x+2y-4=0$. Distance from $P(1,-3)$: …

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4. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The shortest distance between the lines $\vec{r} = \hat{i} + \hat{j} + 3\hat{k} + \lambda(2\hat{i} + 2\hat{j} + \hat{k})$ and $\vec{r} = (2\mu + 1)\hat{i} + (2\mu - 1)\hat{j} + (\mu + 1)\hat{k}$, where $\lambda$ and $\mu$ are parameters, is (A) 1 (B) 6 (C) 3 (D) 4 (E) 2

   ›Reveal solutionSolution

   Recognize the lines as parallel, then apply the parallel-line distance formula.

   Line 1: point $(1,1,3)$, direction $(2,2,1)$. Line 2: $(2\mu+1,2\mu-1,\mu+1)$ has point $(1,-1,1)$ at $\mu=0$ and direction $(2,2,1)$ — same direction, so parallel.

   Joining vector $=(1-1,-1-1,1-3)=(0,-2,-2)$. …

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5. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.Suppose P is the point on the line joining $(-9,4,5)$ and $(11,0,-1)$ that lies closest to the origin O. Then $|OP|^2$ equals to (A) $3$ (B) $4$ (C) $2$ (D) $9$ (E) $1$

   ›Reveal solutionSolution

   $|OP|^2=9$.

   Concept and Intuition

   The closest point on a line to the origin is the foot of the perpendicular from $O$; find the parameter making $\vec{OP}\perp$ the direction vector.

   Step-by-Step Solution

   1. Line: $P=(-9+20t,\,4-4t,\,5-6t)$, direction $\vec d=(20,-4,-6)$.
   2. $\vec{OP}\cdot\vec d=0$: $20(-9+20t)-4(4-4t)-6(5-6t)=-226+452t=0\Rightarrow t=\tfrac12$. …

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6. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.Let $\overrightarrow{OP} = 2\hat{j}$ be the position vector a point P. Let $\vec{r} = \hat{j} + \lambda(\hat{i} + \hat{j})$ be a straight line. The distance of the point P from the line is (A) $\dfrac{\sqrt{2}}{2}$ (B) $\dfrac{\sqrt{3}}{3}$ (C) $\dfrac{\sqrt{6}}{3}$ (D) $\dfrac{\sqrt{2}}{3}$ (E) $\dfrac{\sqrt{2}}{4}$

   ›Reveal solutionSolution

   Point $P=(0,2)$; line passes through $(0,1)$ with direction $(1,1)$. Distance $=\dfrac{|(P-A)\times \hat d|}{1}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$.

   Here $\overrightarrow{OP}=2\hat j$ gives $P=(0,2)$. The line $\vec r=\hat j+\lambda(\hat i+\hat j)$ passes through $A=(0,1)$ with direction $\vec d=(1,1)$. …

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7. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The nearest point on the line $x+2y=5$ from the point $P(7,9)$ is equal to (A) $(6,1)$ (B) $(7,6)$ (C) $(2,3)$ (D) $(8,3)$ (E) $(3,1)$

   ›Reveal solutionSolution

   The nearest point is the foot of the perpendicular from $P$ to the line. Parametrize along the normal direction $(1,2)$: point $(7+t,9+2t)$ on the line gives $t=-4$, so the foot is $(3,1)$.

   The nearest point on $x+2y=5$ from $P(7,9)$ is the foot of the perpendicular. The line's normal direction is $(1,2)$, so points on the perpendicular through $P$ are

   $(7+t,\;9+2t).$

   Require this to lie on the line: …

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8. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The shortest distance between the straight line $\vec{r}=3\hat{i}+12\hat{j}+5\hat{k}+\lambda(2\hat{i}),\ \lambda\in\mathbb{R}$ and the $x$-axis is (A) $7$ (B) $12$ (C) $3\sqrt{5}$ (D) $5\sqrt{3}$ (E) $13$

   ›Reveal solutionSolution

   The line is parallel to the $x$-axis; the shortest distance is $\sqrt{12^2+5^2}=13$.

   The line $\vec r=3\hat i+12\hat j+5\hat k+\lambda(2\hat i)$ has direction $(1,0,0)$, the same as the $x$-axis, so the two lines are parallel. …

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9. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The shortest distance from the point $(-10, 10, -10)$ to the $z$-axis, is (A) $2\sqrt{10}$ (B) $10\sqrt{3}$ (C) $10$ (D) $3\sqrt{10}$ (E) $10\sqrt{2}$

   ›Reveal solutionSolution

   Distance from a point to the $z$-axis is $\sqrt{x^2+y^2}$. …

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10. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.The point with integral coordinates on the line $x + y = 1$, that lie at a distance 2 units from the line $5x + 12y = 0$, is (A) $(-7, 8)$ (B) $(-1, 2)$ (C) $(-12, 13)$ (D) $(-2, 3)$ (E) $(-3, 4)$

    ›Reveal solutionSolution

    Parametrize the point on $x+y=1$ and impose the distance condition; the integer solution is $(-2,3)$.

    A point on $x+y=1$ is $(a,1-a)$. Its distance from $5x+12y=0$ is

    $\frac{|5a+12(1-a)|}{\sqrt{5^2+12^2}}=\frac{|12-7a|}{13}=2.$ …

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11. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.If the distance of the line $4x - 3y + k = 0$ from the point $(1, 2)$ is $5$ units, then the values of $k$ are (A) $27,-23$ (B) $-27,23$ (C) $29,-24$ (D) $-29,24$ (E) $-28,-25$

    ›Reveal solutionSolution

    Distance formula: $\frac{|4(1)-3(2)+k|}{5}=5\Rightarrow|k-2|=25\Rightarrow k=27,\,-23$.

    Apply the point-line distance. $d=\frac{|4x_0-3y_0+k|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{|4(1)-3(2)+k|}{5}=\frac{|k-2|}{5}$. …

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12. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.The distance from the point $(2, 2, 2)$ to the plane $2x - y + 3z = 5$ is equal to (A) $\frac{3\sqrt{7}}{2}$ (B) $\frac{\sqrt{3}}{2}$ (C) $\frac{3\sqrt{14}}{7}$ (D) $\frac{3\sqrt{14}}{14}$ (E) $\frac{\sqrt{3}}{3}$

    ›Reveal solutionSolution

    The distance is $\frac{3\sqrt{14}}{14}$.

    Concept and Intuition

    Distance from point $(x_0,y_0,z_0)$ to plane $ax+by+cz=d$ is $\frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$.

    Step-by-Step Solution

    1. Plug in: $2(2)-1(2)+3(2)-5=4-2+6-5=3$.
    2. Denominator $\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}$.
    3. Distance $=\frac{3}{\sqrt{14}}=\frac{3\sqrt{14}}{14}$.

    Common Mistakes …

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