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 19 on this concept.

1. COMEDK 2025Set 2025-A1 markMCQ
   Q.The length of the perpendicular from the point $P(1,-1,2)$ to the given line $\frac{x+1}{2}=\frac{y-2}{-3}=\frac{z+2}{4}$ is (A) $\sqrt{29}$ units (B) $\sqrt{21}$ units (C) $\sqrt{6}$ units (D) 0 units

   ›Reveal solutionSolution

   The shortest distance from a point to a line in 3D is found by projecting the vector from a point on the line to the given point onto the direction vector, then using the Pythagorean theorem. The perpendicular distance is $\sqrt{6}$ units.

   Concept & Intuition

   In 3D geometry, the perpendicular distance from a point to a line is the length of the line segment that meets the given line at a right angle. Instead of solving for the foot of the perpendicular directly, we can use vector projection:

   • Pick any convenient point $A$ on the line.
   • Form the vector $\overrightarrow{AP}$ from $A$ to the given point $P$.
   • The component of $\overrightarrow{AP}$ parallel to the line’s direction vector $\vec{d}$ gives the “along-the-line” part.
   • The component perpendicular to $\vec{d}$ is the shortest distance. By the Pythagorean theorem:

   $\text{distance} = \sqrt{|\overrightarrow{AP}|^2 - \left(\text{proj}_{\vec{d}} \overrightarrow{AP}\right)^2}.$

   Step-by-step solution

   1. Identify a point on the line and the direction vector The line is given in symmetric form:

   $\frac{x+1}{2} = \frac{y-2}{-3} = \frac{z+2}{4} = t.$

   Setting $t=0$ gives a convenient point $A(-1, 2, -2)$.

   The direction vector is $\vec{d} = (2, -3, 4)$.

   2. Form the vector from $A$ to $P$ $P = (1, -1, 2)$, so

   $\overrightarrow{AP} = P - A = (1 - (-1),\; -1 - 2,\; 2 - (-2)) = (2, -3, 4).$

   3. Compute the length of $\overrightarrow{AP}$

   $|\overrightarrow{AP}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}.$

   4. Compute the projection of $\overrightarrow{AP}$ onto $\vec{d}$ The scalar projection (component along $\vec{d}$) is

   $\text{comp}_{\vec{d}} \overrightarrow{AP} = \frac{\overrightarrow{AP} \cdot \vec{d}}{|\vec{d}|}.$

   Dot product:

   $(2)(2) + (-3)(-3) + (4)(4) = 4 + 9 + 16 = 29.$

   Length of $\vec{d}$:

   $|\vec{d}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29}.$

   Hence

   $$$$

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2. COMEDK 2023Set 2023-E1 markMCQ
   Q.The distance of the point $(2,3,4)$ from the line $1-x=\frac{y}{2}=\frac{1}{3}(1+z)$ is (A) $\frac{2}{7} \sqrt{35}$ (B) $\frac{1}{7} \sqrt{35}$ (C) $\frac{4}{7} \sqrt{35}$ (D) $\frac{3}{7} \sqrt{35}$

   ›Reveal solutionSolution

   Perpendicular distance^2 = 35 - 200/7 = (245 - 200)/7 = 45/7. distance = sqrt(45/7) = 3 sqrt5 / sqrt7 = 3 sqrt(35) / 7.

   Concept: perpendicular distance from a point to a line in 3-D, d = |AP|^2 - (projection of AP on d-hat)^2, i.e. d = sqrt( |AP|^2 - (AP . d)^2/|d|^2 ).

   Put the line in symmetric form. 1 - x = y/2 = (1 + z)/3 becomes

   (x - 1)/(-1) = (y - 0)/2 = (z + 1)/3.

   So a point on the line is A = (1, 0, -1) and the direction is d = (-1, 2, 3), with |d|^2 = 1 + 4 + 9 = 14.

   Given point P = (2, 3, 4). Then

   AP = P - A = (1, 3, 5), |AP|^2 = 1 + 9 + 25 = 35.

   Projection of AP along d: …

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3. KCET 2019Set A-11 markMCQ
   Q.Foot of the perpendicular drawn from the point $(1, 3, 4)$ to the plane $2x - y + z + 3 = 0$ is (A) $(-1, 4, 3)$ (B) $(0, -4, -7)$ (C) $(1, 2, -3)$ (D) $(-3, 5, 2)$

   ›Reveal solutionSolution

   The foot of the perpendicular from a point to a plane is found by moving along the plane's normal vector from the point until the plane equation is satisfied. For point $(1,3,4)$ and plane $2x - y + z + 3 = 0$, the foot is $(-1,4,3)$, which is option (A).

   The key idea: the shortest line from a point to a plane is along the normal vector of the plane. The foot of the perpendicular is the point where this normal line through the given point meets the plane.

   The plane $2x - y + z + 3 = 0$ has normal vector $\vec{n} = (2, -1, 1)$. Any line through $(1,3,4)$ parallel to $\vec{n}$ has parametric form:

   $(x, y, z) = (1 + 2t, \; 3 - t, \; 4 + t)$

   where $t$ is a real parameter. The foot of the perpendicular is the particular point on this line that lies on the plane — we just need to find $t$.

   1. Substitute the parametric coordinates into the plane equation:

   $2(1 + 2t) - (3 - t) + (4 + t) + 3 = 0$

   2. Simplify step by step:

   $2 + 4t - 3 + t + 4 + t + 3 = 0$

   Combine constants: $2 - 3 + 4 + 3 = 6$

   Combine $t$ terms: $4t + t + t = 6t$

   So we get:

   $6t + 6 = 0$

   3. Solve for $t$:

   $6t = -6 \quad\Rightarrow\quad t = -1$

   4. Plug $t = -1$ back into the parametric line:

   $x = 1 + 2(-1) = -1$

   $y = 3 - (-1) = 4$

   $z = 4 + (-1) = 3$

   Thus the foot of the perpendicular is $(-1, 4, 3)$. …

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4. COMEDK 2024Set 2024-E1 markMCQ
   Q.The points on the $x$-axis whose perpendicular distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units are (A) $(8,0)$ and $(-2,0)$ (B) $(-8,0)$ and $(-2,0)$ (C) $(8,0)$ and $(2,0)$ (D) $(-8,0)$ and $(2,0)$

   ›Reveal solutionSolution

   Writing the line as $4x+3y-12=0$ and setting the distance of $(x,0)$ equal to $4$ gives $|4x-12|=20$, so $x=8$ or $x=-2$; the points are $(8,0)$ and $(-2,0)$ — option (A).

   Concept

   A point on the $x$-axis has the form $(x,0)$. The perpendicular distance from $(x_1,y_1)$ to the line $Ax+By+C=0$ is

   $d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$

   Setting $d=4$ produces an absolute-value equation with two solutions — one on each side of the line.

   Solution

   1. Standard form: multiply $\dfrac{x}{3}+\dfrac{y}{4}=1$ by $12$:

   $4x+3y=12\ \Rightarrow\ 4x+3y-12=0,$

   so $A=4,\ B=3,\ C=-12$.

   2. Distance of $(x,0)$:

   $d=\frac{|4x+3(0)-12|}{\sqrt{4^2+3^2}}=\frac{|4x-12|}{5}.$

   3. Set $d=4$:

   $\frac{|4x-12|}{5}=4\ \Rightarrow\ |4x-12|=20.$

   4. Solve both cases: …

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5. COMEDK 2025Set 2025-A1 markMCQ
   Q.The point on the line $x+y=4$ that lie at a unit distance from the line $4 x+3 y=10$ is (A) $(2,2)$ (B) $(3,-1)$ (C) $(5,-1)$ (D) $(-7,11)$

   ›Reveal solutionSolution

   [!TLDR]

   Parametrise the point on $x+y=4$, apply the point-to-line distance formula to $4x+3y=10$, set it to $1$, and the valid option is $(-7,11)$.

   Concept

   The perpendicular distance from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ — the standard CBSE Class-11 Straight Lines formula.

   Solution

   Let the required point lie on $x+y=4$, so it is $(x,\,4-x)$.

   Distance from the line $4x+3y-10=0$ (here $\sqrt{a^2+b^2}=\sqrt{16+9}=5$):

   $d=\frac{|4x+3(4-x)-10|}{5}=\frac{|4x+12-3x-10|}{5}=\frac{|x+2|}{5}.$

   Set $d=1$:

   $|x+2|=5 \;\Rightarrow\; x+2=\pm 5 \;\Rightarrow\; x=3 \text{ or } x=-7.$

   The corresponding points are $(3,\,1)$ and $(-7,\,11)$.

   Checking the options against both conditions (on the line AND unit distance):

   • (A) $(2,2)$: on line, but distance $=|2+2|/5=4/5\neq1$.
   • (B) $(3,-1)$: $3+(-1)=2\neq4$, not even on the line. …

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6. COMEDK 2022Set 20221 markMCQ
   Q.If 2 and 3 are intercepts of a line L = 0, then the distance of L $\equiv$ 0 from the origin is (A) ${5 \over {\sqrt {13} }}$ (B) ${{\sqrt {13} } \over 6}$ (C) ${6 \over {\sqrt {13} }}$ (D) 1

   ›Reveal solutionSolution

   Distance from origin = |3(0) + 2(0) − 6| / √(3² + 2²) = 6/√13.

   Concept: Intercept form + perpendicular distance from origin.

   Intercepts 2 and 3 → x/2 + y/3 = 1 → 3x + 2y − 6 = 0. …

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7. COMEDK 2025Set 2025-A1 markMCQ
   Q.If $Q(1,0,1)$ is the image of the point $P(a, b, c)$ in the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ then $a+b+c$ is equal to : (A) 4 (B) 2 (C) 0 (D) $-$2

   ›Reveal solutionSolution

   Reflecting $Q$ across the line gives $P=(1,2,-3)$, so $a+b+c=0$.

   Since $Q$ is the image of $P$ in the line, $P$ is likewise the reflection of $Q$ in the line. Take the line point $A=(-1,3,0)$ and direction $\vec d=(2,-2,-1)$, with $|\vec d|^2=4+4+1=9$.

   Foot of perpendicular $F$ from $Q(1,0,1)$: with $\vec{AQ}=(2,-3,1)$,

   $t=\frac{\vec{AQ}\cdot\vec d}{|\vec d|^2}=\frac{(2)(2)+(-3)(-2)+(1)(-1)}{9}=\frac{4+6-1}{9}=1.$ …

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8. COMEDK 2023Set 2023-M1 markMCQ
   Q.The distance of the point $(3,4)$ from the line $3 x+2 y+7=0$ measured along the line parallel to $y-2 x+7=0$ is equal to (A) $\frac{24 \sqrt{5}}{7}$ (B) $3 \sqrt{5}$ (C) $\frac{23 \sqrt{5}}{7}$ (D) $4 \sqrt{5}$

   ›Reveal solutionSolution

   Parametrise the point $(3,4)$ along the unit direction of the parallel line and find where it meets $3x+2y+7=0$; the parameter value equals the required distance, $\dfrac{24\sqrt5}{7}$.

   The line $y-2x+7=0$ has slope $2$, so a direction vector is $(1,2)$, of length $\sqrt5$. The unit direction is $\left(\tfrac1{\sqrt5},\tfrac2{\sqrt5}\right)$.

   A point at arc-length $t$ from $(3,4)$ along this direction is $\left(3+\tfrac{t}{\sqrt5},\,4+\tfrac{2t}{\sqrt5}\right)$. It lies on $3x+2y+7=0$ when:

   $3\left(3+\tfrac{t}{\sqrt5}\right)+2\left(4+\tfrac{2t}{\sqrt5}\right)+7=0$ …

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9. COMEDK 2022Set 20221 markMCQ
   Q.The distance of the point $(1,2)$ from the line $x+y+2=0$ measured along the line parallel to $2 x-y=5$ is equal to (A) $\frac{125}{3}$ (B) $\frac{125}{\sqrt{3}}$ (C) $\frac{5 \sqrt{5}}{3}$ (D) $\frac{5 \sqrt{3}}{3}$

   ›Reveal solutionSolution

   The required distance is |t| = 5√5/3.

   Concept: Distance measured along a given direction — parametrise from the point along that direction and find where it meets the line.

   Direction of 2x − y = 5 → slope 2 → unit vector (1, 2)/√5.

   Points on the line through (1, 2) in that direction: (1 + t/√5, 2 + 2t/√5).

   Substitute into x + y + 2 = 0: …

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10. COMEDK 2021Set 20211 markMCQ
    Q.The distance of the point (1, 2) from the line $x+y+5=0$ measured along the line parallel to $3x-y=7$ is equal to (A) $4\sqrt{10}$ (B) $40$ (C) $\sqrt{40}$ (D) $10\sqrt2$

    ›Reveal solutionSolution

    The required distance = |r| = 2 sqrt(10) = sqrt(40).

    Concept: distance measured along a given direction. Parametrise the line through (1, 2) with the direction of 3x - y = 7 (slope 3) and find the parameter where it meets x + y + 5 = 0.

    Slope 3 => direction unit vector = (1, 3)/sqrt(10).

    Parametric point: (1 + r/sqrt(10), 2 + 3r/sqrt(10)).

    Substitute into x + y + 5 = 0: …

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11. COMEDK 2026Set 2026-A1 markMCQ
    Q.A line L passes through the point of intersection of the lines $3 x+y-10=0$ and $x-y-2=0$. If the perpendicular distance of the line $L$ from the point $(5,1)$ is exactly $\frac{2}{\sqrt{5}}$ units, which of the following represents the correct equation for line L ? (A) $x+2 y-5=0$ (B) $2 x+y-7=0$ (C) $x-2 y+1=0$ (D) $2 x-y-5=0$

    ›Reveal solutionSolution

    The line L must pass through the intersection of the two given lines and be at a perpendicular distance of $2/\sqrt{5}$ from (5,1). Solving the family of lines through that intersection and applying the distance formula yields two possible lines; only one matches the given options, which is option (B).

    We start by finding the fixed point through which L must pass — the intersection of the two given lines. Then we write the general equation of any line through that point (using a parameter for slope). The condition on perpendicular distance from (5,1) gives an equation in the slope; solving it yields two slopes, hence two possible lines. Finally we check which of the given options matches one of these.

    1. Find the intersection point of the two given lines. Solve

    $3x + y - 10 = 0 \quad\text{and}\quad x - y - 2 = 0.$

    Adding the equations eliminates $y$:

    $(3x + y - 10) + (x - y - 2) = 0 \;\Rightarrow\; 4x - 12 = 0 \;\Rightarrow\; x = 3.$

    Substitute $x = 3$ into $x - y - 2 = 0$:

    $3 - y - 2 = 0 \;\Rightarrow\; y = 1.$

    So the intersection point is $P(3,1)$.

    2. Write the family of lines through $P(3,1)$. Any non-vertical line through $P$ can be written as

    $y - 1 = m(x - 3),$

    where $m$ is the slope. Rearranging into standard form:

    $mx - y + (1 - 3m) = 0.$

    (We will also consider the vertical line $x = 3$ separately later.)

    3. Apply the perpendicular distance condition. The distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is

    $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.$

    For our line $mx - y + (1 - 3m) = 0$ and point $(5,1)$:

    $A = m,\; B = -1,\; C = 1 - 3m.$

    The distance is

    $\frac{|m\cdot 5 + (-1)\cdot 1 + (1 - 3m)|}{\sqrt{m^2 + 1}} = \frac{|5m - 1 + 1 - 3m|}{\sqrt{m^2 + 1}} = \frac{|2m|}{\sqrt{m^2 + 1}}.$

    We are told this equals $\frac{2}{\sqrt{5}}$. So

    $\frac{|2m|}{\sqrt{m^2 + 1}} = \frac{2}{\sqrt{5}}.$

    4. Solve for $m$. Cancel the factor 2 (valid since both sides are positive):

    $\frac{|m|}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{5}}.$

    Square both sides:

    $\frac{m^2}{m^2 + 1} = \frac{1}{5}.$

    Cross-multiply:

    $5m^2 = m^2 + 1 \;\Rightarrow\; 4m^2 = 1 \;\Rightarrow\; m^2 = \frac{1}{4} \;\Rightarrow\; m = \pm \frac{1}{2}.$

    5. Write the two possible lines.
       • For $m = \frac{1}{2}$:

    $y - 1 = \frac{1}{2}(x - 3) \;\Rightarrow\; 2y - 2 = x - 3 \;\Rightarrow\; x - 2y - 1 = 0.$

    • For $m = -\frac{1}{2}$:

    $y - 1 = -\frac{1}{2}(x - 3) \;\Rightarrow\; 2y - 2 = -x + 3 \;\Rightarrow\; x + 2y - 5 = 0.$

    Also check the vertical line $x = 3$: its distance from $(5,1)$ is $|5-3| = 2$, not $2/\sqrt{5}$, so it is not a solution.

    6. Match with the given options. The options are: …

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12. COMEDK 2024Set 2024-M1 markMCQ
    Q.The perpendicular distance of a line from the origin is 5 units and its slope is $-1$. The equation of the line is (A) $x+y \pm 5 \sqrt{2}=0$ (B) $x-y \pm 2 \sqrt{5}=0$ (C) $x+y \pm 2 \sqrt{5}=0$ (D) $x-y \pm 5 \sqrt{2}=0$

    ›Reveal solutionSolution

    A slope of $-1$ gives lines $x + y = c$; requiring the distance $\dfrac{|c|}{\sqrt{2}} = 5$ gives $|c| = 5\sqrt{2}$, so the line is $x + y \pm 5\sqrt{2} = 0$ — option (A).

    Concept

    The perpendicular distance from the origin to $Ax + By + C = 0$ is $\dfrac{|C|}{\sqrt{A^2 + B^2}}$. A line of slope $-1$ has the form $x + y = c$ (since $-A/B = -1$ with $A = B = 1$), so we only need to fix the constant $c$ from the distance condition.

    Solution

    1. Standard form: $x + y - c = 0$, so $A = 1,\, B = 1,\, C = -c$.
    2. Distance from origin:

    $\frac{|1\cdot 0 + 1\cdot 0 - c|}{\sqrt{1^2 + 1^2}} = \frac{|c|}{\sqrt{2}}.$

    3. Apply the condition: $\frac{|c|}{\sqrt{2}} = 5 \Rightarrow |c| = 5\sqrt{2}.$ …

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