Question 161 markMCQSection A

The length of the perpendicular drawn from the point (2,5,7)(2, 5, 7) on the line x1=y0=z0\frac{x}{1} = \frac{y}{0} = \frac{z}{0} is

Three Dimensional Geometry
Distance of a Point from a Line in 3D (Foot of Perpendicular)

Options

(A)2
(B)5
(C)74\sqrt{74}
(D)78\sqrt{78}

Official answer

Correct option: (C) 74\sqrt{74}


The line x1=y0=z0\frac{x}{1}=\frac{y}{0}=\frac{z}{0} is the x-axis, direction (1,0,0)(1,0,0), through the origin. The foot of perpendicular from (2,5,7)(2,5,7) onto this line is (2,0,0)(2,0,0). The distance is (2−2)2+(5−0)2+(7−0)2=25+49\sqrt{(2-2)^2+(5-0)^2+(7-0)^2} = \sqrt{25+49} = 74\sqrt{74}.

Answer keywords:foot of perpendiculardistance from point to line3D geometryx-axisdirection ratiosperpendicular distance formula

Marking scheme

  1. 1.1 mark: awarded only for selecting option (C) 74\sqrt{74}; no partial credit.

Hint

Recognize x1=y0=z0\frac{x}{1}=\frac{y}{0}=\frac{z}{0} as the x-axis; the foot of perpendicular from (2,5,7)(2,5,7) is simply (2,0,0)(2,0,0), then apply the distance formula.

Quick oral answer

The line is the x-axis, so the foot of perpendicular from (2,5,7)(2,5,7) is (2,0,0)(2,0,0); the distance is 02+52+72=74\sqrt{0^2+5^2+7^2}=\sqrt{74}.

Common mistakes

  1. 1.Forgetting to include the z-coordinate difference and only computing a partial distance.
  2. 2.Wrongly assuming the foot of perpendicular is the origin instead of correctly projecting onto the axis to get (2,0,0).
  3. 3.Arithmetic slips while adding squares under the root, leading to close but wrong values like 78\sqrt{78}.

Analysis and explanation

Concept: Foot of perpendicular and distance formula in 3D coordinate geometry.


The given line has direction ratios (1,0,0)(1,0,0) and passes through the origin — this is simply the x-axis. Any general point on this line is (t,0,0)(t,0,0). The foot of the perpendicular from P(2,5,7)P(2,5,7) is found by projecting OP onto the direction vector: t=2×1+5×0+7×012+02+02=2t = \frac{2\times1+5\times0+7\times0}{1^2+0^2+0^2} = 2. So the foot is F(2,0,0)F(2,0,0).


The perpendicular distance PF=(2−2)2+(5−0)2+(7−0)2=0+25+49=74PF = \sqrt{(2-2)^2 + (5-0)^2 + (7-0)^2} = \sqrt{0+25+49} = \sqrt{74}.


Why the distractors fail:

  • (A) 2 is simply the x-coordinate of the point/foot, not the actual 3D distance — a common error of stopping at the projection value instead of computing the full distance.
  • (B) 5 picks out just the y-coordinate, ignoring the z-coordinate entirely.
  • (D) sqrt(78) comes from an arithmetic slip, e.g., mistakenly including an extra unit (25+49+4=78) as if the foot were at the origin instead of (2,0,0).

Exam tip: For a coordinate axis line like the x-axis, the foot of perpendicular from any point is just that point's coordinates with the other two coordinates set to zero — a quick shortcut instead of full vector projection.

Interesting facts

When a line coincides with a coordinate axis, finding the foot of perpendicular from any point becomes trivial — it is simply that point's coordinate along the axis, with the other two coordinates zeroed out.

This perpendicular-distance concept generalizes directly to skew lines in 3D, forming the basis of the shortest-distance formula tested in higher-mark Class 12 problems like Q35 in this very paper.

Question details

Q16 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Three Dimensional Geometry
Topic
Distance of a Point from a Line in 3D (Foot of Perpendicular)
Bloom's level
apply
Syllabus topic
Unit 4: Three Dimensional Geometry — Perpendicular Distance of a Point from a Line
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do you recognize that x/1=y/0=z/0 represents the x-axis?

The line passes through the origin (since numerators are x, y, z with no constant offset) with direction ratios (1,0,0), meaning it moves only along the x-direction — this is exactly the definition of the x-axis.

Is there a faster way than full vector projection here?

Yes — since the line is a coordinate axis, the foot of perpendicular from any point is just that point's own x-coordinate (here, 2) with y and z set to 0, i.e., (2,0,0)(2,0,0), avoiding the general projection formula.