Q16

1 markMCQSection A

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

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

**Correct option: (C) $\sqrt{74}$** The line $\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$ is the x-axis, direction $(1,0,0)$, through the origin. The foot of perpendicular from $(2,5,7)$ onto this line is $(2,0,0)$. The distance is $\sqrt{(2-2)^2+(5-0)^2+(7-0)^2} = \sqrt{25+49}$ = $\sqrt{74}$.

Three Dimensional Geometry

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

Options

(A)2

(B)5

(C)

\sqrt{74}

(D)

\sqrt{78}

Official Answer

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

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

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

Marking Scheme

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

Hint

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

Quick Oral Answer

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

Analysis & Explanation

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

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

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

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.

Common Mistakes

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

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.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

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., $(2,0,0)$ , avoiding the general projection formula.