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CBSE Sample Paper 2025-26 · Section A · 1 mark

Q14: Distance of 3i^+4j^+5k^3\hat i + 4\hat j + 5\hat k from the y-axis

Sample 2025-26Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

The distance of the point with position vector 3i^+4j^+5k^3\hat i + 4\hat j + 5\hat k from the y-axis is

(A) 44 units

(B) 34\sqrt{34} units

(C) 55 units

(D) 525\sqrt{2} units

Exam answer

The foot of the perpendicular from P(3,4,5)P(3, 4, 5) to the y-axis is M(0,4,0)M(0, 4, 0)⁠.

Why?

Every point of the y-axis has x=0x = 0 and z=0z = 0⁠. The nearest one to PP keeps the y-coordinate of PP⁠, which is 4.

Check: MP→=3i^+5k^\overrightarrow{MP} = 3\hat i + 5\hat k⁠, and (3i^+5k^)⋅j^=0(3\hat i + 5\hat k) \cdot \hat j = 0⁠, so MPMP is perpendicular to the y-axis.

PM=32+02+52=34 units\begin{aligned} PM &= \sqrt{3^2 + 0^2 + 5^2} \\ &= \sqrt{34} \text{ units} \end{aligned}
Why?

The distance formula between P(3,4,5)P(3, 4, 5) and M(0,4,0)M(0, 4, 0)⁠:

PM2=(3−0)2+(4−4)2+(5−0)2=32+02+52=9+0+25=34PM=34\begin{aligned} PM^2 &= (3 - 0)^2 + (4 - 4)^2 \\ &\quad + (5 - 0)^2 \\ &= 3^2 + 0^2 + 5^2 \\ &= 9 + 0 + 25 = 34 \\ PM &= \sqrt{34} \end{aligned}

In general, the distance of (a,b,c)(a, b, c) from the y-axis is a2+c2\sqrt{a^2 + c^2}⁠: leave out the y-coordinate.

Watch out: 32+42=5\sqrt{3^2 + 4^2} = 5 is the distance from the z-axis, and 9+16+25=52\sqrt{9 + 16 + 25} = 5\sqrt{2} is the distance from the origin.

9+16+25=50=25×2=52\begin{aligned} \sqrt{9 + 16 + 25} &= \sqrt{50} \\ &= \sqrt{25 \times 2} = 5\sqrt{2} \end{aligned}

Answer: (B) 34\sqrt{34} units

Where this comes from

Chapter 11, Three Dimensional Geometry: Distance of a point from a coordinate axis.

Question type: the foot of a perpendicular

How to spot it: click to see

Type 10. Beyond the book, but in CBSE papers: the foot of a perpendicular

Spot it: "Find the foot of the perpendicular from a point to a line", "the image of a point in a line", or "do the lines meet?". This book has none of these, but recent CBSE sample papers ask all three.

Here: The line is the y-axis, so the foot from (3,4,5)(3, 4, 5) keeps only the y-coordinate: M(0,4,0)M(0, 4, 0)⁠. Then PM=32+52=34PM = \sqrt{3^2 + 5^2} = \sqrt{34}⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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