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CBSE Board Examination 2024 · Section A · 1 mark

Q16: Distance of P(a,b,c)P(a, b, c) from the y-axis

Board 2024Ch 11 · Three Dimensional Geometry
The question

The distance of point P(a,b,c)P(a, b, c) from y-axis is :

(A) bb

(B) b2b^2

(C) a2+c2\sqrt{a^2 + c^2}

(D) a2+c2a^2 + c^2

Exam answer

The foot of the perpendicular from P on the y-axis is (0,b,0)(0, b, 0)⁠.

Why?

Every point of the y-axis has the form M(0,t,0)M(0, t, 0)⁠. The foot is the point where MP→\overrightarrow{MP} is perpendicular to the axis direction j^\hat j⁠:

MP→=ai^+(b−t)j^+ck^MP→⋅j^=b−t=0\begin{aligned} \overrightarrow{MP} &= a\hat i + (b - t)\hat j + c\hat k \\ \overrightarrow{MP} \cdot \hat j &= b - t = 0 \end{aligned}

So t=bt = b⁠, and the foot is (0,b,0)(0, b, 0)⁠.

The distance of P from the y-axis is the distance from P(a,b,c)P(a, b, c) to (0,b,0)(0, b, 0)⁠:

d2=(a−0)2+(b−b)2+(c−0)2d=a2+02+c2=a2+c2\begin{aligned} d^2 &= (a - 0)^2 + (b - b)^2 \\ &\quad + (c - 0)^2 \\ d &= \sqrt{a^2 + 0^2 + c^2} \\ &= \sqrt{a^2 + c^2} \end{aligned}
Why?

This is the distance formula between two points (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2)⁠:

d2=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} d^2 &= (x_2 - x_1)^2 \\ &\quad + (y_2 - y_1)^2 \\ &\quad + (z_2 - z_1)^2 \end{aligned}

Distance is never negative, so we take the positive square root.

Watch out: (D) is d2d^2⁠, not dd⁠. And ∣b∣|b| is the distance of P from the xz-plane, not from the y-axis.

Answer: (C) a2+c2\sqrt{a^2 + c^2}

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 P(a,b,c)P(a, b, c) keeps only the y-coordinate: it is (0,b,0)(0, b, 0)⁠. The distance from P to this foot is a2+c2\sqrt{a^2 + c^2}⁠.

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