Question

The ratio of the distances from the points (1,–1,3) and (3,3,3) to plane \(5x+2y-7z+9=0\) is

• A. 1:3
• B. 1:3
• C. 1:1
• D. 3:2

Hint:

Same magnitude in numerator → equal distances.

Answer 1

The Correct Option is C

Step 1: Use point–plane distance: \[ D=\frac{|5x+2y-7z+9|}{\sqrt{25+4+49}}=\frac{|5x+2y-7z+9|}{\sqrt{78}}. \]
Step 2: For first point: \[ D_1=\frac{|5(1)+2(-1)-7(3)+9|}{\sqrt{78}} =\frac{|5-2-21+9|}{\sqrt{78}} =\frac{| -9|}{\sqrt{78}}. \] For second: \[ D_2=\frac{|15+6-21+9|}{\sqrt{78}} =\frac{|9|}{\sqrt{78}}. \]
Step 3: \[ D_1=D_2 \Rightarrow \text{ratio }1:1. \] Hence → (C).