The distance of the point (i + 2j + 5k) from the plane r.(i + j + k) + 17 = 0. is ​

Question

The distance of the point (\(\hat{i}\) + 2\(\hat{j}\) + 5\(\hat{k}\)) from the plane \(\bar{r}\).(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) + 17 = 0. is

A. \(\frac{25}{\sqrt2}\)units 

B. \(\frac{25}{\sqrt3}\)units 

C. 25√2 units 

D. 25√3 units

Answer 1

Given: Point is at (\(\hat{i}\) + 2\(\hat{j}\) + 5\(\hat{k}\)) and equation of plane is \(\bar{r}\).(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) + 17 = 0

To find: distance of point from plane 

Formula Used: Perpendicular distance from (x₁, y₁, z₁) to the plane ax + by + cz + d = 0 is

Explanation: 

The point is at (1, 2, 5) and equation of plane is x + y + z + 17 = 0

= \(\frac{25}{\sqrt3}\)

Therefore, distance = \(\frac{25}{\sqrt3}\) units