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

Question

Find 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

Answer 1

Formula : Distance = \(\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}\)

where (x₁,y₁,z₁) is point from which distance is to be calculated 

Therefore , 

Plane r.(i + j + k) + 17 = 0 can be written in cartesian form as 

x + y + z + 17 = 0 

Point = (i + 2j + 5k) 

Which can be also written as 

Point = (1 , 2 , 5)