Find the distance of the point (2i - j - 4k) from the plane r.(3i - 4j + 12k) = 9 ​

Question

Find the distance of the point (2\(\hat{i}\) - \(\hat{j}\) - 4\(\hat{k}\)) from the plane \(\bar{r}\).(3\(\hat{i}\) - 4\(\hat{j}\) + 12\(\hat{k}\)) = 9

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.(3i - 4j + 12k) = 9 can be written in cartesian form as

3x - 4y + 12z = 9 

3x - 4y + 12z - 9 = 0 

Point = (2i - j - 4k) 

Which can be also written as 

Point = ( 2 , - 1 , - 4 )