​ Find the equation of the plane through the line of intersection of the planes r.(2i - 3j + 4k) = 1 and r.(i - j) + 4 = 0 ​

Question

Find the equation of the plane through the line of intersection of the planes \(\bar{r}\).(2\(\hat{i}\) - 3\(\hat{j}\) + 4\(\hat{k}\)) = 1 and \(\bar{r}\)(\(\hat{i}\) - \(\hat{j}\)) + 4 = 0 and perpendicular to the plane \(\bar{r}\).(2\(\hat{i}\) - \(\hat{j}\) + \(\hat{k}\)) + 8 = 0

Answer 1

Equation of plane through the line of intersection of two planes in vector form is

Given a plane perpendicular to this plane, So if n₁ and n₂ are normal 

vectors of planes

2.(2 + λ) + (-1).(-3-λ) + 1.4 = 0 

4 + 2λ + 3 + λ + 4 = 0 

11 + 3λ = 0

λ = -11/3

Putting the value of λ in equation (2)