Question

Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati + hatj + hatk) = 1` and `vecr.(2hati + 3hatj -hatk) + 4 = 0` and parallel to x-axis.

Answer 1

The given planes are

The equation of any plane passing through the line of intersection of these planes is
r→.i⏜+j⏜+k⏜-1+λr→.2i⏜+3j⏜-k⏜+4=0

r→.2λ+1i⏜+3λ+1j⏜+1-λk⏜+4λ-1=0     ...(1)

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.