Question

Find the equation of the plane passing through the line of intersection of the planes $\vec r.(\hat i+\hat j+\hat k)=1$ and $\vec r.(2\hat i+3\hat j-\hat k)+4=0$ and parallel to x-axis.

Answer 1

The given planes are

$\vec r.(\hat i+\hat j+\hat k)=1$

⇒$\vec r.(\hat i+\hat j+\hat k)-1=0$

$\vec r.(2\hat i+3\hat j-\hat k)+4=0$

The equation of any plane passing through the intersection of these planes is

$[\vec r.(\hat i+\hat j+\hat k)-1]$ + λ$$[\vec r.(2\hat i+3\hat j-\hat k)+4] $= 0$

$\vec r.[(2λ+1)\hat i+(3λ+1)\hat j+(1-λ)\hat k]+(4λ+1)=0$ ...(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.

$∴1.(2+λ+1)+0(3λ+1)+0(1-λ)=0$

⇒ $2λ+1=0 ⇒ λ=-\frac 12$

Substituting, $λ=-\frac 12$ in equation (1), we obtain

⇒$\vec r.[-\frac 12\hat j+\frac 32\hat k]+(-3)=0$

⇒$\vec r.(\hat j-3\hat k)+6=0$

Therefore, its cartesian equation is $y-3z+6=0$

This is the equation of the required plane.