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.