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.
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.
Solution and Explanation
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.
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