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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.
Solution
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.
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