Q.Find the equation of the plane which passes through the point î+2ĵ−k̂ and is perpendicular to the line of intersection of the planes r̄·(3î−ĵ+k̂)=1 and r̄·(î+4ĵ−2k̂)=2. OR Prove that the lines r̄ = î+ĵ−k̂+λ(3î−ĵ) and r̄ = 4î−k̂+μ(2î+3k̂) intersect. Also find the point of intersection.

A plane 'perpendicular to a line' has that line's direction vector as its normal. The line of intersection's direction is the cross product of the two planes' normals.

Given planes: $\bar r\cdot(3\hat i-\hat j+\hat k)=1$ (normal $\vec n_1=(3,-1,1)$) and $\bar r\cdot(\hat i+4\hat j-2\hat k)=2$ (normal $\vec n_2=(1,4,-2)$).

Direction of their line of intersection:

$\vec d = \vec n_1\times\vec n_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\3&-1&1\\1&4&-2\end{vmatrix}$

$\hat i: (-1)(-2)-(1)(4) = 2-4=-2$

$\hat j: -[(3)(-2)-(1)(1)] = -[-6-1] = 7$

$\hat k: (3)(4)-(-1)(1) = 12+1=13$

$\vec d = (-2,7,13)$

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