If the line drawn from the point (-2,-1,-3) meets a plane at right angle at the point (1,-3,3), then find the equation of the plane.

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Solution

Since, the line drawn from the point (-2,-1,-3) meets a plane at right angle at the point (1,-3,3). So, the plane passes through the point (1,-3,3) and normal to plane is $(- 3 ˆ i + 2 ˆ j - 6 ˆ k)$ .
$\Rightarrow \to a = ˆ i - 3 ˆ j + 3 ˆ k$
and $\to N = - 3 ˆ i + 2 ˆ j - 6 ˆ k$
So, the equation of required plane is $(\to r - \to a) . \to N = 0 \Rightarrow [(x ˆ i + y ˆ j + z ˆ k) - (ˆ i - 3 ˆ j + 3 ˆ k)] . (- 3 ˆ i + 2 ˆ j - 6 ˆ k) = 0 \Rightarrow [(x - 1) ˆ i + (y + 3) ˆ j + (z - 3) ˆ k)] . (- 3 ˆ i + 2 ˆ j - 6 ˆ k) = 0 \Rightarrow - 3 x + 3 + 2 y + 6 - 6 z + 18 = 0 \Rightarrow - 3 x + 2 y - 6 z = - 27 ∴ 3 x - 2 y + 6 z - 27 = 0$

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