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Question
If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane
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Solution
Since, the line drawn from the point B(–2, –1, –3) meets a plane a plane at right angle at the point 4(1, –3, 3).
So, the plane passes through the point 4(1, –3, 3)
Also normal to plane is `(vecr - veca) * vecn` = 0
Where `veca = hati - 3hatj + 3hatk`
⇒ `[(xhati + yhatj + zhatk) - (hati - 3hatj + 3hatk)] * (-3hati + 2hatj - 6hatk)` = 0
⇒ `[(x - 1)hati + (y + 3)hatj + (z - 3)hatk] * (-3hati + 2hatj - 6hatk)` = 0
⇒ `-3x + 3 + 2y + 6 - 6z + 18` = 0
∴ 3x – 2y + 6z – 27 = 0
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