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Question
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
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Solution
Given plane is 2x – 2y + 4z + 5 = 0 and given point is `(1, 3/2, 2)`
D’ratios of the normal to the plane are 2, – 2, 4
So, the equation of the line passing through `(1, 3/2, 2)` and whose d’ratios are equal to the d’ratios of the normal to the plane i.e., 2, – 2, 4 is
`(x - 1)/2 = (y - 3/2)/(-2) = (z - 2)/4 = lambda`
∴ Any point in the plane is `2lambda + 1, -2lambda + 3/2, 4lambda + 2`
Since, the point lies in the plane, then
`2(2lambda + 1) - 2(-2lambda + 3/2) + 4(4lambda + 2) + 5` = 0
⇒ 4λ + 2 + 4λ – 3 + 16λ + 8 + 5 = 0
⇒ 24λ + 12 = 0
∴ `lambda = - 1/2`
So, the coordinates of the point in the plane are `2(-1/2) + 1`
`-2(-1/2) + 3/2`
`4(- 1/2) + 2`
i.e., 0, `5/2`, 0
Hence, the foot of the perpendicular is `(0, 5/2, 0)` and the required length
= `sqrt((1 - 0)^2 + (3/2 - 5/2)^2 + (2 - 0)^2)`
= `sqrt(1 + 1 + 4)`
= `sqrt(6)` units
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