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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.
Solution
Given plane is 2x – 2y + 4z + 5 = 0 and point `(1, 3/2, 2)`
The direction ratios of the normal to the plane are 2, –2, 4
So, the equation of the line passing through `(1, 3/2, 2)` and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, –2, 4 is
`(x - 1)/2 = (y - 3/2)/(-2) = (z - 2)/4 = lambda`
Now, any point in the plane is 2λ + 1, –2λ + `3/2`, 4λ + 2
Since, the point lies in the plane, then
2(2λ + 1) – 2(–2λ + `3/2`) + 4(4λ + 2) + 5 = 0
4λ + 2 + 4λ – 3 + 16λ + 8 + 5 = 0
24λ + 12 = 0λ = `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 = 0, 5/2, 0`
Thus, 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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