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Question
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 6y – 3z = 63.
Solution
Given equation of plane is 2x + 6y – 3z = 63
∴ The direction ratios of the normal to the plane 2x + 6y – 3z = 63 are 2, 6, – 3
∴ Direction cosines are,
l = `2/sqrt(2^2 + 6^2 + (-3)^2`
m = `6/sqrt(2^2 + 6^2 + (-3)^2`
n = `(-3)/sqrt(2^2 + 6^2 + (-3)^2`
∴ l = `2/7, "m" = 6/7, "n" = (-3)/7`
The normal form of the plane is
`2/7x + 6/7y - 3/7z = 63/7`
∴ `2/7x + 6/7y - 3/7z` = 9
The co-ordinates of the foot of the perpendicular are
(lp, mp, np) = `[(2/7)9, (6/7)9, ((-3)/7)9]`
= `(18/7, 54/7, (-27)/7)`
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