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Question
Solve the following :
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.
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Solution
The equation of the plane is 2x + 3y + 6z = 49.
Dividing each term by
`sqrt(2^2 + 3^2 + (-6)^2)`
= `sqrt(49)`
= 7,
we get
`(2)/(7)x + (3)/(7)y - (6)/(7)z = (49)/(7)` = 7
This is the normal form of the equation of plane.
∴ the direction cosines of the perpendicular drawn from the origin to the plane are
l = `(2)/(7), m = (3)/(7), n = (6)/(7)`
and length of perpendicular from origin to the plane is p = 7.
∴ the coordinates of the foot of the perpendicular from the origin to the plane are
`(lp, mp, np) "i.e." (2, 3, 6)`
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