In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

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#### Solution

Let the coordinates of the foot of perpendicular P from the origin to the plane be (*x*_{1}, *y*_{1}, *z*_{1}).

2*x* + 3*y* + 4*z* − 12 = 0

⇒ 2*x* + 3*y* + 4*z* = 12 … (1)

The direction ratios of normal are 2, 3, and 4

`:. sqrt((2)^2 + (3)^2 + (4)^2) = sqrt29`

Dividing both sides of equation (1) by sqrt29, we obtain

This equation is of the form *lx* + *my* + *nz* = *d*, where *l*, *m*, *n* are the direction cosines of normal to the plane and *d* is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(*ld*, *md*, *nd*).

Therefore, the coordinates of the foot of the perpendicular are

Concept: Vector and Cartesian Equation of a Plane

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