Question

the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x +Y- 2Z = 18 is

Answer 1

Given Question :-

• The coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + y - 2z = 18 is _____

ANSWER

GIVEN

• The equation of plane is 2x + y - 2z = 18.

• The point (0, 0, 0)

TO FIND :-

• Co-ordinates of foot of perpendicular drawn from (0, 0, 0) to the plane 2x + y - 2z = 18.

CALCULATION :-

Since,

• the equation of plane is 2x + y - 2z = 18.

So,

• the normal vector or direction ratios of the plane is (2, 1, -2).

• Let the coordinates (0, 0, 0) be represented by O.

• Let OP be the perpendicular drawn to the plane from (0, 0, 0) intersecting plane at P.

So,

• Equation of line OP passes through (0, 0, 0) having direction ratios (2, 1, -2) is given by

$\sf \: \dfrac{x - 0}{2} = \dfrac{y - 0}{1} = \dfrac{z - 0}{ - 2} = k$

$\rm :\implies\:\: \dfrac{x }{2} = \dfrac{y }{1} = \dfrac{z }{ - 2} = k$

So,

Co-ordinates of any point on the line, say P is given by

$\longmapsto \boxed{ \green{ \bf \:P(2k, \: k, \: - 2k) }}$

Now,

• P lies on the plane 2x + y - 2z = 18

Therefore,

$\rm :\implies\:2(2k) + k - 2( - 2k) = 18$

$\rm :\implies\:4k + k + 4k = 18$

$\rm :\implies\:9k = 18$

$\longmapsto \boxed{ \green{ \bf \: k \: = \: 2}}$

Hence,

Coordinates of P becomes (4, 2, - 4)

$\boxed{ \green{ \bf \: coordinates \: of \: foot \: of \: perpendicular \: is \: P(4,2, - 4)}}$