Find the coordinates of the foot of the perpendicular drawn from the origin to the plane
(i) 2x + 3y + 4z -12 = 0
(ii) 5y + 8 = 0
(i) 2x + 3y + 4z - 12 = 0
Given :
Equation of plane : 2x + 3y + 4z + 12 = 0
To Find :
coordinates of the foot of the perpendicular
Note :
If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then
From the given equation of the plane
2x + 3y + 4z – 12 = 0
⇒ 2x + 3y + 4z = 12
Direction ratios of the vector normal to the plane are (2, 3, 4)
Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.
Therefore perpendicular drawn is
Let direction ratios of are (x, y, z)
As normal vector and are parallel
⇒x = 2k, y = 3k, z = 4k
As point P lies on the plane, we can write
2(2k) + 3(3k) + 4(4k) = 12
⇒ 4k + 9k + 16k = 12
⇒ 29k = 12
,
Therefore co-ordinates of the foot of perpendicular are
P(x, y, z) =
P =
(ii) Given :
Equation of plane : 5y + 8 = 0
To Find :
coordinates of the foot of the perpendicular
Note :
If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then
From the given equation of the plane
5y + 8 = 0
⇒ 5y = - 8
Direction ratios of the vector normal to the plane are (0, 5, 0)
Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.
Therefore perpendicular drawn is
Let direction ratios of are (x, y, z)
As normal vector and are parallel
⇒x = 0, y = 5k, z = 0
As point P lies on the plane, we can write
5(5k) = -8
⇒ 25k = -8
,
Therefore co-ordinates of the foot of perpendicular are
P(x, y, z) =
P =