From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y - 2z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.
Given :
Equation of plane : 2x + y – 2z + 3 = 0
P = (1, 2, 4)
To Find :
i) Equation of perpendicular
ii) Length of perpendicular = d
iii) coordinates of the foot of the perpendicular
Formulae :
1) Unit Vector :
Let be any vector
Then unit vector of is
Where,
2) Length of perpendicular :
The length of the perpendicular from point A with position vector to the plane is given by,
Note :
If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then
Given equation of the plane is
2x + y – 2z + 3 = 0
⇒2x + y – 2z = -3 ………..eq(1)
From eq(1) direction ratios of normal vector of the plane are
(2, 1, -2)
Therefore, equation of normal vector is
= 3
From eq(1), p = -3
Given point P = (1, 2, 4)
Position vector of A is
Here,
Now,
= (1×2) + (2×1) + (4×(-2))
= 2 + 2 - 8
= -4
Length of the perpendicular from point A to the plane is
Let Q be the foot of perpendicular drawn from point P to the given plane,
Let Q = (x, y, z)
As normal vector and are parallel, we can write,
This is the equation of perpendicular.
⇒x = 2k+1, y = k+2, z = -2k+4
As point Q lies on the plane, we can write
2(2k+1) + (k+2) - 2(-2k+4) = -3
⇒ 4k + 2 + k + 2 + 4k - 8 = -3
⇒ 9k = 1
,
Therefore co-ordinates of the foot of perpendicular are
Q(x, y, z) =
Q ≡