Find the length and the foot of the perpendicular drawn from the point (1, 1, 2) to the plane
Given :
Equation of plane :
A = (1, 1, 2)
To Find :
i) Length of perpendicular = d
ii) 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
………..eq(1)
As
Therefore equation of plane is
2x – 2y + 4z = -5 ……… eq(2)
From eq(1) normal vector of the plane is
From eq(1), p = -5
Given point A = (1, 1, 2)
Position vector of A is
Now,
= (1×2) + (1×(-2)) + (2×4)
= 2 – 2 + 8
= 8
Length of the perpendicular from point A to the plane is
Let P be the foot of perpendicular drawn from point A to the given plane,
Let P = (x, y, z)
As normal vector and are parallel
⇒x = 2k+1, y = -2k+1, z = 4k+2
As point P lies on the plane, we can write
2(2k+1) – 2(-2k+1) + 4(4k+2) = -5
⇒ 4k + 2 + 4k – 2 + 16k + 8 = -5
⇒ 24k = -13
,
Therefore co-ordinates of the foot of perpendicular are
P(x, y, z) =
P ≡