Find the length and the foot of perpendicular drawn from the point (2, 3, 7) to the plane 3x – y – z = 7.
Given :
Equation of plane : 3x – y – z = 7
A = (2, 3, 7)
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
3x – y – z = 7 ………..eq(1)
Therefore direction ratios of normal vector of the plane are
(3, -1, -1)
Therefore normal vector of the plane is
From eq(1), p = 7
Given point A = (2, 3, 7)
Position vector of A is
Now,
= (2×3) + (3×(-1)) + (7×(-1))
= 6 – 3 – 7
= -4
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 = 3k+2, y = - k+3, z = -k+7
As point P lies on the plane, we can write
3(3k+2) - (- k+3) - (-k+7) = 7
⇒ 9k + 6 + k – 3 + k – 7 = 7
⇒ 11k = 11
,
Therefore co-ordinates of the foot of perpendicular are
P(x, y, z) =
P = (5, 2, 6)