. Find the length and the foot of the perpendicular from the point (1, 3/2, 2) to the plane 2x - 2y + 4z + 5 = 0.
Given, point P (1, 3/2, 2)
The plane is 2x - 2y + 4z + 5 = 0
We must find the foot of the perpendicular from the point P to the equation of the given plane.
Also, we must find the distance from the point P to the plane.
Let us consider the foot of the perpendicular from point P to be Q.
Let Q be Q (x1 , y1 , z1)
So, the direction ratio of PQ is given by
(x1 - 1, y1 - 3/2, z1 - 2)
Now, let us consider the normal to the plane 2x - 2y + 4z + 5 = 0:
It is obviously parallel to PQ, since a normal is a line or vector that is perpendicular to a given object. The direction ratio simply states the number of units to move along each axis.
For any plane, ax + by + cz = d, where, a, b, and c are normal vectors to the plane.
Hence, the direction ratios are (a, b, c).
Therefore, the direction ratio of the normal = (2, -2, 4) for plane 2x - 2y + 4z + 5 = 0.
The Cartesian equation of the line PQ, where P(1, 3/2, 2) and Q (x1 , y1 , z1) is:
To find any point on this line,
Any point on the line is
This point is Q.
And, it was assumed that is lies on the given plane. Substituting x1, y1, and z1 in the plane equation, we get:
2x1 - 2y1 + 4z1 + 5 = 0
Simplifying to find the value of
Since Q is the foot of the perpendicular from the point P,
We substitute the value of in equation (i) to get:
Then, to find
Where, P = (1, 3/2, 2) and Q = (0, 5/2, 0)
Thus, the foot of the perpendicular from the given point to the plane is (0, 5/2, 0) and the distance is units.