Find equation of the plane parallel to x + 3y – 2z + 7 = 0 and passing through the origin.
The equation of any plane through (a, b, c) is
A (x – a) + B (y – b) + C (z – c) = 0 ...(1)
Consider the plane
∴ normal to the plane is the vector
∴ normal to the plane has direction ratios 1, 1, 1.
∴ A = k, B = k, C = k
Putting values of A, B. C in (1), we get,
k (x – a) + k (y – b) + k (z – c) = 0
or x – a + y – b + z – c = 0
or x + y + z – a –b – c = 0
which is required equation of plane.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z = 1