Find the equation of the plane which is at a distance of 5 units from

1.	Find the equation of the plane which is at a distance of 5 units from the origin & is perpendicular to 2i - 3j + 6k.
Answer» Let n be the unit vector in the direction of 2i - 3j + 6k ∴ n = (2i - 3j + 6k)/√(2² + (-3)² + 6²) = (2i - 3j + 6k)/7 = (2/7)i - (3/7)j + (6/7)k Let r = xi + yj + zk be the position vector of any point (x,y,z) on the plane. Then the equation of the plane is n.vector r = P or, (xi + yj + zk).((2/7)i - (3/7)j + (6/7)k) = 5 or, 2x - 3y + 6z = 35

Find the equation of the plane which is at a distance of 5 units from the origin & is perpendicular to 2i - 3j + 6k.

Answer»

Let n be the unit vector in the direction of 2i - 3j + 6k

∴ n = (2i - 3j + 6k)/√(2² + (-3)² + 6²) = (2i - 3j + 6k)/7 = (2/7)i - (3/7)j + (6/7)k

Let r = xi + yj + zk be the position vector of any point (x,y,z) on the plane. Then the equation of the plane is

n.vector r = P

or, (xi + yj + zk).((2/7)i - (3/7)j + (6/7)k) = 5

or, 2x - 3y + 6z = 35