Write the value of k for which the planes 2x – 5y + kz = 4 and x + 2y – z = 6 are perpendicular to each other.
Given : equations of perpendicular planes-
2x – 5y + kz = 4
x + 2y – z = 6
To Find : k
Formulae :
Normal vector to the plane :
If equation of the plane is ax + by + cz = d then,
Vector normal to the plane is given by,
Answer :
For given planes –
2x – 5y + kz = 4
x + 2y – z = 6
normal vectors are
As given vectors are perpendicular, hence their normal vectors are also perpendicular to each other.
(2×1) + (-5×2) + (k×(-1)) = 0
2 – 10 – k = 0
- 8 – k = 0
k = -8