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


1