# Find a vector perpendicular to the plane 2*x-y+2z=5 ?

Sep 17, 2016

The Rqd. Vector is $\left(2 , - 1 , 2\right)$.

#### Explanation:

If the eqn. of a plane $\pi$ is $\pi : a x + b y + c z + d = 0$, then, the the

vector $\bot$ to $\pi$, called the normal vector of $\pi$ is the

vector $\left(a , b , c\right)$.

In our case, it is the vector $\left(2 , - 1 , 2\right)$.