# How do you find the vector perpendicular to the plane containing (-2,1,0), (-3,0,0), and (5,2,0)?

Aug 1, 2016

the Reqd. $\bot$ Vector$= \left(0 , 0 , - 6\right) = - 6 \hat{k}$.

#### Explanation:

Let us name the given pts. contained in the plane, say $\pi$,

$A \left(- 3 , 0 , 0\right) , B \left(- 2 , 1 , 0\right) \mathmr{and} C \left(5 , 2 , 0\right)$

Then, clearly, both $\vec{A B} \mathmr{and} , \vec{A C} \in \pi$.

From Vector Geometry , we remember that the Cross Product

of two vectors is perpendicular to each of them.

Accordingly, (vec(AB) xx vec(AC)) bot vec(AB) &, vec (AC), and,

hence, $\left(\vec{A B} \times \vec{A C}\right) \bot \pi$.

so, the Reqd. $\bot$ Vector$= \left(\vec{A B} \times \vec{A C}\right)$

$= \left(1 , 1 , 0\right) \times \left(8 , 2 , 0\right) = \left(0 , 0 , - 6\right) = - 6 \hat{k}$.

It is note-worthy that all the pts. $A , B , \mathmr{and} , C$ have $0$ as their z-co-ordinate, so they all are in $X Y$-plane, [i.e., plane $\pi$ is $X Y$-plane], and, so, the reqd. vector $\bot$ to the plane $\pi$ has to be along $Z - a \xi s$, i.e., along the vector$\hat{k}$!