# How do you find a vector perpendicular to the vectors v=(1 3 -2) and u=(1 2 -2)?

Aug 17, 2016

The Reqd. Vector$= \left(- 2 , 0 , - 1\right)$.

#### Explanation:

It is known from the Vector Geometry that the Cross Product

or Vector Product of given two vectors $\vec{x}$ and $\vec{y}$,

denoted by $\vec{x} \times \vec{y}$ is a vector $\bot$ to both $\vec{x}$ and

$\vec{y}$.

Hence, the Reqd. Vector$= \vec{v} \times \vec{u}$

$= \det | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(1 , 3 , - 2\right) , \left(1 , 2 , - 2\right) |$

$= \left(- 6 + 4\right) \hat{i} - \left(- 2 + 2\right) \hat{j} + \left(2 - 3\right) \hat{k}$

$= - 2 \hat{i} - \hat{k}$

$= \left(- 2 , 0 , - 1\right)$.