# How do you find a unit vector perpendicular to the vectors 6i + 2j - k and -2i + 3j + 4k?

Jul 25, 2016

$\left(\pm \frac{1}{3}\right) \left(1 , - 2. 2\right)$. These two are in opposite directions.

#### Explanation:

If

$a = a {1}_{i} + {a}_{2} j + {a}_{3} k = \left(a {1}_{i} , {a}_{2} , {a}_{3}\right)$ and

 b = b_1 i + b_2 j + b_3 k=(b_1 i, b_2 , b_3)

then

+-(a X b )/} a X b }

=(+- b_2a_2 b_3=-a_3, a_3 b_1-a_1 b_3, a_1 b_2-a_2 b_1)/}a X b }

is a unit vector perpendicular to both $a \mathmr{and} b$.

Here, $a = \left(6 , 2 , - 1\right) \mathmr{and} b = \left(- 2 , 3 , 4\right) .$

So,  a X b =(11, -22, 22) and } a X b }= sqrt(11^2+(-22)^2+22^2)=33.

And so, the answer is $\pm \left(\frac{1}{33}\right) \left(11\right) \left(1 , - 2 , 2\right) = \left(\pm \frac{1}{3}\right) \left(1 , - 2. 2\right)$.

These two are in opposite directions,

like ,$\pm k$ perpendicular to both $i \mathmr{and} j$.