# How do you find a unit vector that is orthogonal to a and b: a = 7 i − 4 j + 8 k and b = −7 i + 9 j + 4 k?

Sep 27, 2016

$\text{The reqd. unit vector} = - \frac{88}{\sqrt{16025}} i - \frac{84}{\sqrt{16025}} j + \frac{35}{\sqrt{16025}} k$

$\approx \frac{1}{126.59} \left(- 88 i - 84 j + 35 k\right)$

#### Explanation:

We know from Vector Geometry that the Vector or Outer

Product of veca & vecb, i.e., $\vec{a} \times \vec{b}$ is orthogonal

to both of them.

The Unit Vector, then, is, $\frac{\vec{a} \times \vec{b}}{|} | \left(\vec{a} \times \vec{b}\right) | |$.

Now, $\vec{a} \times \vec{b} = \left(7 , - 4 , 8\right) \times \left(- 7 , 9 , 4\right)$

$= | \left(i , j , k\right) , \left(7 , - 4 , 8\right) , \left(- 7 , 9 , 4\right) |$

$= \left(- 16 - 72\right) i - \left(28 + 56\right) j + \left(63 - 28\right) k$

$= - 88 i - 84 j + 35 k$

$\therefore | | \left(\vec{a} \times \vec{b}\right) | | = \sqrt{{\left(- 88\right)}^{2} + {\left(- 84\right)}^{2} + {35}^{2}}$

$= \sqrt{7744 + 7056 + 1225} = \sqrt{16025} \approx 126.59$.

Hence, the reqd. unit vector$= - \frac{88}{\sqrt{16025}} i - \frac{84}{\sqrt{16025}} j + \frac{35}{\sqrt{16025}} k$

or, $\approx \frac{1}{126.59} \left(- 88 i - 84 j + 35 k\right)$

Enjoy Maths.!