# How do you find two unit vectors orthogonal to a=⟨−1,2,1⟩ and b=⟨−4,0,−5⟩?

May 25, 2017

$\pm \left(- \frac{10}{7 \sqrt{5}} , - \frac{9}{7 \sqrt{5}} , \frac{8}{7 \sqrt{5}}\right) .$

#### Explanation:

We know that, $\vec{x} \times \vec{y}$ is $\bot$to $\vec{x} \mathmr{and} \vec{y} .$

Further, for $\vec{v} \ne \vec{0} , \pm \frac{\vec{v}}{|} | \vec{v} | |$ are the unit vectors along $\vec{v} .$

Now, $\vec{a} \times \vec{b} = | \left(i , j , k\right) , \left(- 1 , 2 , 1\right) , \left(- 4 , 0 , - 5\right) |$

=(-10-0)i-(5-(-4))j+(0-(-8)k,

$\therefore \vec{a} \times \vec{b} = - 10 i - 9 j + 8 k = \left(- 10 , - 9 , 8\right) .$

$\therefore | | \vec{a} \times \vec{b} | | = \sqrt{{\left(- 10\right)}^{2} + {\left(- 9\right)}^{2} + {8}^{2}} = \sqrt{245} = 7 \sqrt{5.}$

Therefore, $\pm \left(- \frac{10}{7 \sqrt{5}} , - \frac{9}{7 \sqrt{5}} , \frac{8}{7 \sqrt{5}}\right)$ are the desiredunit vectors.

May 25, 2017

$\hat{c} = \pm \frac{\sqrt{5}}{35} \left(\begin{matrix}- 10 \\ - 9 \\ 8\end{matrix}\right)$

#### Explanation:

We need to find a vector, $\vec{c}$, such that $\vec{c} \bot \vec{a}$ and $\vec{c} \bot \vec{b}$.

As such $\vec{c} \cdot \vec{a} = 0$ and $\vec{c} \cdot \vec{b} = 0$

Let $\vec{c} = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$

Then $- x + 2 y + z = 0$ and $- 4 x - 5 z = 0$

Let $z = 1$

$- x + 2 y = - 1$

$- 4 x = 5$

$x = - \frac{5}{4}$

$- \left(- \frac{5}{4}\right) - 2 y = - 1$

$2 y = - \frac{9}{4}$

$y = - \frac{9}{8}$

$\vec{c} = \pm \frac{1}{8} \left(\begin{matrix}- 10 \\ - 9 \\ 8\end{matrix}\right)$

$c = \sqrt{{x}^{2} + {y}^{2} + {z}^{z}}$

$c = \sqrt{{\left(- \frac{5}{4}\right)}^{2} + {\left(- \frac{9}{8}\right)}^{2} + 1} = \frac{7 \sqrt{5}}{8}$

$\hat{c} = \frac{1}{c} \vec{c}$

$\hat{c} = \frac{8}{7 \sqrt{5}} \times \pm \frac{1}{8} \left(\begin{matrix}- 10 \\ - 9 \\ 8\end{matrix}\right)$

$\hat{c} = \pm \frac{\sqrt{5}}{35} \left(\begin{matrix}- 10 \\ - 9 \\ 8\end{matrix}\right)$