# How do you find two unit vectors orthogonal to A=(1, 3, 0) B =(2, 0, 5)?

Nov 15, 2016

#### Explanation:

Begin by computing the cross product. I use a determinant:

barA xx barB = | (hati, hatj, hatk), (1, 3, 0), (2,0, 5) |

$\overline{A} \times \overline{B} = \hat{i} | \left(3 , 0\right) , \left(0 , 5\right) | + \hat{j} | \left(0 , 1\right) , \left(5 , 2\right) | + \hat{k} | \left(1 , 3\right) , \left(2 , 0\right) |$

$\overline{A} \times \overline{B} = 15 \hat{i} - 5 \hat{j} - 6 \hat{k}$

Let $\overline{C} = 15 \hat{i} - 5 \hat{j} - 6 \hat{k}$

The unit vector, $\hat{C} = \frac{\overline{C}}{|} \overline{C} |$

$| \overline{C} | = \sqrt{{15}^{2} + \left(- {5}^{2}\right) + {\left(- 6\right)}^{2}}$

$| \overline{C} | = \sqrt{286}$

$\hat{C} = \frac{15}{\sqrt{286}} \hat{i} - \frac{5}{\sqrt{286}} \hat{j} - \frac{6}{\sqrt{286}} \hat{k}$

The only other vector that can be orthogonal to $\overline{A} \mathmr{and} \overline{B}$ is:

$\overline{B} \times \overline{A}$

Because $\overline{A} \times \overline{B} = - \left(\overline{B} \times \overline{A}\right)$, the only other unit vector orthogonal to $\overline{A} \mathmr{and} \overline{B}$ is:

$- \hat{C} = - \frac{15}{\sqrt{286}} \hat{i} + \frac{5}{\sqrt{286}} \hat{j} + \frac{6}{\sqrt{286}} \hat{k}$