# How do you find the unit vector which bisects the angle AOB given A and B are position vectors a=2i-2j-k and b=3i+4k?

Dec 8, 2016

#### Explanation:

Given: $\overline{a} = 2 \hat{i} - 2 \hat{j} - \hat{k} \mathmr{and} \overline{b} = 3 \hat{i} + 4 \hat{k}$

Here is a reference regarding how to find an Angle Bisector Vector , $\overline{c}$:

$\overline{c} = | | \overline{a} | | \overline{b} + | | \overline{b} | | \overline{a}$

Compute the magnitude of $\overline{a}$:

$| | \overline{a} | | = \sqrt{{2}^{2} + {\left(- 2\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{9} = 3$

Compute the magnitude of $\overline{b}$:

$| | \overline{b} | | = \sqrt{{3}^{2} + {4}^{2}} = \sqrt{25} = 5$

$| | \overline{a} | | \overline{b} = 3 \left(3 \hat{i} + 4 \hat{k}\right) = 9 \hat{i} + 12 \hat{k}$

$| | \overline{b} | | \overline{a} = 5 \left(2 \hat{i} - 2 \hat{j} - \hat{k}\right) = 10 \hat{i} - 10 \hat{j} - 5 \hat{k}$

$\overline{c} = 9 \hat{i} + 12 \hat{k} + 10 \hat{i} - 10 \hat{j} - 5 \hat{k}$

$\overline{c} = 19 \hat{i} - 10 \hat{j} + 7 \hat{k}$

However, this is not the unit vector, $\hat{c}$. To make is at unit vector we must divide vector $\overline{c}$ by its magnitude:

||barc|| = sqrt(19^2 + (-10)^2 + 7^2) = sqrt(510)

$\hat{c} = \frac{19}{\sqrt{510}} \hat{i} - \frac{10}{\sqrt{510}} \hat{j} + \frac{7}{\sqrt{510}} \hat{k}$