How would you find the unit vector along the line joining point (2, 4, 4) to point (-3, 2, 2)?

Nov 28, 2016

$\left(\frac{1}{\sqrt{33}}\right) \left(\begin{matrix}- 5 \\ - 2 \\ - 2\end{matrix}\right)$

Explanation:

If point P is (2,4,4) and Q is (-3,2,2), the vector PQ would be (-5,-2,-2). To find the unit vector, divide vector PQ by its magnitude. $| | \vec{P Q} | |$ would be sqrt((-5)^2 +(-2)^2 + (-2)^2)) =sqrt33. Hence unit vector would be
$\left(\frac{1}{\sqrt{33}}\right) \left(\begin{matrix}- 5 \\ - 2 \\ - 2\end{matrix}\right)$

Nov 28, 2016

Explanation:

Given the points $\left({x}_{0} , {y}_{0} , {z}_{0}\right) \mathmr{and} \left({x}_{1} , {y}_{1} , {z}_{1}\right)$

You make a vector in the direction from one point to another by subtracting each starting coordinate from its respective ending coordinate:

$\overline{V} = \left({x}_{1} - {x}_{0}\right) \hat{i} + \left({y}_{1} - {y}_{0}\right) \hat{j} + \left({z}_{1} - {z}_{0}\right) \hat{k}$

To make it a unit vector, $\hat{V}$, you divide each component of the vector by the magnitude, $| \overline{V} |$:

$| \overline{V} | = \sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}$

Therefore the general form for a unit vector is:

$\hat{V} = \frac{{x}_{1} - {x}_{0}}{\sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}} \hat{i} + \frac{{y}_{1} - {y}_{0}}{\sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}} \hat{j} + \frac{{z}_{1} - {z}_{0}}{\sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}} \hat{k}$

For the given points $\left(2 , 4 , 4\right)$ and $\left(- 3 , 2 , 2\right)$

Compute the magnitude

$| \overline{V} | = \sqrt{{\left(x 1 - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}} =$

$\sqrt{{\left(- 3 - 2\right)}^{2} + {\left(2 - 4\right)}^{2} + {\left(2 - 4\right)}^{2}} =$

$\sqrt{33}$

We need to divide which it the same as multiplying by the reciprocal, $\frac{\sqrt{33}}{33}$

Substitute into the unit vector general form:

$\hat{V} = \left(- 3 - 2\right) \frac{\sqrt{33}}{33} \hat{i} + \left(2 - 4\right) \frac{\sqrt{33}}{33} \hat{j} + \left(2 - 4\right) \frac{\sqrt{33}}{33} \hat{k}$

Simplify:

$\hat{V} = - \frac{5 \sqrt{33}}{33} \hat{i} - \frac{2 \sqrt{33}}{33} \hat{j} - \frac{2 \sqrt{33}}{33} \hat{k}$