Question

Vector u has initial point at (8, 6) and terminal point at (–6, 12). Which are the magnitude and direction of u?

Answer 1

SOLUTION

Write out the given point

$(8,6)\text{ and (-6,12)}$

The magnitude of the vertor u is the distance between the two point.

$\begin{gathered} \text{distance = }\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{Where } \\ x_2=-6,x_1=8 \\ y_2=12,y_1=6 \end{gathered}$

Substitute into the formula, we have

$\begin{gathered} \mleft\Vert u\mleft\Vert=\sqrt[]{(-6-8)^2+(12-6)^2}\mright?\mright? \\ \mleft\Vert u\mleft\Vert=\sqrt[]{(-14)^2+6^2}\mright?\mright? \\ \mleft\Vert u\mleft\Vert=\sqrt[]{169+36}=\sqrt[]{232}\mright?\mright? \end{gathered}$

Hence

|| u || =15.23

The magnitude of the vector is 15.232

Then the direction is obtain by using the formula

$\begin{gathered} \tan \theta=\frac{y_2-y_1}{x_2-x_1} \\ \text{Then } \\ \tan \theta=\frac{12-6}{-6-8}=\frac{6}{-14}=-0.4286 \end{gathered}$

Then we have

$\tan \theta=-0.4286$

take inverse tan of the equation above, we have

$\begin{gathered} \theta=\tan ^{-1}(-0.4286) \\ \theta=156.801^0 \end{gathered}$

Hence

The direction is of u is 156.801°

Answer: Second Option