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

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=(y_2-y_1)/(x_2-x_1) \\ \text{Then } \\ \tan \theta=(12-6)/(-6-8)=(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