Question

Find the magnitude and direction of vector RS if R is (-4, -2) and S is (3, 7).

Answer 1

\[\text{magnitude} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
\[\text{direction} = \tan^{-1} (\frac{y_2 - y_1}{x_2 - x_1})\]

Answer 2

a.
magnitude: 9.1
direction: 50.8°
c.
magnitude: 11.4
direction: 52.1°
b.
magnitude: 5.1
direction: 59.2°
d.
magnitude: 8.6
direction: 60.0°

Answer 3

The magnitude is the straight length of the vector, found using the distance formula.
The direction is based on the angle between the horizontal and the vector extended to it. So you can think of the angle made being arctan(gradient) of the line
\[\tan^{-1} \frac{ y2-y1 }{x2-x1}\]. so find the gradient and take its inverse tangent.

Answer 4

So how would I find that? And which one is it in the multiple choice?

Answer 5

Your points are R (-4, -2) and S (3, 7).
Substitute into the formulae given by abb50.
The distance formula\[\sqrt{(3 -(-4))^{2} + (7-(-2))^{2}} = \sqrt{(7^{2}+9^{2})} = \sqrt{49+81} = \sqrt{130}\]
Then you'd use your calculator to express it to decimal form. 11.4 seems like it
Then the gradient is \[\frac{ 7-(-2) }{ 3 - (-4) } = \frac{ 9 }{ 7 }\]
Take the inverse tangent of this.\[\tan^{-1} (9/7)\] = 52.1 degrees