Question

What is the magnitude and direction of line GH if G(1, -6) and H(8, 3) ?

A) magnitude: 11.4 units: direction: 37.87
B) magnitude: 11.4 units; direction: 52.13°
C) magnitude: 7.62 units; direction 37.87°
D) magnitude: 7.62 units: direction: 52.13

Answer 1

1. This is the wrong section, this should be in Physics, not Chemistry.

2. I'll help! :-)




Think of it like this:
G makes the point move 1 to the right and 6 down
H makes the point move 8 to the right and 3 up


So for down:
-6 + 3 = -3
1 + 8 = 9
Either of the two ways you draw this you get the same ending point.


To find the distance:

The distance is the Pythagorean Theorem (remember that?) with the absolute value of the displacement being the distance (which is the magnitude here).

\[displacement = \pm \sqrt{(x_2-x_1)^2+(x_2-x_1)^2}\]
Distance is never negative (see this and scroll to the end to see why: http://openstudy.com/updates/4ff7436de4b01c7be8ca117d ), let x\(_1\) and y\(_1\) be the origin (0,0)
\[distance = |displacement| = \sqrt{(x_2-x_1)^2+(x_2-x_1)^2}\]



To find the angle:

The angle however is a bit tricky at first glance, but it's actually kind of simple with a trig function. You have an angle of declination here, which is what I would want in the answer. That's what I labeled with \(\theta\) above.

\(tangent (\theta) = \tan\theta\)
For how we labeled things in the picture above, theta is just \(\large\frac{rise}{run}\)
\(\large \theta = \tan^{-1}\huge(\frac{y_2}{x_2})\)

Now this will give you a negative angle, and angle of declination. That's ok.



This is all well and good except that the correct answer doesn't match anything you've written which means G & H are not vectors like your question was indicating.