# Find angle between vectors..given 4 points..no idea what to do

• Apr 8th 2008, 05:43 PM
Find angle between vectors..given 4 points..no idea what to do
Hey just really need some help with this one I have no idea where to start i want to draw it out but I don't now how to draw xyz graphs (Worried)

The question is as follows:

A(-1,2,-2), B(2,14,2), C(2,-3,4) and D(6,-3,7) are four points in space. Find the angle between AB and CD.

Thanks heaps(Yes)
• Apr 8th 2008, 06:05 PM
Mathstud28
Ok
Quote:

Hey just really need some help with this one I have no idea where to start i want to draw it out but I don't now how to draw xyz graphs (Worried)

The question is as follows:

A(-1,2,-2), B(2,14,2), C(2,-3,4) and D(6,-3,7) are four points in space. Find the angle between AB and CD.

Thanks heaps(Yes)

You dont need to per say...just use the three dimensional distance equation $D=\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$...and then apply the converse of the law of cosines $arccos\bigg[\frac{c^2-a^2-b^2}{-2ab}=\theta\bigg]$
• Apr 9th 2008, 01:19 AM
i can get to the distance but where do i plug them in the cosine arccos rule?
Thanks so much for the reply... I plugged the numbers on the 3-d distance rule and got AB distance = 13 and CD distance = 5... I'm just not familiar with this cosine arccos rule.

Where do i plug the numbers?
And since there is 3 distances ( I'm guessing... a,b,c) do I then need to work out distance BC or am I missing the point?

Thanks again (Rofl)
• Apr 9th 2008, 01:44 AM
earboth
Quote:

Thanks so much for the reply... I plugged the numbers on the 3-d distance rule and got AB distance = 13 and CD distance = 5... I'm just not familiar with this cosine arccos rule.

Where do i plug the numbers?
And since there is 3 distances ( I'm guessing... a,b,c) do I then need to work out distance BC or am I missing the point?

Thanks again (Rofl)

1. Calculate the vector $\overrightarrow{AB}$:

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2,14,2) - (-1,2,-2) = (3,12,4)$

and

$\overrightarrow{CD}=(6,-3,7) - (2, -3, 4) = (4, 0, 3)$

2. Calculate the absolute value (=length) of the vectors:

$|\overrightarrow{AB}| = \sqrt{3^2+12^2+4^2}=13$

$|\overrightarrow{CD}| = \sqrt{4^2+3^2}=5$

3. Now apply the formula Mathstud28 has posted:

$\cos(\theta) = \frac{(3, 12, 4) \cdot (4, 0, 3)}{13 \cdot 5} = \frac{24}{65}~\implies~ \theta \approx 68.33^\circ$
• Apr 9th 2008, 04:04 AM