3-D vector problem: "A 3-D box has the dimensions 6, 8 and 4 respectively...."

[enott312]

3-D vector problem: "A 3-D box has the dimensions 6, 8 and 4 respectively...."

I don't know if this is in the correct category, however it falls within the specialist mathematics field of Australian curriculum and is different to any other Math that I have done before.
Here is my problem:
A 3-D box has the dimensions 6, 8 and 4 respectively. A midpoint, N occurs at (6, 4, 2). Find the angle PNQ from point P (0, 0, 4) to Point Q (0, 8, 0).
Using pythagoras and trigonometry I have found the angle as 73.4 degrees. However, I am supposed to use vectors to solve this problem. Any ideas?

 

[Deleted member 4993]

I don't know if this is in the correct category, however it falls within the specialist mathematics field of Australian curriculum and is different to any other Math that I have done before.
Here is my problem:
A 3-D box has the dimensions 6, 8 and 4 respectively. A midpoint, N occurs at (6, 4, 2). Find the angle PNQ from point P (0, 0, 4) to Point Q (0, 8, 0).
Using pythagoras and trigonometry I have found the angle as 73.4 degrees. However, I am supposed to use vectors to solve this problem. Any ideas?

First determine the position vectors NP and NQ IN Cartesian System.

Then use Vector dot product:

\(\displaystyle NP\cdot NQ = |NP|*|NQ|cos(\theta)\)

 

[enott312]

First determine the position vectors NP and NQ IN Cartesian System.

Then use Vector dot product:

\(\displaystyle NP\cdot NQ = |NP|*|NQ|cos(\theta)\)

Thank you for your help. This is what I did

P =(0,0,4) and N = (6,4,2) then NP = (6, 4, 2) - (0, 0, 4) = (6, 4,-2)
Q = (0, 8, 0) so NQ = (0, 8, 0) - (6, 4,2) = (-6, 4, -2)

Pythagoras gave me |NP| = SQRT(56), as does |NQ|

So using the dot product, I get (-16) =56 cos(\theta), which gives me an angle of (-73.4) degrees. I assume that this is negative because the angle is clockwise from P to Q as per the diagram

Thanks again for your help.