My mind is a total blank on this ...

[val1]

Please can you help with what appears to be a simple question but I just have blanked out on it completely.

Triangle ABC has a right angle at A where \(\displaystyle \L AB = \left[ \begin{array}{r}
- 3 \\
3 \\
- 2 \\
\end{array} \right]\) and where \(\displaystyle \L CA = \left[ \begin{array}{r}
- 1 \\
3 \\
6 \\
\end{array} \right]\)

a) Calculate the acute angle at B. (I think it's 45 degs because it's a right angled triangle, but I don't know how to prove that using vector information)

b) Find the length of AB

I have tried to solve this one for hours but even though I guess it will be pretty straight forward for someone who knows how to handle this, I'm just getting a headache looking at it. I've checked my textbook, but for this kind of blank I'm not getting anywhere there either.

Any help will be greatly appreciated. :roll:

 

[pka]

The angle between AB and AC is \(\displaystyle \arccos \left( {\frac{{AB \cdot AC}}{{\left\| {AB} \right\|\left\| {CA} \right\|}}} \right)\).

 

[val1]

Thanks for your reply.

I'm a little confused with the answer because the dot product of AB.CA equals 0. They are the two perpendicular sides of a right angle triangle, where the right angle is at A, and I need to find the acute angle at B.

The numerator of the supplied formula will be zero, which doesn't seem right....

 

[pka]

You need to know BA and BC.
Apply the same formula.

 

[val1]

Hi again
I'm really stuck with this one.
I think that BA =\(\displaystyle \
\left[ \begin{array}{r}
3 \\
- 3 \\
2 \\
\end{array} \right]\)but I didn't apply a formula to get that - just changed the signs of the numbers for AB. Is that correct for BA?
How would I find BC?

Which formula do you think I should apply?

I'm sorry if this seems an obvious question but for some reason I'm stumped and I really would appreciate any help with this one.

Thanks

 

[pka]

Sorry to be a day late. I was too tired last evening to think clearly.

From the given, \(\displaystyle \L
AC = \left[ \begin{array}{r}
1 \\
- 3 \\
- 6 \\
\end{array} \right]\) so we get \(\displaystyle \L
BC = AC - AB = \left[ \begin{array}{r}
4 \\
- 6 \\
- 4 \\
\end{array} \right]\) .

The angle at B would be \(\displaystyle \L
\arccos \left( {\frac{{BA \cdot BC}}{{\left\| {BC} \right\|\left\| {BA} \right\|}}} \right)\) .

\(\displaystyle \L
\left\| {BA} \right\| = \sqrt {9 + 9 + 4}\)