Angles between vectors (parallel, orthogonal, ...)

**saeros** · January 18th 2010, 09:46 PM

p = (2 , k)
q = (3 , 5)

Find such k that:
a) p and q are orthogonal
b) the angle between p and q is $\frac {\pi}{4}$

I was using:
$ \cos \phi = \frac {p * q}{\|p\|*\|q\|} $
but somehow I can't solve it right.

Thanks!

**Prove It** · January 18th 2010, 10:35 PM

Originally Posted by saeros

p = (2 , k)
q = (3 , 5)

Find such k that:
a) p and q are orthogonal
b) the angle between p and q is $\frac {\pi}{4}$

I was using:
$ \cos \phi = \frac {p * q}{\|p\|*\|q\|} $
but somehow I can't solve it right.

Thanks!

If the vectors are orthogonal, then $\mathbf{p}\cdot \mathbf{q} = 0$ .

If the vectors are parallel, then $\mathbf{p}\cdot \mathbf{q} = |\mathbf{p}||\mathbf{q}|$ .

**saeros** · January 19th 2010, 12:43 AM

I understand, but then I get to a equation I obviously can't solve

!

It's like that:
$ 5k - \sqrt{34\cdot(4+k^2)} + 6 = 0 $

I guess I'm not wrong at this point.. but I have no clue how to get k out of it. Any clues please?

**Grandad** · January 19th 2010, 01:28 AM

Hello saeros

Originally Posted by saeros

I understand, but then I get to a equation I obviously can't solve

!

It's like that:
$ 5k - \sqrt{34\cdot(4+k^2)} + 6 = 0 $

I guess I'm not wrong at this point.. but I have no clue how to get k out of it. Any clues please?

I assume it's part (b) that you're stuck with. So here it is: