# Thread: Angles between vectors (parallel, orthogonal, ...)

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

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

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

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

Thanks!

2. 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 $\displaystyle \frac {\pi}{4}$

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

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

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

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

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

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

4. Hello saeros
Originally Posted by saeros
I understand, but then I get to a equation I obviously can't solve !

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

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:
$\displaystyle \textbf{p}.\textbf{q}=|\textbf{p}||\textbf{q}|\cos \phi$
$\displaystyle \Rightarrow 5k+6 = \sqrt{34(4+k^2)}.\frac{1}{\sqrt2}$, when $\displaystyle \phi = 45^o$
Square both sides:
$\displaystyle \Rightarrow 25k^2+60k+36 = 17(4+k^2)$
$\displaystyle =68+17k^2$
$\displaystyle \Rightarrow 8k^2 + 60k -32 = 0$

$\displaystyle \Rightarrow 2k^2+15k-8=0$

$\displaystyle \Rightarrow (2k-1)(k+8)=0$

$\displaystyle \Rightarrow k = \tfrac12,\;-8$
When you substitute back in the original equation, the negative root is invalid. So the answer is:
$\displaystyle k = \tfrac12$