1. ## Vectors

Let p=(2,k) and q=(3,5), find k such that the angle between p and q is π/4.

The question is under the vector section of the course elementary linear algebra. It's the first math course I've taken at university level. According to the key, the answer is 1/2,
but I see no way of getting there. Any help is appreciated.

2. ## Re: Vectors

If the angle between the two vectors is $\pi/4$, then the dot product of the two vectors is $|p||q|\sin(\pi/4)$. Do you know how to calculate the dot product of vectors (2,k) and (3,5)?

3. ## Re: Vectors

Hello, Cloudhorn!

$\text{Let } \vec{p}=\langle 2,k\rangle\,\text{ and }\,\vec{q} = \langle3,5\rangle$
$\text{Find }k\text{ such that the angle between }\vec{p}\text{ and }\vec{q}\text{ is }\tfrac{\pi}{4}$
You are expected to know this formula.
The angle $\theta$ between vectors $\vec{u}$ and $\vec{v}$ is given by: . $\cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}| }$

In our problem: . $\frac{\vec{p}\cdot\vec{q}}{|\vec{p}||\vec{q}|} \;=\;\cos\tfrac{\pi}{4}$

We have: . $\begin{Bmatrix}\vec{p}\cdot\vec{q} &=&\langle2,k\rangle\cdot\langle3,5\rangle &=& 6 + 5k \\ |\vec{p}| &=& \sqrt{2^2+k^2 &=& \sqrt{k^2+4} \\ |\vec{q}| &=& \sqrt{3^2+5^2} &=& \sqrt{34} \end{Bmatrix}$

Substitute: . $\dfrac{5k+6}{\sqrt{k^2+4}\sqrt{34}} \:=\:\dfrac{1}{\sqrt{2}} \quad\Rightarrow\quad \sqrt{2}(5k+6) \;=\;\sqrt{34}\sqrt{k^2+4}$

Square both sides: . $2(5k+6)^2 \;=\;34(k^2+4) \quad\Rightarrow\quad (5k+6)^2 \;=\;17(k^2+4)$

. . . . . $25k^2 + 60k + 36 \:=\:17k^2 + 68 \quad\Rightarrow\quad 8k^2 + 60k - 32 \:=\:0$

. . . . . $2k^2 + 15k - 8 \:=\:0 \quad\Rightarrow\quad (2k-1)(k+8) \:=\:0$

We have: . $\begin{Bmatrix}2k-1 \:=\:0 & \Rightarrow & k \:=\:\frac{1}{2} \\ k+8 \:=\:0 & \Rightarrow & k \:=\:\text{-}8 \end{Bmatrix}$

If $k = \text{-}8$, we find that $\theta \,=\,\tfrac{3\pi}{4}$

Therefore: . $k \,=\,\tfrac{1}{2}$

4. ## Re: Vectors

Originally Posted by Cloudhorn
Let p=(2,k) and q=(3,5), find k such that the angle between p and q is π/4.
The question is under the vector section of the course elementary linear algebra. It's the first math course I've taken at university level. According to the key, the answer is 1/2, but I see no way of getting there. Any help is appreciated.
The angle between to vectors is $\cos \left( {\frac{\pi }{4}} \right) = \frac{{p \cdot q}}{{\left\| p \right\|\left\| q \right\|}}$.

5. ## Re: Vectors

This is perfectly clear to me now, thank you all for your answers.