# Vectors

• Jan 30th 2013, 06:01 AM
Cloudhorn
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.

• Jan 30th 2013, 06:21 AM
ebaines
Re: Vectors
If the angle between the two vectors is $\displaystyle \pi/4$, then the dot product of the two vectors is $\displaystyle |p||q|\sin(\pi/4)$. Do you know how to calculate the dot product of vectors (2,k) and (3,5)?
• Jan 30th 2013, 06:50 AM
Soroban
Re: Vectors
Hello, Cloudhorn!

Quote:

$\displaystyle \text{Let } \vec{p}=\langle 2,k\rangle\,\text{ and }\,\vec{q} = \langle3,5\rangle$
$\displaystyle \text{Find }k\text{ such that the angle between }\vec{p}\text{ and }\vec{q}\text{ is }\tfrac{\pi}{4}$

You are expected to know this formula.
The angle $\displaystyle \theta$ between vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ is given by: .$\displaystyle \cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}| }$

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

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

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

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

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

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

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

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

Therefore: .$\displaystyle k \,=\,\tfrac{1}{2}$
• Jan 30th 2013, 06:53 AM
Plato
Re: Vectors
Quote:

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 $\displaystyle \cos \left( {\frac{\pi }{4}} \right) = \frac{{p \cdot q}}{{\left\| p \right\|\left\| q \right\|}}$.
• Jan 30th 2013, 08:31 AM
Cloudhorn
Re: Vectors
This is perfectly clear to me now, thank you all for your answers.