Question: **Question**: Find a vector that is perpendicular to both $2\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}$ and $\underset{\sim}{i}-2\underset{\sim}{j}+\underset{\sim}{k}$.

This is for the 3D topic of the Extension 2 HSC maths course. This belongs to a section introducing the scaler product.

I have tried using creating simultaneous equations with it:

> 
> let $\underset{\sim}{u}=2\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}$ and $\underset{\sim}{v}=\underset{\sim}{i}-2\underset{\sim}{j}+\underset{\sim}{k}$
> 
> 
> let $\underset{\sim}{w}$ be the vector that I'm trying to find, therefore $\underset{\sim}{w}=a\underset{\sim}{i}+b\underset{\sim}{j}+c\underset{\sim}{k}$.
> 
> 
> * $\underset{\sim}{u} \times \underset{\sim}{w}=2a+b-c=0$
> * $\underset{\sim}{v} \times \underset{\sim}{w}=a-2b+c=0$
> 
> 
> 

but that doesn't seem to get me anywhere.

Answer: 
This just means that teh third order is not sufficient in particular because $-1.09$ is very far away from $-2$.

To make the story short, what you have is
$$x^2 \cos(x)=\sum\_{n=0}^\infty \frac{\left(-n^2+n+4\right) \cos \left(2-\frac{\pi n}{2}\right)+4 n \sin \left(2-\frac{\pi n}{2}\right)}{n!}(x+2)^n$$ So, if you truncate to $O((x+2)^{p+1})$ you will have the following results
$$\left(
\begin{array}{cc}
 2 & 0.4926841712 \\
 3 & 0.0938985194 \\
 4 & 0.6047229336 \\
 5 & 0.5723469909 \\
 6 & 0.5466012534 \\
 7 & 0.5489493571 \\
 8 & 0.5495411050 \\
 9 & 0.5494858556 \\
 10 & 0.5494781119 \\
 11 & 0.5494788050 \\
 12 & 0.5494788702 \\
 13 & 0.5494788647 \\
 14 & 0.5494788644
\end{array}
\right)$$
Expand just the cosine series around $x=-2$ and you will have:
$$x^2cos x = x^2\left(cos(2) + sin(2)[x+2] - \frac{cos(2)[x+2]^2}{2} - \frac{sin(2)[x+2]^3}{6}\right )$$
$$\\$$
(Expanding the $x^2$ in a Taylor polynomial and multiplying everything by brute force also works, but the Taylor series for $x^2$ will just reduce back to $x^2$ anyway, so it is not necessary)

$$\\$$
Evaluated at $x=-1.09$: 

$$ (-1.09)^2cos(-1.09) \approx 0.549479$$
$$\\$$
$$(-1.09)^2\left(cos(2) + sin(2)[0.91] - \frac{cos(2)[0.91]^2}{2} - \frac{sin(2)[0.91]^3}{6}\right )\approx 0.557713$$