Originally Posted by

**brd_7** I got 28 for the norm of 4x-4y.. I hope thats right.. I then gained th result of 66.4 for the theta angle in part b.. Thanks for the help.

As the angle you've found is correct, I think you mistyped your answer to the first question : $\displaystyle \|4x-4y\|=20$.

I think im going to need a litrle bit more help with the last part of the question...

The dot product is a bilinear function : $\displaystyle \begin{cases}\langle u, v+w\rangle = \langle u, v\rangle + \langle u, w\rangle & (1)\\\langle u+ v, w\rangle=\langle u,w\rangle +\langle v,w\rangle &(2)\\

\langle \alpha u,v\rangle=\langle u, \alpha v\rangle = \alpha\langle u, v\rangle & (3)\end{cases}$ so the LHS of $\displaystyle \langle 2x-y,\lambda x-y\rangle=0$ can be expanded.

$\displaystyle \begin{aligned}

\langle 2x-y,\lambda x-y\rangle &=0\\

\langle 2x-y,\lambda x\rangle+\langle 2x-y,-y\rangle &=0 \text{\,\,\,\,using (1)}\\

\lambda \langle 2x-y, x\rangle-\langle 2x-y,y\rangle &=0 \text{\,\,\,\,using (3)}\\

\end{aligned}$

This can be further expanded using (2) and (3) to make appear $\displaystyle \|x\|^2,\,\|y\|^2$ and $\displaystyle \langle x ,y\rangle$...