No idea how to do this, all I know is you need the dot product.
Find two vectors V1 and V2 whose sum is <-4,3,0>, where V1 is parallel to and V2 is perpendicular to <-3,-1,5>.
Understand, I will not do this for you.
Here is the set up.
$\displaystyle \left\{ \begin{gathered} (\alpha + 1)a = - 4 \hfill \\ (\alpha + 1)b = - 1 \hfill \\ (\alpha + 1)c = 0 \hfill \\ - 3a - b + 5c = 0 \hfill \\ \end{gathered} \right.$
Solve for $\displaystyle a,~b,~c,~\&,\alpha$
Again, do not ask me to do it.
How does he not expect help..? You could help him understand the question of course...
Okay so I'm assuming you know how to do cross products.
Let $\displaystyle \vec{a} = \vec{V_1}$ and $\displaystyle \vec{b} = \vec{V_2}$ and $\displaystyle \vec{a} = <a_1,a_2,a_3> $ and $\displaystyle \vec{b}=<b_1,b_2,b_3>$
Then it says the sum of the two vectors is <-4,3,0>. so
$\displaystyle <a_1 + b_1,a_2 + b_2,a_3 + b_3> = <-4,3,0>$
$\displaystyle a_1+b_1=-4$
$\displaystyle a_2+b_2=3$
$\displaystyle a_3+b_3=0$
Then it says the first vector is parallel to <-3,-1,5>, so we use the dot product: $\displaystyle \vec{b} \bullet <-3,-1,5> = 0$
$\displaystyle \vec{b} * <-3,-1,5> = 0$
$\displaystyle -3b_1-b_2+5b_3=0$
Now the cross product. The theory is that $\displaystyle \vec{a} \times <-3,-1,5> = \vec{0}$
Doing that will give you three more equations. Solve for $\displaystyle a_1$ for $\displaystyle a_2$ and $\displaystyle a_3$ for $\displaystyle a_2$. Finally, plug those values in the first 3 equations. Then, solve for $\displaystyle b_1$, $\displaystyle b_2$, and $\displaystyle b_3$ and plug into the equation we got from the dot product.