Find vectors $v_1$ and $v_2$, whose sum is $u$ with $v_1$ parallel to $\langle 4,-3\rangle $, with $v_2$ perpendicular to it

Question

Find two vectors $v_1$ and $v_2$, whose sum is $\langle 0,-2\rangle$ where $v_1$ is parallel to $\langle 4,-3\rangle $, while $v_2$ is perpendicular to $\langle 4,-3\rangle$.

To start, $v_1= \langle x_1,y_1\rangle$ which is parallel to $\langle 4,-3\rangle$ can be written as $\langle 4a,-3a\rangle$ and $v_2=\langle x_2,y_2\rangle$ is perpendicular to $\langle 4,-3\rangle$ has to equal zero, and then $\langle x_1,y_1\rangle+\langle x_2,y_2\rangle=\langle 0,-2\rangle$. From here I am very lost.

Answer 1

It is not true that being perpendicular to $\langle 4,-3\rangle$ means it must equal zero; the dot product with $\langle 4,-3\rangle$ must equal zero, which tells you that $v_2=\langle x_2,y_2\rangle$ must be of the form $\langle 3b,4b\rangle$ (since $4x_2-3y_2 = 0$, so $4x_2=3y_2$).

So now you know that one vector is of the form $\langle 4a,-3a\rangle$; and the other vector is of the form $\langle 3b,4b\rangle$, and you know their sum must be $\langle 0,-2\rangle$. That should set you up to a nice system of two linear equations in two unknowns...

Answer 2

Let $v_1 = \langle 4t, -3t \rangle, v_2 = \langle 3s,4s \rangle\implies v_1 \parallel \langle 4,-3\rangle, v_2 \perp \langle 4,-3\rangle$.We solve for $s,t$: $v_1+v_2=u\implies \langle 4t+3s, -3t+4s\rangle=\langle 0,-2\rangle\implies 3s+4t=0, 4s-3t=-2$. From this you can solve for $s,t$ and substitute them back to $v_1, v_2$ to find $v_1,v_2$.