If $a=\langle12,5\rangle$ and $b=\langle6,8\rangle$, give orthogonal vectors $u_1$ and $u_2$ that $u_1$ lies on a and $u_1+u_2=b$

Question

Question

If $a=\langle12,5\rangle$ and $b=\langle6,8\rangle$, give orthogonal vectors $u_1$ and $u_2$ that $u_1$ lies on a and $u_1+u_2=b$

My steps:

I am not sure whether or not $u_1$ has to lie in the same direction as vector a. but I know that it has to intersect it

then I recognize that $u_1$ and $u_2$ has to lie perpendicular on vector b,

but I do not know what to do from here

a detailed explanation is much appreciated.

Answer 1

If $u_1$ lies on $a$, then $u_1 = t\langle5,12\rangle=\langle 5t,12t\rangle$ for some $t$. Now, $u_2$ needs to be orthogonal to $u_1,$ that is, $u_1\cdot u_2=0,$ which implies $u_2=\langle 12w,-5w\rangle$ for some $w$. Then we need $u_1+u_2=\langle5t+12w,12t-5w\rangle=\langle6,8\rangle,$ which you can easily solve from there.

Answer 2

Assuming $u_1$ lies on $a$ means that $u_1$ is parallel to $a$: solve $$\begin{align} 12x+5y+0t& =0 \\ x+0y+12t&=6 \\ 0x+y+5t&=8\end{align}$$ The first equation makes sure the $u_1\cdot u_2=0$, the second and third make sure they add to $b$, they represent the components of $u_1+u_2$.

Answer 3

"$u_1$ lies on $a$" doesn't make sense. However I suspect it means "$u_1$ is parallel to $a$". In that case, this is asking for the orthogonal projection of $b$ onto $a$ and the orthogonal rejection of $b$ from $a$.

The formulas for those things are

$$u_1 = \operatorname{proj}_a(b) = \frac{b\cdot a}{a\cdot a}a \\ u_2 = \operatorname{rej}_a(b) = b-u_1$$