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If $a=(12,5)$ and $b=(6,8)$ give two orthogonal vectors $U_1$ and $U_2$ such that:

$1)$ $U_1$ lies in $a$

$2)$ $U_1+U_2=b$

I am not really sure what the problem is even asking and I would like some assistance in understanding what the problem is asking about.

I have a vague idea on the problem as it is supposed to require some sort of usage of projections but I do not know how projections apply in this problem.

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  • $\begingroup$ Have you learned about orthogonal projections? $\endgroup$ – Jean-François Gagnon Mar 24 '17 at 14:22
  • $\begingroup$ @Jean-FrançoisGagnon yes $\endgroup$ – John Rawls Mar 24 '17 at 14:31
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Here is my guess:

$U_1$ lies in $a$ would means that $U_1=ka$, $k\in \Bbb R^*$, so $U_2=b-ka$.

Once $U_1$ and $U_2$ are orthogonal then $U_1\cdot U_2=0$ so,

$$U_1\cdot U_2=ka\cdot(b-ka)=ka\cdot b-k^2|a|^2=0\to k=\frac{a\cdot b}{|a|^2}$$

but

$$a\cdot b=(12,5)\cdot (6,8)=72+40=132\\ |a|^2=12^2+5^2=169$$

Now you are able to find $U_1$ and $U_2$.

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Just to give you a pictorial view of the problem. ${e}$ is the error. Just to give you a pictorial view of your problem.

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