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Determine the vector(s) whose vector projection on $u = \langle1,2,2\rangle$, is $v =\langle 3,6,6 \rangle$ and its scalar projection on $w =\langle1,1,1\rangle$ is $√3.$

I am really stuck on this question. The only thing I know is to start off with the unknown vector to be $\langle a,b,c\rangle$. Any help would be appreciated. Thanks!

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The vector projection $v$ on $u$ of a vector $x$ is given by the formula $$ v=\frac{(x,u)}{\|u\|^2}\;u=(3,6,6). $$ By scalar projection of $x$ on $w$, you probably mean $$ \frac{(x,w)}{\|w\|}=\sqrt{3}. $$ Now writing $x=(a,b,c)$ and computing the above formulae will lead you to a set of two linear equations in $a,b,c$.

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So I did that and got $a+b+c = 3$ for one equation using the scalar projection, but I'm not too sure for the vector projection one – user61653 Mar 10 '13 at 2:33
Simply, $(x,u)=a+2b+2c$ and $\|u\|^2=9$. Looking at one coordinate is enough. Say the first one. On the left, you have $\frac{a+2b+2c}{9}\cdot 1$. On the right, you have $3$. Just equate them. – 1015 Mar 10 '13 at 2:36
Would the other one be $a + 2b + 2c = 0$? – user61653 Mar 10 '13 at 2:36
No, I think it is $a+2b+2c=27$. – 1015 Mar 10 '13 at 2:37
SO $a = -24$?, but how would $b$ and $c$ be found since we can't isolate them because they have the same coefficient – user61653 Mar 10 '13 at 2:40

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