Given the vector u = $<-2, 6, 4>$ and a vector v such that the vector projection of u onto v is $<2,4,4>$, and the vector projection of v onto u is $<−8,24,16>$. What is the vector v?
I tried it like this but can't reach to final answer.
2 Answers
$\begingroup$The projection of $\def\v#1{{\bf#1}}\v u$ onto $\v v$ is a scalar times $\v v$. So from the given information we have $$\v v=\lambda(2,4,4)\ .$$ Hence $${\rm proj}_{\v u}\v v=\frac{\v v\cdot\v u}{\v u\cdot\v u}\,\v u =\lambda\frac{9}{14}(-2,6,4)\ .$$ It is given that this projection is $(-8,24,16)$, so $\lambda=\frac{56}9$ and hence $$\v v=\tfrac{56}9(2,4,4)\ .$$
$\endgroup$ $\begingroup$Here is a more tangent way to solve this problem:
Notice $v$ and the projection of $u$ onto $v$ must have the same direction, therefore we can assume $$v = \lambda(2, 4, 4),$$ in which $\lambda$ is a constant to be determined. Now use the other condition to establish the equation $$(-8, 24, 16) = \frac{\langle v, u \rangle}{\langle u, u \rangle}u = \frac{36\lambda}{56}(-2, 6, 4).$$ Solve this to get $\lambda = \frac{56}{9}$. Therefore, $$v = \frac{56}{9}(2, 4, 4).$$
$\endgroup$
