 # Projection of u onto v and v onto u Given the vector u = <−2,6, stop2dance3l 2021-12-14 Answered
Projection of u onto v and v onto u
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?
Let $\stackrel{\to }{v}=$
Projection of is given by:
$pro{j}_{u}v=\frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{{|u|}^{2}}\stackrel{\to }{u}=\frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{{\left(-2\right)}^{2}+{6}^{2}+{4}^{2}}<-2,6,4>$
$<-8,24,16\ge \frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{{\left(-2\right)}^{2}+{6}^{2}+{4}^{2}}<-2,6,4>$
$⇒4<-2,6,4\ge \frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{4+36+16}<-2,6,4>$
$⇒<-2,6,4\ge \frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{224}<-2,6,4>$
On comparing $\frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{224}=1$
$⇒\stackrel{\to }{u}.\stackrel{\to }{v}=224$
$pro{j}_{v}u=\frac{\stackrel{\to }{u}.\stackrel{\to }{v}}{{|v|}^{2}}\stackrel{\to }{v}=\frac{224}{{|v|}^{2}}$
$⇒<2,4,4\ge \frac{224}{{|v|}^{2}}$
Dividing both sides by 2 we get:
$⇒<1,2,2\ge \frac{112}{{|v|}^{2}}$
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The projection of u onto v is a scalar times v. So from the given information we have $v=\lambda \left(2,4,4\right)$.
Hence $pro{j}_{u}v=\frac{v\cdot u}{u\cdot u}u=\lambda \frac{9}{14}\left(-2,6,4\right)$.
It is given that this projection is and hence
$v=\frac{56}{9}\left(2,4,4\right)$

We have step-by-step solutions for your answer! Karen Robbins
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 $v=\lambda \left(2,4,4\right)$,
in which $\lambda$ is a constant to be determined. Now use the other condition to establish the equation
$\left(-8,24,16\right)=\frac{\left(v,u\right)}{\left(u,u\right)}u=\frac{36\lambda }{56}\left(-2,6,4\right)$.
Solve this to get $\lambda =\frac{56}{9}$. Therefore, $v=\frac{56}{9}\left(2,4,4\right)$.

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