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I would like to know if my problem is correct. It is a homework from my university course.

The questions are:

a) Find the projection of $y$ on $x$
b) Find the vector projection of $y$ on $x$
c) Find the vector projection of $y$ on the normal of $x$
where $x = [5,2,-1]$, $y = [0,1,1]$

a)

$P = \frac{[0,1,1] \cdot [5,2,-1]}{\sqrt{30}} = √30 / 30 *(-5) = -3√30 / 18$

b)

$yx = -3√30 / 18 * √30 / 30 [5,2,-1] = -3 / 18 [5,2,-1] = [-15/18,-6/18,3/18]$

c)

$y \perp x = [0,1,1] - [-15/18,-6/18,3/18] = [15/18,4/3,15/18]$

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    Just to way to rewrite it2012-11-01

1 Answers 1

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Well, $y$ should be composed as $y=y_1+y_2$ where $y_1||x$ and $y_2\perp x$. By scalar multiplication by $x$ we get $x\cdot y = x\cdot y_1 + 0 $ and $y_1=\alpha\cdot x$ for some $\alpha$ (because it is parallel to $x$). So, $x\cdot y=x\cdot\alpha x \implies \alpha=\displaystyle\frac{x\cdot y}{x\cdot x}$, then you get $y_1$ which answers a). For this, you will indeed need $x\cdot y = 5* 0 + 2*1 + (-1)*1 = 1 \ \text{ and }\ x\cdot x = 30. $

For c), I assume the normal of $x$ is meant the plane of all vectors that are orthogonal to $x$. If you draw it, you will see that $y_2\ \ (=y-y_1)$ of the same decomposition gives the answer.