Composition of vector projection

Suppose we have a vector $a \in \mathcal{R}^2$ and $x,y \in \mathcal{R}^2$ and let $\alpha$ is the angle between $x$ and $y$. Define: $$z = (a \cdot x) x + (a \cdot y)y$$

If $\alpha = \pi /2$ then $z = x$, if $\alpha < \pi /2$ $z$ is longer than $x$ otherwise it is shorter.

How is the length of $z$ related with $\alpha$? Does it depends also on the relative angle between $x$ and $a$ or $y$ and $a$?

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What does $ax$ mean? – Phira Oct 30 '12 at 17:22

If we talk about the length of z, then it is determined by

$$|z|^2 = |x|^2 + |y|^2 + 2 |x||y| \cos\alpha$$

which clearly shows that it is dependent on the length of vectors $x, y$ and angle between the two vectors $\alpha$.

For $0<\alpha<\pi/2$, $z$ is greater than $x$.

For $\pi/2<\alpha<\pi$, $z$ is less than $x$.

For $\alpha = \pi/2$, $z$ is greater than $x$ unless the length of vector $y$ is zero.

Since the product is not defined by you in the case of $ax$, then one can make assumptions about the definition of product $ax$ and $ay$. Please clarify about it.

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You forgot the square on $|z|^2$. – xavierm02 Oct 30 '12 at 17:37
Thanks for pointing it out. It is now corrected. – neeraj t Oct 30 '12 at 17:46