Suppose $P$ is a plane and $x$ is a vector (both in $\mathbb{R^3}$), can we say that
$$x \cdot \text{proj} _{P}x = 0$$
For the dot product, must it always be $0$?
Dot product of projection and vector?
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linear-algebra
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1Let $x$ be a vector on the plane $P$. Then the projection of $x$ on $P$ must be equal to $x$. Hence $x \cdot \text{proj}_Px = x \cdot x$. But $x \cdot x$ is never zero if $x \neq 0$. So, what you are saying is not always true. In fact, it is usually false. – 2017-02-06
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1The only case I can think of that this would be true is if $x$ was perpendicular to $P$. – 2017-02-06
1 Answers
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Think about it. If p is some non-zero vector, the projection of u onto p is usually given by $$\frac{u\cdot{p}}{|p|^2}p$$ then, $$ u\cdot{\frac{u\cdot{p}}{|p|^2}p} $$ But we can see that $\frac{u\cdot{p}}{|p|^2}$is just a scalar that we can factor out of the dot product, thus we are left with $$ \frac{u\cdot{p}}{|p|^2}u\cdot{p}=\frac{(u\cdot{p})^2}{|p|^2} $$ Which is zero if and only if $u\perp p$.