# Projections

A projection is like the shadow of a vector, say $$\mathtt{u}$$, on another vector, say $$\mathtt{v}$$, if light rays were coming in across $$\mathtt{u}$$ and perpendicular to $$\mathtt{v}$$. For the vectors at the right, imagine a light source above and to the left of the illustration, perpendicular to the vector $$\mathtt{v}$$. The projection, which we’ll call $$\mathtt{p}$$, will be a vector that will extend from the point shown to where the end of the shadow of $$\mathtt{u}$$ touches $$\mathtt{v}$$.

If you take a moment maybe to read that description twice (because it’s kind of dense), you may be able to tell that the vector $$\mathtt{p}$$ that we’re looking for will be some scalar multiple of vector $$\mathtt{v}$$, since it will lie exactly on top of $$\mathtt{v}$$. In fact, given the picture, the projection vector $$\mathtt{p}$$ will have the opposite sign as $$\mathtt{v}$$ and will have a scale factor pretty close to 0, since the projection vector looks like it will be much smaller than $$\mathtt{v}$$.

That information is shown at the right. Helpfully, the vector $$\mathtt{u-p}$$ is perpendicular to $$\mathtt{v}$$, so we know that $$\mathtt{\left(u-p\right)\cdot v=0}$$. And, using the Distributive Property, we get $$\mathtt{u\cdot v-p\cdot v=0}$$.

Since we know that our sought-after vector $$\mathtt{p}$$ will be some scalar multiple of $$\mathtt{v}$$, we can substitute, say, $$\mathtt{cv}$$ for $$\mathtt{p}$$ in the above to get $$\mathtt{u\cdot v-cv\cdot v=0}$$. And a property of dot product multiplication allows us to rewrite that as $$\mathtt{u\cdot v-c\left(v\cdot v\right)=0}$$.

This means that $$\mathtt{u\cdot v=c\left(v\cdot v\right)}$$, which means that the scale factor $$\mathtt{c}$$ that we’re after—the factor we can multiply by $$\mathtt{v}$$ to produce $$\mathtt{p}$$—is $\mathtt{c=\frac{u\cdot v}{v\cdot v}\quad\rightarrow\quad p=\left(\frac{u\cdot v}{v\cdot v}\right)v}$