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 \[\mathtt{c=\frac{u\cdot v}{v\cdot v}\quad\rightarrow\quad p=\left(\frac{u\cdot v}{v\cdot v}\right)v}\]

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Josh Fisher

Instructional designer, software development in K-12 mathematics education.