Fact 3.12 Projection Formula.
\begin{equation}
\text{proj}_{\mathbf v}\mathbf u = \left(\frac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\right)\mathbf v.\tag{3.8}
\end{equation}
Section 3.2 Projection of \(\mathbf u\) in the Direction of \(\mathbf v\)
The dot product measures how much two vectors point in the same direction. This section turns that measurement into a vector: given \(\mathbf u\) and \(\mathbf v\text{,}\) we ask how much of \(\mathbf u\) lies along \(\mathbf v\text{.}\) The answer is the projection of \(\mathbf u\) onto \(\mathbf v\text{,}\) and it lets us split any vector into a part that points along \(\mathbf v\) and a part perpendicular to it. That decomposition is what makes the dot product useful in practice—it is how we find the component of a force along a direction of motion, as in Section 3.3, or the component of gravity down a slope.
The projection of \(\mathbf u\) in the direction of \(\mathbf v\) is denoted \(\text{proj}_{\mathbf v}\mathbf u\text{,}\) and is shown in Figure 3.11.
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The length of \(\text{proj}_{\mathbf v}\mathbf u\) is \(\left\|\text{proj}_{\mathbf v}\mathbf u\right\| = \|\mathbf u\|\cos\theta\text{,}\) and since its direction is the same as \(\mathbf v\text{,}\) we have
\begin{equation}
\text{proj}_{\mathbf v}\mathbf u
= (\|\mathbf u\|\cos\theta)\,\frac{\mathbf v}{\|\mathbf v\|}
= \left(\|\mathbf u\|\,\frac{\mathbf u\cdot\mathbf v}{\|\mathbf u\|\|\mathbf v\|}\right)\frac{\mathbf v}{\|\mathbf v\|}
= \left(\frac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\right)\mathbf v.\tag{3.7}
\end{equation}
Writing \(\mathbf u\) as Two Vectors, One Parallel and One Perpendicular to \(\mathbf v\).
As can be seen in Figure 3.11, \(\text{proj}_{\mathbf v}\mathbf u\) is parallel to \(\mathbf v\text{,}\) and the vector \(\mathbf u - \text{proj}_{\mathbf v}\mathbf u\) is perpendicular to it. Also, the sum of these two vectors equals \(\mathbf u\text{.}\) This means we can write \(\mathbf u\) as
\begin{equation}
\begin{aligned}
\mathbf u \amp = \text{proj}_{\mathbf v}\mathbf u + \left(\mathbf u - \text{proj}_{\mathbf v}\mathbf u\right) \\
\amp = \underbrace{\left(\frac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\right)\mathbf v}_{\text{parallel to } \mathbf v}
\;+\; \underbrace{\left(\mathbf u - \left(\frac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\right)\mathbf v\right)}_{\text{orthogonal to } \mathbf v}.
\end{aligned}\tag{3.9}
\end{equation}
Example 3.14 Decomposing a vector into parallel and perpendicular parts.
Consider the two vectors \(\mathbf v = \mathbf i + \mathbf j\) and \(\mathbf u = 2\mathbf i + 2\mathbf j + 2\mathbf k\text{.}\) Write \(\mathbf u\) as a sum of two vectors, one of which is parallel to \(\mathbf v\) and the other perpendicular to it.
Solution.
Since the projection of \(\mathbf u\) in the direction of \(\mathbf v\) is parallel to \(\mathbf v\text{,}\) we have
\begin{equation*}
\begin{aligned}
\mathbf u \cdot \mathbf v \amp = (2\times1)+(2\times1)+(2\times0) = 4 \\
\|\mathbf v\| \amp = \sqrt{1+1} = \sqrt2 \\
\mathbf u_{\parallel} = \text{proj}_{\mathbf v}\mathbf u
\amp = \left(\frac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\right)\mathbf v
= \frac{4}{2}(\mathbf i+\mathbf j) = 2\mathbf i + 2\mathbf j.
\end{aligned}
\end{equation*}
For the perpendicular component of \(\mathbf u\) to \(\mathbf v\text{,}\) we have
\begin{equation}
\mathbf u_{\perp} = \mathbf u - \text{proj}_{\mathbf v}\mathbf u
= (2\mathbf i+2\mathbf j+2\mathbf k) - (2\mathbf i+2\mathbf j) = 2\mathbf k.\tag{3.10}
\end{equation}
To achieve a visual understanding of this decomposition, watch the video below.
Checkpoint 3.16 Concept Check.
Everything in (3.9) rests on the claim read off from Figure 3.11: that \(\text{proj}_{\mathbf v}\mathbf u\) is parallel to \(\mathbf v\text{,}\) and that \(\mathbf u - \text{proj}_{\mathbf v}\mathbf u\) is perpendicular to \(\mathbf v\text{.}\) A picture is not a proof. Explain why each half of the claim is true.
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For the parallel half, look at the formula for \(\text{proj}_{\mathbf v}\mathbf u\) in Fact 3.12 and say what kind of expression it is. Why does that observation alone settle the question?
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For the perpendicular half, compute \(\left(\mathbf u - \text{proj}_{\mathbf v}\mathbf u\right)\cdot\mathbf v\text{.}\) Justify each step by naming the property of the dot product from Fact 3.10 that you used, and then say what the value you obtain tells you about the two vectors.
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