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Section 3.3 Projection of \(\mathbf u\) in the Direction of \(\mathbf v\)

Figure 3.11. The projection of \(\vec{u}\) onto \(\vec{v}\) as a shadow cast by light from above.
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 3.12.
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Figure 3.12. The projection \(\text{proj}_{\mathbf v}\mathbf u\) of \(\mathbf u\) in the direction of \(\mathbf v\text{,}\) together with the perpendicular component \(\mathbf u - \text{proj}_{\mathbf v}\mathbf u\text{.}\)
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 3.12, \(\text{proj}_{\mathbf v}\mathbf u\) 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} \mathbf u = \text{proj}_{\mathbf v}\mathbf u + \left(\mathbf u - \text{proj}_{\mathbf v}\mathbf u\right) = \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}.\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.
Figure 3.15. Decomposing \(\mathbf u = 2\mathbf i + 2\mathbf j + 2\mathbf k\) into \(\mathbf u_{\parallel} = 2\mathbf i + 2\mathbf j\text{,}\) lying in the \(xy\)-plane along \(\mathbf v\text{,}\) and \(\mathbf u_{\perp} = 2\mathbf k\text{,}\) pointing straight up.
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