Resolve parallel and perpendicular components

Resolve parallel and perpendicular components

Start with a nonzero target direction and turn it into a unit vector. Then decide whether the question asks for a signed scalar, a nonnegative length, or a vector component. The same dot product appears in each, but the final objects are different.

Let $\widehat{\mathbf b}=\mathbf b/|\mathbf b|$, with $\mathbf b\ne\mathbf0$. In the source diagrams, $\overrightarrow{PR}=\mathbf a$ and $Q$ is the perpendicular foot from $R$ to the line through $P$ in direction $\mathbf b$. Distinguish three different results:

$$\begin{aligned} PQ&=|\mathbf a\cdot\widehat{\mathbf b}| &&\text{(projection length: scalar)},\\ \overrightarrow{PQ}&=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b} &&\text{(parallel component: vector)},\\ \overrightarrow{QR}&=\mathbf a-(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b} &&\text{(perpendicular component: vector)}. \end{aligned}$$

The signed scalar component $\mathbf a\cdot\widehat{\mathbf b}$ may be negative; a length cannot. The source assumes nonparallel vectors to draw a genuine triangle. The formulas also work for parallel or perpendicular $\mathbf a$ and for $\mathbf a=\mathbf0$; only the target $\mathbf b$ must be nonzero.

Acute case: the projection follows b

Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$. In the acute diagram, it is also $\angle QPR$. The complete length proof is

$$\frac{PQ}{PR}=\cos\theta,\qquad PQ=PR\cos\theta=|\mathbf a|\cos\theta=|\mathbf a||\widehat{\mathbf b}|\cos\theta=\mathbf a\cdot\widehat{\mathbf b}.$$

Since $\overrightarrow{PQ}$ and $\mathbf b$ have the same unit direction,

$$\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{\mathbf b}{|\mathbf b|} \quad\Longrightarrow\quad \overrightarrow{PQ}=|\overrightarrow{PQ}|\frac{\mathbf b}{|\mathbf b|}=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}.$$

Obtuse case: the projection opposes b

In the source’s second drawing, $\mathbf b$ points left while $\overrightarrow{PR}$ points up-right. The angle between the vectors is obtuse $\theta$, while the triangle’s angle $\angle QPR=\pi-\theta$ is acute. This distinction corrects the source’s unqualified opening “let angle QPR be θ” for the second case. Now

$$\begin{aligned} \frac{PQ}{PR}&=\cos(\pi-\theta),\\ PQ&=-PR\cos\theta\\ &=-|\mathbf a|\cos\theta\\ &=-|\mathbf a||\widehat{\mathbf b}|\cos\theta\\ &=-\mathbf a\cdot\widehat{\mathbf b}. \end{aligned}$$

The unit directions are opposite, so

$$\begin{aligned} \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}&=-\frac{\mathbf b}{|\mathbf b|},\\ \overrightarrow{PQ}&=-|\overrightarrow{PQ}|\frac{\mathbf b}{|\mathbf b|}\\ &=-(-\mathbf a\cdot\widehat{\mathbf b})\frac{\mathbf b}{|\mathbf b|}\\ &=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}. \end{aligned}$$

The Teacher asks why the length formulas have opposite signs: they are the positive and negative cases of the same absolute-value expression. The vector formula is unchanged in both cases. Finally, the source triangle-law proof gives

$$\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR},\qquad \mathbf a=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}+\overrightarrow{QR},\qquad \overrightarrow{QR}=\mathbf a-(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}.$$

Example 15 applies all three formulas and labels the parallel and perpendicular arrows (ii) and (iii). Its Teacher note stresses that projection of a on b follows b, whereas projection of b on a follows a. Their vectors generally differ even if their lengths happen to agree, as they do in this example. Self-Practice 5 keeps a negative projection coefficient; do not put absolute-value bars into that vector answer.

The earlier scalar-product table refers forward to this projection section for a distributivity proof. The supplied projection pages derive these formulas; a separate proof of distributivity is not printed there.

Reverse the direction and track what changes

Let $\mathbf a=(-3,4,0)$ and $\widehat{\mathbf n}=(1,0,0)$. Find the signed scalar component, projection length, parallel vector component and perpendicular vector component of $\mathbf a$ relative to this direction. Predict which results change when $\widehat{\mathbf n}$ is replaced by $-\widehat{\mathbf n}$, then verify.

After your attempt: compare the reasoning

Initially the signed component is $-3$, the projection length is 3, the parallel component is $(-3,0,0)$ and the perpendicular component is $(0,4,0)$. With the reversed unit direction the signed component becomes 3, but

$[\mathbf a\cdot(-\widehat{\mathbf n})](-\widehat{\mathbf n})=(\mathbf a\cdot\widehat{\mathbf n})\widehat{\mathbf n}.$

Both vector components and the projection length stay unchanged. Also $|\mathbf a\times\widehat{\mathbf n}|=4$ measures the perpendicular length. A signed scalar and a vector component are different objects: reversing the measuring direction changes the scalar’s sign without moving the foot.