Multiplying Vectors — Question 3

Question 3

Split a vector into parallel and perpendicular parts

A displacement is $\vec A=(4,3)$ m. The reference direction is the dimensionless unit vector $\hat u=(3/5,4/5)$. Define $\vec A_\parallel=(\vec A\cdot\hat u)\hat u$ and $\vec A_\perp=\vec A-\vec A_\parallel$.

1. Find the signed scalar projection and both vector parts.

2. Check that the perpendicular part has zero dot product with u and recover the original magnitude using the two part lengths.

3. Would replacing u by its negative change the projection vector? Distinguish it from the signed scalar projection.

Question 3 - Solution

Split a vector into parallel and perpendicular parts

(a) Project, then restore a vector direction.

$$\vec A\cdot\hat u=\frac{12+12}{5}=\frac{24}{5}\,\mathrm m,\quad\boxed{\vec A_\parallel=\frac{(72,96)}{25},\quad\vec A_\perp=\frac{(28,-21)}{25}}\quad\mathrm m$$ $\vec A\cdot\hat u=(12+12)/5=24/5$ m. Therefore $\vec A_\parallel=(72/25,96/25)$ m and $\vec A_\perp=(28/25,-21/25)$ m. Their component sum is exactly $(4,3)$ m.

(b) Check the defining geometry.

$$\vec A_\perp\cdot\hat u=\frac{84-84}{125}=0,\quad A^2=(24/5)^2+(7/5)^2=25\,\mathrm{m^2}$$ $\vec A_\perp\cdot\hat u=(84-84)/125=0$ m. The lengths are $24/5$ m and $7/5$ m, respectively. Their squared sum is $(576+49)/25=25\ \mathrm{m^2}$, recovering $|\vec A|=5$ m. The two parts form a right triangle.

(c) Reverse both factors in the projection formula.

$$[\vec A\cdot(-\hat u)](-\hat u)=(\vec A\cdot\hat u)\hat u,\qquad\vec A\cdot(-\hat u)=-24/5\,\mathrm m$$ The scalar projection changes to $-24/5$ m, but $[\vec A\cdot(-\hat u)](-\hat u)=(\vec A\cdot\hat u)\hat u$. The projection vector is unchanged because it projects onto the same undirected line. Its coefficient changes sign with the chosen unit direction.