Compare parallel and perpendicular lengths

Compare parallel and perpendicular lengths

Use a unit target direction before comparing lengths. The absolute dot product measures the parallel length; the cross-product magnitude measures the perpendicular length. Both answers are scalars and should satisfy the Pythagorean check below.

Let $\mathbf a,\mathbf b$ be nonparallel nonzero vectors, $\widehat{\mathbf b}=\mathbf b/|\mathbf b|$, and $\widehat{\mathbf n}$ their right-hand unit normal. In the source right triangle $PQR$, $\overrightarrow{PR}=\mathbf a$ and $Q$ is the perpendicular foot on the line through $P$ parallel to $\mathbf b$. The acute drawing has $\theta$ at $P$. Its parallel length is $PQ=|\mathbf a\cdot\widehat{\mathbf b}|$, and the complete source height proof is

$$\begin{aligned} \frac{QR}{PR}&=\sin\theta,\\ QR&=|\mathbf a|\sin\theta\\ &=|\mathbf a||\widehat{\mathbf b}|\sin\theta|\widehat{\mathbf n}|\\ &=\big|\big(|\mathbf a||\widehat{\mathbf b}|\sin\theta\big)\widehat{\mathbf n}\big|\\ &=|\mathbf a\times\widehat{\mathbf b}|. \end{aligned}$$

Thus the dot-product magnitude gives the parallel length, and the cross-product magnitude gives the perpendicular length. The cross-product vector itself points normal to their plane; it is not the in-plane perpendicular component. For an obtuse vector angle, the triangle angle is $\pi-\theta$ and $\sin(\pi-\theta)=\sin\theta$, giving the same height. The length formulas extend to parallel directions and $\mathbf a=\mathbf0$ without choosing a unit normal there; $\mathbf b$ must remain nonzero. Pythagoras supplies a useful check:

$$|\mathbf a\cdot\widehat{\mathbf b}|^2+|\mathbf a\times\widehat{\mathbf b}|^2=|\mathbf a|^2.$$

Choose between the two products

The source closes with six comparisons. The geometric “if and only if” statements below assume both vectors nonzero; the algebraic laws also hold for zero vectors. Here $t$ is any real scalar. Here “iff” means “if and only if”.

Self-product

Scalar product:

$$\mathbf a\cdot\mathbf a=\lvert\mathbf a\rvert^2$$

Vector product:

$$\mathbf a\times\mathbf a=\mathbf0$$

Greatest possible magnitude

Scalar product:

$$\lvert\mathbf a\cdot\mathbf b\rvert=\lvert\mathbf a\rvert\lvert\mathbf b\rvert$$

iff parallel

Vector product:

$$\lvert\mathbf a\times\mathbf b\rvert=\lvert\mathbf a\rvert\lvert\mathbf b\rvert$$

iff perpendicular

Zero product

Scalar product:

$$\mathbf a\cdot\mathbf b=0$$

iff perpendicular

Vector product:

$$\mathbf a\times\mathbf b=\mathbf0$$

iff parallel

Reverse order

Scalar product:

$$\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$$

Vector product:

$$\mathbf a\times\mathbf b=-(\mathbf b\times\mathbf a)$$

Scalar compatibility

Scalar product:

$$\mathbf a\cdot(t\mathbf b)=t(\mathbf a\cdot\mathbf b)=(t\mathbf a)\cdot\mathbf b$$

Vector product:

$$t(\mathbf a\times\mathbf b)=(t\mathbf a)\times\mathbf b=\mathbf a\times(t\mathbf b)$$

Distribute in either position

Scalar product:

$$\mathbf a\cdot(\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c$$ $$(\mathbf b+\mathbf c)\cdot\mathbf a=\mathbf b\cdot\mathbf a+\mathbf c\cdot\mathbf a$$

Vector product:

$$\mathbf a\times(\mathbf b+\mathbf c)=\mathbf a\times\mathbf b+\mathbf a\times\mathbf c$$ $$(\mathbf b+\mathbf c)\times\mathbf a=\mathbf b\times\mathbf a+\mathbf c\times\mathbf a$$

The source calls scalar compatibility an “associative law”; it does not assert that a cross product of three vectors is associative. Use these length relationships later for distances to lines and planes.