Exercises — Cross product — formula, geometric meaning (area), right-hand rule

What: the cross product outputs a vector (three numbers), not a scalar. One guaranteed fact: that vector is perpendicular to both $\vec{a}$ and $\vec{b}$. That is the definition — we do not need to compute anything to know it. Its length would additionally equal the area of the parallelogram the two vectors span, but the "perpendicular vector" fact is the surest instant answer.

Cyclic order $i\to j\to k\to i$ (see figure): going forward around the cycle gives a plus sign.

What: $\vec{a}\times \vec{a}=\vec{0}=(0,0,0)$. Why: the angle between a vector and itself is $\theta =0$, so $\sin \theta =\sin 0=0$, giving magnitude $|\vec{a}{|}^2\cdot 0=0$. Geometrically, two identical arrows span a parallelogram of zero width — no area, no flap, nothing to stand up out of.

What: apply the component formula, one slot at a time.

• First slot: $a_2{b}_3-a_3{b}_2=(3)(4)-(-2)(-1)=12-2=10$.
• Middle slot (watch the sign): $a_3{b}_1-a_1{b}_3=(-2)(2)-(1)(4)=-4-4=-8$.
• Third slot: $a_1{b}_2-a_2{b}_1=(1)(-1)-(3)(2)=-1-6=-7$. $\vec{a}\times \vec{b}=(10,-8,-7).$ Verify (why?): a genuine cross product must be perpendicular to $\vec{a}$, i.e. the dot product is zero: $\vec{a}\cdot (10,-8,-7)=(1)(10)+(3)(-8)+(-2)(-7)=10-24+14=0.✓$

What/Why: area $=|\vec{u}\times \vec{v}|$, so first cross, then take length.

• First: $(0)(0)-(1)(2)=-2$.
• Middle: $(1)(1)-(2)(0)=1$.
• Third: $(2)(2)-(0)(1)=4$. $\vec{u}\times \vec{v}=(-2,1,4),|\vec{u}\times \vec{v}|=\sqrt{(-2{)}^2+1^2+4^2}=\sqrt{4+1+16}=\sqrt{21}.$ Area $=\sqrt{21}\approx 4.583$.

Why edges from one vertex? A triangle is exactly half the parallelogram built on two of its edges (see figure). Pick $A$ as the corner.

Which tool and why? We are not given components, so the component formula is useless. But we are given lengths and a dot product — exactly the ingredients of Lagrange's identity, which links the two products: $|\vec{a}\times \vec{b}{|}^2=|\vec{a}{|}^2|\vec{b}{|}^2-(\vec{a}\cdot \vec{b}{)}^2.$ This is the tool because it converts "how parallel" (dot) into "how perpendicular" (cross) using only magnitudes. $|\vec{a}\times \vec{b}{|}^2=(5^2)(4^2)-10^2=400-100=300.$ $|\vec{a}\times \vec{b}|=\sqrt{300}=10\sqrt{3}\approx \mathrm{17.32.}$ Sanity: $\cos \theta =\frac{10}{5\cdot 4}=\frac{1}{2}$, so $\theta =60°$, $\sin 60°=\frac{\sqrt{3}}{2}$, and $|\vec{a}||\vec{b}|\sin \theta =20\cdot \frac{\sqrt{3}}{2}=10\sqrt{3}.$ ✓

What we use: the cross product distributes over addition (it is linear in each slot), and $\vec{b}\times \vec{b}=\vec{0}$. $(\vec{a}+\vec{b})\times \vec{b}=(\vec{a}\times \vec{b})+(\vec{b}\times \vec{b})=(\vec{a}\times \vec{b})+\vec{0}=\vec{a}\times \vec{b}.$ Geometric meaning: sliding the tip of one edge along the direction of the other edge (that is, replacing $\vec{a}$ by $\vec{a}+\vec{b}$) does not change the parallelogram's area — it is a shear. Base and height are untouched (see figure), so the flap has the same area, hence the same cross product.

Why: $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta =0$. Since $|\vec{a}|\ne 0$ and $|\vec{b}|\ne 0$, the only way the product vanishes is $\sin \theta =0$. $\sin \theta =0$ happens at $\theta =0°$ and at $\theta =180°$ — both cases:

• $\theta =0°$: the vectors point the same way (parallel).
• $\theta =180°$: they point opposite ways (antiparallel). Either way the two arrows lie on one line — they are parallel (collinear) and span no area, so there is no flap and the perpendicular arrow has zero length.

Which tool and why? Volume of a box on three edges is $|\vec{a}\cdot (\vec{b}\times \vec{c})|$: the cross $\vec{b}\times \vec{c}$ gives a vector whose length is the base area and whose direction is perpendicular to the base; dotting with $\vec{a}$ then measures how far $\vec{a}$ sticks out along that perpendicular — i.e. the height. Base area × height = volume. $\vec{b}\times \vec{c}=((1)(1)-(0)(1),(0)(1)-(1)(1),(1)(1)-(1)(1))=(1,-1,0).$ $\vec{a}\cdot (\vec{b}\times \vec{c})=(1)(1)+(0)(-1)+(0)(0)=1.$ Volume $=|1|=1$. (Makes sense: these are three unit "staircase" edges enclosing a unit box.)

Plan: cross gives a perpendicular vector; divide by its length to make it unit-long (Orthogonality). $\vec{a}\times \vec{b}=((-1)(-1)-(1)(0),(1)(1)-(2)(-1),(2)(0)-(-1)(1))=(1,3,1).$ $|\vec{a}\times \vec{b}|=\sqrt{1+9+1}=\sqrt{11}.$ $\hat{n}=\frac{1}{\sqrt{11}}(1,3,1).$ Note: $-\hat{n}$ is also perpendicular and unit-length — there are exactly two answers, one for each side of the plane. The cross product's right-hand rule picks the one written above.

Why cross product? Torque and Angular Momentum: torque measures the twisting effect, which depends on the perpendicular part of the force and the direction of the rotation axis — precisely what a cross product encodes. $\vec{\tau }=\vec{r}\times \vec{F}=((0)(0)-(0)(3),(0)(0)-(2)(0),(2)(3)-(0)(0))=(0,0,6)\text{N⋅m}.$ Direction: along $+\hat{k}$ (out of the $xy$-plane). By the right-hand rule, curling from $\vec{r}$ (along $+x$) toward $\vec{F}$ (along $+y$) points the thumb up — the object spins counterclockwise in the $xy$-plane. Magnitude $6$ N·m $=|\vec{r}||\vec{F}|\sin 90°=2\cdot 3\cdot 1$.

What we do: compute $\vec{b}\times \vec{a}$ by swapping the roles of $a$ and $b$ in each slot, and compare. By the formula, the first component of $\vec{b}\times \vec{a}$ is $b_2{a}_3-b_3{a}_2=-(a_2{b}_3-a_3{b}_2),$ which is exactly $-1$ times the first component of $\vec{a}\times \vec{b}$. The same swap flips the middle slot: $b_3{a}_1-b_1{a}_3=-(a_3{b}_1-a_1{b}_3)$, and the third: $b_1{a}_2-b_2{a}_1=-(a_1{b}_2-a_2{b}_1)$. Every component negates, so $\vec{b}\times \vec{a}=-(\vec{a}\times \vec{b}).■$ Why it matches geometry: swapping the two vectors reverses the curl direction of the right hand, so the thumb flips — same area, opposite direction. Algebra and picture agree.

Left side: $\vec{a}\times \vec{b}=((2)(0)-(2)(-1),(2)(2)-(1)(0),(1)(-1)-(2)(2))=(2,4,-5).$ $|\vec{a}\times \vec{b}{|}^2=4+16+25=45.$ Right side: $|\vec{a}{|}^2=1+4+4=9$, $|\vec{b}{|}^2=4+1+0=5$, $\vec{a}\cdot \vec{b}=2-2+0=0$. $|\vec{a}{|}^2|\vec{b}{|}^2-(\vec{a}\cdot \vec{b}{)}^2=9\cdot 5-0=45.✓$ Angle: $\vec{a}\cdot \vec{b}=0$ means $\cos \theta =0$, so $\theta =90°$ and $\sin \theta =1$. Check: $|\vec{a}||\vec{b}|\sin \theta =3\cdot \sqrt{5}\cdot 1=\sqrt{45}=|\vec{a}\times \vec{b}|$. ✓

Why it lies in that plane: the identity writes $\vec{a}\times (\vec{b}\times \vec{c})$ as $(\text{number})\vec{b}-(\text{number})\vec{c}$ — a combination of only $\vec{b}$ and $\vec{c}$. Any such combination lives in the plane those two span. So the triple product is coplanar with $\vec{b}$ and $\vec{c}$, never poking out of their plane. Numeric test:

• Inner cross: $\vec{b}\times \vec{c}=(0,1,0)\times (0,0,1)=\hat{j}\times \hat{k}=\hat{i}=(1,0,0)$.
• Outer cross: $\vec{a}\times (\vec{b}\times \vec{c})=(1,0,0)\times (1,0,0)=\vec{0}$ (parallel!).
• Right side: $\vec{a}\cdot \vec{c}=0$, $\vec{a}\cdot \vec{b}=0$, so $\vec{b}(0)-\vec{c}(0)=\vec{0}.✓$ Both sides give $\vec{0}$, and indeed $\vec{0}$ trivially lies in the $\vec{b},\vec{c}$ plane.