Exercises — Cross product — formula, direction (right-hand rule), torque - area calculation

Recall the cycle $\hat{i}\to \hat{j}\to \hat{k}\to \hat{i}$. Going forward along this loop is a $+$; going backward is a $-$; a vector crossed with itself is $\vec{0}$.

• (a) $\hat{j}\times \hat{k}=\hat{k}$? No — check the loop. $\hat{j}\to \hat{k}$ is forward, and the next letter after $\hat{k}$ is $\hat{i}$. So $\hat{j}\times \hat{k}=+\hat{i}$.
• (b) $\hat{k}\times \hat{j}$: this is $\hat{j}\times \hat{k}$ with the order swapped, so it flips sign: $-\hat{i}$.
• (c) $\hat{i}\times \hat{i}=\vec{0}$: the angle between a vector and itself is $0$, and $\sin 0=0$. No spread, no area.
• (d) $\hat{k}\times \hat{i}$: is $\hat{k}\to \hat{i}$ forward on the loop? $\dots \hat{k}\to \hat{i}$ yes (the loop wraps around). Next letter is $\hat{j}$. So $\hat{k}\times \hat{i}=+\hat{j}$.

WHAT: plug straight into tool 1. WHY: we are given lengths and the angle, which is exactly what $AB\sin \theta$ eats. $|\vec{A}\times \vec{B}|=AB\sin \theta =(6)(5)\sin 30^{\circ }=30\times \frac{1}{2}=15$ That $15$ is the area of the slanted patch the two arrows span.

Read the components: $A_x=1,A_y=2,A_z=3$ and $B_x=4,B_y=5,B_z=6$.

• $\hat{i}$: $A_y{B}_z-A_z{B}_y=(2)(6)-(3)(5)=12-15=-3$
• $\hat{j}$: $-(A_x{B}_z-A_z{B}_x)=-[(1)(6)-(3)(4)]=-(6-12)=+6$. The middle term always carries a minus — that's the $+,-,+$ cofactor pattern.
• $\hat{k}$: $A_x{B}_y-A_y{B}_x=(1)(5)-(2)(4)=5-8=-3$ $\vec{A}\times \vec{B}=-3\hat{i}+6\hat{j}-3\hat{k}$ Quick sanity check: this result must be perpendicular to both. Dot with $\vec{A}$: $(-3)(1)+(6)(2)+(-3)(3)=-3+12-9=0$. ✓

$\vec{r}$ points along $x$, $\vec{F}$ along $y$. Only the $\hat{k}$ term survives because $\hat{i}\times \hat{j}=\hat{k}$. $\vec{\tau }=(r_y{F}_z-r_z{F}_y)\hat{i}+(r_z{F}_x-r_x{F}_z)\hat{j}+(r_x{F}_y-r_y{F}_x)\hat{k}$ $=(0)\hat{i}+(0)\hat{j}+(3\cdot 4-0)\hat{k}=12\hat{k}\text{N⋅m}$. Magnitude $=12$ N·m, pointing $+\hat{k}$ (out of the page). By the right-hand rule that means counter-clockwise turning. See Torque and rotational equilibrium.

WHAT: build two edge vectors from the same vertex $P$. WHY: the cross product needs both arrows sharing a tail (look at the figure — both green edges start at $P$). $\overrightarrow{PQ}=Q-P=(3,0,0),\overrightarrow{PR}=R-P=(0,4,0)$ Both lie in the $z=0$ plane, so only the $\hat{k}$ component can survive: $\overrightarrow{PQ}\times \overrightarrow{PR}=(0)\hat{i}+(0)\hat{j}+(3\cdot 4-0\cdot 0)\hat{k}=12\hat{k}$ Area of parallelogram $=|12\hat{k}|=12$. Triangle is half of it: $\text{Area}=\frac{1}{2}(12)=6\text{ square units}$ Check by base × height: $PQ$ lies along $x$ with length $3$ (the base); $R$ sits at height $4$ above that base line. Triangle area $=\frac{1}{2}(3)(4)=6$. ✓

WHY the cross product for an angle? Because $|\vec{A}\times \vec{B}|=AB\sin \theta$, so once we know the three lengths we can solve for $\sin \theta$. Cross product (only $\hat{k}$ survives in a plane): $A_x{B}_y-A_y{B}_x=(2)(3)-(1)(1)=5$, so $|\vec{A}\times \vec{B}|=5$. Lengths: $A=\sqrt{2^2+1^2}=\sqrt{5}$, $B=\sqrt{1^2+3^2}=\sqrt{10}$. $\sin \theta =\frac{|\vec{A}\times \vec{B}|}{AB}=\frac{5}{\sqrt{5}\sqrt{10}}=\frac{5}{\sqrt{50}}=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}$ So $\theta =45^{\circ }$. Note: $\sin$ alone can't distinguish $45^{\circ }$ from $135^{\circ }$. Here a quick dot-product check ($\vec{A}\cdot \vec{B}=2+3=5>0$) confirms the angle is acute, so $45^{\circ }$ is correct. See Dot product — formula, projection, work calculation.

$\vec{r}$ along $y$, $\vec{p}$ along $z$. Cross terms:

• $\hat{i}$: $r_y{p}_z-r_z{p}_y=(2)(3)-(0)(0)=6$
• $\hat{j}$: $-(r_x{p}_z-r_z{p}_x)=-(0-0)=0$
• $\hat{k}$: $r_x{p}_y-r_y{p}_x=0-0=0$ $\vec{L}=6\hat{i}{\text{kg⋅m}}^2/\text{s}$ Consistent with $\hat{j}\times \hat{k}=\hat{i}$. See Angular momentum — L = r × p.

First the cross product $\vec{v}\times \vec{B}$: $\vec{v}$ along $x$, $\vec{B}$ along $z$.

• $\hat{i}$: $v_y{B}_z-v_z{B}_y=0-0=0$
• $\hat{j}$: $-(v_x{B}_z-v_z{B}_x)=-[(1)(2)-0]=-2$
• $\hat{k}$: $v_x{B}_y-v_y{B}_x=0-0=0$ So $\vec{v}\times \vec{B}=-2\hat{j}$. Multiply by $q$: $\vec{F}=q(\vec{v}\times \vec{B})=2(-2\hat{j})=-4\hat{j}\text{N}$ The force is perpendicular to both motion and field — that's why magnetic forces curve paths rather than speed things up. See Magnetic force — F = qv × B.

WHAT: set $\vec{B}=\vec{A}$ in the component formula, so $B_x=A_x$, $B_y=A_y$, $B_z=A_z$.

• $\hat{i}$: $A_y{A}_z-A_z{A}_y=0$
• $\hat{j}$: $A_z{A}_x-A_x{A}_z=0$
• $\hat{k}$: $A_x{A}_y-A_y{A}_x=0$ Every component is a quantity minus itself. So $\vec{A}\times \vec{A}=\vec{0}$ for all $\vec{A}$. WHY it must be so: a vector makes angle $0$ with itself, $\sin 0=0$ — the two pictures agree.

Both vectors lie in the plane, so only the $\hat{k}$ component of the cross product is nonzero: $\vec{A}\times \vec{B}=(A_x{B}_y-A_y{B}_x)\hat{k}=(a\cdot b\sin \varphi -0\cdot b\cos \varphi )\hat{k}=ab\sin \varphi \hat{k}$ Its magnitude is $S=|ab\sin \varphi |=ab\sin \varphi$ (for $0\le \varphi \le 180^{\circ }$, $\sin \varphi \ge 0$). This is exactly tool 1: the second vector's height above the base $\vec{A}$ is $b\sin \varphi$ (look at the pink dashed height in the figure), base is $a$, so area $=a\cdot b\sin \varphi$. The determinant and the geometry are the same statement. For $a=4,b=3,\varphi =90^{\circ }$: $S=(4)(3)\sin 90^{\circ }=12$ square units.

$\vec{A}\times \vec{B}$ returns a vector or a scalar? ::: A vector (magnitude $AB\sin \theta$, direction by right-hand rule). If both vectors lie in the $xy$-plane, which component survives? ::: Only the $\hat{k}$ component, equal to $A_x{B}_y-A_y{B}_x$. Two triangle edges must share what? ::: A common tail (start from the same vertex).