AHL 3.16—Vector product

For $\mathbf{u}=(u_1,u_2,u_3)$ and $\mathbf{v}=(v_1,v_2,v_3)$, the cross product is $\mathbf{u}\times \mathbf{v}=(u_2{v}_3-u_3{v}_2,u_3{v}_1-u_1{v}_3,u_1{v}_2-u_2{v}_1)$. Here $\mathbf{u}=(2,-1,3)$ and $\mathbf{v}=(-1,4,2)$.

$i\text{-component}=u_2{v}_3-u_3{v}_2=(-1)(2)-(3)(4)=-2-12=-14$

$j\text{-component}=u_3{v}_1-u_1{v}_3=(3)(-1)-(2)(2)=-3-4=-7$

$k\text{-component}=u_1{v}_2-u_2{v}_1=(2)(4)-(-1)(-1)=8-1=7$

Therefore $\mathbf{u}\times \mathbf{v}=\left(\begin{array}{c}-14\\ -7\\ 7\end{array}\right)$. We can quickly verify: $\mathbf{u}\cdot (\mathbf{u}\times \mathbf{v})=(2)(-14)+(-1)(-7)+(3)(7)=-28+7+21=0$ ✓

All four options differ in signs. The $k$-component is $u_1{v}_2-u_2{v}_1=(2)(4)-(-1)(-1)=8-1=7$, which is positive. This eliminates any option with a negative $k$-component.

The option $\left(\begin{array}{c}-14\\ -7\\ -7\end{array}\right)$ has $k=-7$, which contradicts our calculation. Eliminate this option.

The $i$-component is $(-1)(2)-(3)(4)=-2-12=-14$, which is negative. The option $\left(\begin{array}{c}14\\ -7\\ 7\end{array}\right)$ has $i=+14$. Eliminate this option.

The $j$-component is $(3)(-1)-(2)(2)=-3-4=-7$. The option $\left(\begin{array}{c}-14\\ 7\\ 7\end{array}\right)$ has $j=+7$. Eliminate this option.

The only remaining option is $\left(\begin{array}{c}-14\\ -7\\ 7\end{array}\right)$, which matches all three computed components.

By definition, $\mathbf{p}\times \mathbf{q}=|\mathbf{p}||\mathbf{q}|\sin \theta \mathbf{n}$, where $\mathbf{n}$ is the unit vector perpendicular to both $\mathbf{p}$ and $\mathbf{q}$. This is the defining geometric property of the cross product.

Computing: $\mathbf{p}\times \mathbf{q}=\left(\begin{array}{c}(-2)(-1)-(3)(1)\\ (3)(4)-(1)(-1)\\ (1)(1)-(-2)(4)\end{array}\right)=\left(\begin{array}{c}-1\\ 13\\ 9\end{array}\right)$. Check: $\mathbf{p}\cdot (\mathbf{p}\times \mathbf{q})=(1)(-1)+(-2)(13)+(3)(9)=-1-26+27=0$ ✓

The cross product is anti-commutative: $\mathbf{p}\times \mathbf{q}=-(\mathbf{q}\times \mathbf{p})$, so the claim $\mathbf{p}\times \mathbf{q}=\mathbf{q}\times \mathbf{p}$ is false unless both equal the zero vector.

The cross product always returns a vector (not a scalar) that is perpendicular to both input vectors. The correct statement is that $\mathbf{p}\times \mathbf{q}$ is perpendicular to both $\mathbf{p}$ and $\mathbf{q}$.

This tests knowledge of fundamental properties of the cross product: its type (vector vs scalar), its direction, and the commutativity rule.

The dot product returns a scalar; the cross product returns a vector. The statement '$\mathbf{p}\times \mathbf{q}$ is a scalar' is definitionally wrong. Eliminate.

The cross product is anti-commutative: $\mathbf{p}\times \mathbf{q}=-(\mathbf{q}\times \mathbf{p})$. Equality only holds if both equal $0$, which is not the case here since $\mathbf{p}$ and $\mathbf{q}$ are not parallel. Eliminate.

If $\mathbf{p}\times \mathbf{q}$ were parallel to $\mathbf{p}$, then $\mathbf{p}\cdot (\mathbf{p}\times \mathbf{q})\ne 0$ in general. In fact $\mathbf{p}\cdot (\mathbf{p}\times \mathbf{q})=0$ always, meaning the result is perpendicular to $\mathbf{p}$, not parallel. Eliminate.

The only valid statement is that $\mathbf{p}\times \mathbf{q}$ is perpendicular to both $\mathbf{p}$ and $\mathbf{q}$, which is the defining geometric property of the cross product.