Let $\mathbf { v } = 7 \mathbf { j }$ and let $\mathbf { u }$ be a vector with length 5 that starts at the origin and rotates in the $x y$ - plane. Find the maximum value of the length of the vector $| \mathbf { u } \times \mathbf { v } |$.
A) $| \mathbf { u } \times \mathbf { v } | = 5$
B) $| \mathbf { u } \times \mathbf { v } | = 7$
C) $| \mathbf { u } \times \mathbf { v } | = 35$
D) $| \mathbf { u } \times \mathbf { v } | = 157$
E) $| \mathbf { u } \times \mathbf { v } | = 1$

Question

Quizplus · Accepted Answer

Let $\mathbf{u}=\begin{pmatrix}a\b\0\end{pmatrix}$. Then $\mathbf{u}	imes\mathbf{v}=\begin{pmatrix}0\0\7b\end{pmatrix}$, so $\left|\mathbf{u}	imes\mathbf{v}ight|=7|b|$. Clearly, $\left|\mathbf{u}	imes\mathbf{v}ight|$ is maximized when $b$ reaches the largest possible value. Since $\mathbf{u}$ has length $5$, it lies on the circle with radius $5$ centered at the origin, so $a^2+b^2=25$. We want to maximize $|b|$, which is the distance from the $x$-axis. Since this distance is maximized when $\mathbf{u}$ is tangent to the circle, that means $a=\pm 2\sqrt{6}$ and $b=\pm 1$. Plugging in $\mathbf{u}=\begin{pmatrix}2\sqrt{6}\1\0\end{pmatrix}$ or $\mathbf{u}=\begin{pmatrix}-2\sqrt{6}\-1\0\end{pmatrix}$ results in $\boxed{	extbf{(C) } 35}$ as the maximum value.

Let v=7j\mathbf { v } = 7 \mathbf { j }v=7j

Let $\mathbf { v } = 7 \mathbf { j }$