Dot Cross Maximum Value
Let $ \mathbf{b} = \begin{pmatrix} -1 \\ 4 \\ 6 \end{pmatrix} $ and $ \mathbf{c} = \begin{pmatrix} 2 \\ -7 \\ -10 \end{pmatrix} $. Find the maximum value of
\[\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}),\]where $ \mathbf{a} $ is a unit vector.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
By the scalar triple product,
\begin{align*}
\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) &= \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \\
&= \mathbf{a} \cdot \left( \begin{pmatrix} -1 \\ 4 \\ 6 \end{pmatrix} \times \begin{pmatrix} 2 \\ -7 \\ -10 \end{pmatrix} \right) \\
&= \mathbf{a} \cdot \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}.\end{align*}Note that
\[\left| \mathbf{a} \cdot \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} \right| \le \|\mathbf{a}\| \left\| \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} \right\| \le 3.\]Equality occurs when $ \mathbf{a} $ is the unit vector pointing in the direction of $ \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}, $ which is $ \begin{pmatrix} 2/3 \\ 2/3 \\ -1/3 \end{pmatrix}, $ so the maximum value we seek is $ \boxed{3} $.
Freeflows