Vector Cross Product 1
Let $ \mathbf{u} $ and $ \mathbf{v} $ be unit vectors, and let $ \mathbf{w} $ be a vector such that
\[\mathbf{w} + \mathbf{w} \times \mathbf{u} = \mathbf{v}.\]Find the largest possible value of $ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} $.
- 1
- 2
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- +
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- -
- 7
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- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
From $ \mathbf{w} + \mathbf{w} \times \mathbf{u} = \mathbf{v}, $
\[\mathbf{w} \times \mathbf{u} = \mathbf{v} - \mathbf{w}.\]Then
\begin{align*}
\|\mathbf{w} \times \mathbf{u}\|^2 &= \|\mathbf{v} - \mathbf{w}\|^2 \\
&= \|\mathbf{v}\|^2 - 2 \mathbf{v} \cdot \mathbf{w} + \|\mathbf{w}\|^2 \\
&= 1 - 2 \mathbf{v} \cdot \mathbf{w} + \|\mathbf{w}\|^2
\end{align*}Hence,
\[\mathbf{v} \cdot \mathbf{w} = \frac{1 +\|\mathbf{w}\|^2 - \|\mathbf{w} \times \mathbf{u}\|^2}{2}.\quad (*)\]Also from $ \mathbf{w} + \mathbf{w} \times \mathbf{u} = \mathbf{v}, $ we can take the dot product with $ \mathbf{v}, $ to get
\[\mathbf{w} \cdot \mathbf{v} + (\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{v} = 1.\]By the scalar triple product, $ (\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v} = (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}, $ so
\[(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = 1 - \mathbf{v} \cdot \mathbf{w}.\]From equation $ (*), $
\begin{align*}
(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} &= 1 - \frac{1 +\|\mathbf{w}\|^2 - \|\mathbf{w} \times \mathbf{u}\|^2}{2} \\
&= \frac{1}{2} - \frac{1}{2} \|\mathbf{w}\|^2 + \frac{1}{2} \|\mathbf{w} \times \mathbf{u}\|^2
\end{align*}Let $ \theta $ be the angle between $ \mathbf{u} $ and $ \mathbf{w} $. Then
\begin{align*}
(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} &= \frac{1}{2} - \frac{1}{2} \|\mathbf{w}\|^2 + \frac{1}{2} \|\mathbf{w} \times \mathbf{u}\|^2 \\
&= \frac{1}{2} - \frac{1}{2} \|\mathbf{w}\|^2 + \frac{1}{2} \|\mathbf{u}\|^2 \|\mathbf{w}\|^2 \sin^2 \theta \\
&= \frac{1}{2} - \frac{1}{2} \|\mathbf{w}\|^2 + \frac{1}{2} \|\mathbf{w}\|^2 \sin^2 \theta \\
&= \frac{1}{2} - \frac{1}{2} \|\mathbf{w}\|^2 \cos^2 \theta \\
&\le \frac{1}{2}
\end{align*}Equality occurs when $ \mathbf{u} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, $ $ \mathbf{v} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, $ and $ \mathbf{w} = \begin{pmatrix} 0 \\ 1/2 \\ 1/2 \end{pmatrix}, $ so the largest possible value of $ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} $ is $ \boxed{\frac{1}{2}} $.
Freeflows