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Let \(\vec u\) and \(\vec v\) be two vectors in \(\mathbb R^3\text{.}\) Then, \(\vec u \times \text{proj}_{\vec v}\vec u\) is perpendicular to \(\vec u - \text{proj}_{\vec v}\vec u\text{.}\)
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Let \(\vec u\) and \(\vec v\) be two standard unit vectors in \(\mathbb R^3\text{.}\) Then, \(\left(\vec u \cdot \vec v\right)^2 + \left|\vec v \times \vec u\right|^2 = 1\text{.}\)
Solution.
Statement I is true. Note that \(\vec u \times \text{proj}_{\vec v}\vec u\) is perpendicular to both \(\vec u\) and \(\text{proj}_{\vec v}\vec u\text{,}\) which gives:
\begin{equation*}
\begin{aligned}
\left(\vec u \times \text{proj}_{\vec v}\vec u\right) \cdot
\left(\vec u - \text{proj}_{\vec v}\vec u\right)
\amp =
\left(\vec u \times \text{proj}_{\vec v}\vec u\right) \cdot \vec u
- \left(\vec u \times \text{proj}_{\vec v}\vec u\right) \cdot \text{proj}_{\vec v}\vec u\\
= 0 - 0 = 0,
\end{aligned}
\end{equation*}
and therefore \(\vec u \times \text{proj}_{\vec v}\vec u\) is perpendicular to \(\vec u - \text{proj}_{\vec v}\vec u\text{.}\)
Statement II is true. Note that:
\begin{align*}
\left(\vec u \cdot \vec v\right)^2 + \left|\vec v \times \vec u\right|^2
\amp= |\vec u|^2|\vec v|^2\cos^2\theta + |\vec u|^2|\vec v|^2\sin^2\theta\\
\amp= |\vec u|^2|\vec v|^2\left(\cos^2\theta + \sin^2\theta\right)
= |\vec u|^2|\vec v|^2.
\end{align*}
Now, since \(\vec u\) and \(\vec v\) are standard unit vectors, we have \(|\vec u| = |\vec v| = 1\text{,}\) which means:
\begin{equation*}
\left(\vec u \cdot \vec v\right)^2 + \left|\vec v \times \vec u\right|^2
= |\vec u|^2|\vec v|^2 = 1.
\end{equation*}
