Inner Product and Orthogonality
- Given \( \overrightarrow{w}=\left[\begin{array}{r}3 \\ -1 \\
-5\end{array}\right], \: \overrightarrow{x}=\left[\begin{array}{r}6 \\ -2 \\
3\end{array}\right]\). Compute: (a) \(\overrightarrow{w} \cdot
\overrightarrow{w}, \quad \overrightarrow{x} \cdot \overrightarrow{w}, \: \)
and \(\:\dfrac{\overrightarrow{x} \cdot
\overrightarrow{w}}{\overrightarrow{w} \cdot \overrightarrow{w}}\quad \) (b)
\(\left(\dfrac{\overrightarrow{x} \cdot
\overrightarrow{w}}{\overrightarrow{x} \cdot \overrightarrow{x}}\right)
\overrightarrow{x} \quad \) (c) \(\|\overrightarrow{x}\|\) solution
- Find a unit vector in the direction of the given vector. \(\quad \overrightarrow{u}=\left[\begin{array}{r} -6 \\ 4 \\ -3 \end{array}\right] \). solution
- Show that \(\left\{\overrightarrow{u}_{1}, \overrightarrow{u}_{2}\right\}\) is an orthogonal basis for \(\mathbb{R}^{2}\). Then express \(\overrightarrow{x}\) as a linear combination of the \(\overrightarrow{u}\)'s.
\( \quad \overrightarrow{u}_{1}=\left[\begin{array}{r}3 \\ 1\end{array}\right], \overrightarrow{u}_{2}=\left[\begin{array}{r}-2 \\ 6\end{array}\right],\) and \(\overrightarrow{x}=\left[\begin{array}{r} -6\\ 3\end{array}\right].\) solution
- Compute the orthogonal projection of \(\left[\begin{array}{l}1 \\ 7\end{array}\right]\) onto the line through \(\left[\begin{array}{r}-4 \\ 2\end{array}\right]\) and the origin. solution
- Let \(W\) be a subspace of \(\mathbb{R}^{n},\) and let \(W^{\perp}\) be the set of all vectors orthogonal to \(W\). Verify the three conditions to show that \(W^{\perp}\) is a subspace of \(\mathbb{R}^{n}\). solution