91Ó°ÊÓ

Find the spherical coordinates of the Cartesian point \((\sqrt{2},-\sqrt{6},-2 \sqrt{2}).\)

Calculate the dot product of \(\mathbf{x}=(1,-1,0,2) \in \mathbb{R}^{4}\) and \(\mathbf{y}=(1,2,3,4) \in \mathbb{R}^{4}\).

Complete the computations. $$(-21,23)-(?, 6)=(-25, ?)$$

Verify that interchanging the first two rows of the \(3 \times 3\) determinant $$ \left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 0 & 1 \\ 2 & 0 & 2 \end{array}\right| $$ changes the sign of the determinant.

$$\text { Calculate }(3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}) \cdot(\mathbf{i}+2 \mathbf{j}-\mathbf{k})$$

Find the spherical coordinates of the Cartesian point \((\sqrt{6},-\sqrt{2},-2 \sqrt{2}).\)

In \(\mathrm{R}^{n}\) show that (a) \(2\|\mathbf{x}\|^{2}+2\|\mathbf{y}\|^{2}=\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\) (This is known as the parallelogram law.) (b) \(\|\mathbf{x}-\mathbf{y}\|\|\mathbf{x}+\mathbf{y}\| \leq\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}\) (c) \(4\langle\mathbf{x}, \mathbf{y}\rangle=\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\) (This is called the polarization identity.)

Evaluate the determinants (a) \(\left|\begin{array}{rrr}2 & -1 & 0 \\ 4 & 3 & 2 \\ 3 & 0 & 1\end{array}\right|\) (b) \(\left|\begin{array}{ccc}36 & 18 & 17 \\ 45 & 24 & 20 \\ 3 & 5 & -2\end{array}\right|\) (c) \(\left[\begin{array}{ccc}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{array}\right]\) (d) \(\left|\begin{array}{rrr}2 & 3 & 5 \\ 7 & 11 & 13 \\ 17 & 19 & 23\end{array}\right|\)

Complete the computations. $$3(133,-0.33,0)+(-399,0.99,0)=(?, ?, ?)$$

$$\begin{aligned} &\text { Calculate a } \cdot \mathbf{b}, \text { where } \mathbf{a}=2 \mathbf{i}+10 \mathbf{j}-12 \mathbf{k} \text { and }\\\ &\mathbf{b}=-3 \mathbf{i}+4 \mathbf{k} \end{aligned}$$