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{\bf Problem 1.}

$$a=\pmatrix{1\cr-2\cr-1\cr},\quad
b=\pmatrix{2\cr1\cr-2}
$$

$$\langle a,b\rangle=a^\tau b=(1)(2)+(-2)(1)+(-1)(-2)=2-2+2=2$$

$$\langle b,b\rangle=b^\tau b=(2)(2)+(1)(1)+(-2)(-2)=9$$

$$|b|=\sqrt{\langle b,b\rangle}=3$$

$$3a=\pmatrix{3\cr-6\cr-3},\quad a+b=\pmatrix{3\cr-1\cr-3}$$

$$\langle3a,a+b\rangle=(3)(3)+(-6)(-1)+(-3)(-3)=9+6+9=24$$

$$\langle a,a\rangle=(1)(1)+(-2)(-2)+(-1)(-1)=6$$

$$|a|=\sqrt{\langle a,a\rangle}=\sqrt6$$

$$\theta=\cos^{-1}{\langle a,b\rangle\over|a||b|}=\cos^{-1}{2\over3\sqrt6}=
\cos^{-1}{\sqrt2\over3\sqrt3}$$

\vfill\eject
{\bf Problem 2.} The expression $x_1y_1+x_3y_3$ is not an inner product
because for some $a\ne0$ we obtain the result $\langle a,a\rangle=0$.
Specifically, for an $a$ in the set $\{(0,t,0):t\in{\cal R},t\ne0\}$
we obtain
$\langle a,a\rangle=0$.

\vfill\eject

{\bf Problem 3.}

$$
b_1=\pmatrix{0\cr0\cr1\cr0},\quad
b_2=\pmatrix{1\cr1\cr2\cr1},\quad
b_3=\pmatrix{1\cr0\cr1\cr1}
$$

$$|b_1|=1,\quad\hat b_1=|b_1|^{-1}b_1=\pmatrix{0\cr0\cr1\cr0}$$

$$\langle b_2,\hat b_1\rangle=\pmatrix{1&1&2&1}\pmatrix{0\cr0\cr1\cr0}=2$$

$$b_2^\prime=b_2-\langle b_2,\hat b_1\rangle\hat b_1=
\pmatrix{1\cr1\cr2\cr1}-2\pmatrix{0\cr0\cr1\cr0}=\pmatrix{1\cr1\cr0\cr1}$$

$$\hat b_2=|b_2^\prime|^{-1}b_2^\prime={1\over\sqrt3}\pmatrix{1\cr1\cr0\cr1}$$

$$\langle b_3,\hat b_1\rangle=\pmatrix{1&0&1&1}\pmatrix{0\cr0\cr1\cr0}=1$$

$$\langle b_3,\hat b_2\rangle={1\over\sqrt3}\pmatrix{1&0&1&1}\pmatrix{1\cr1\cr0\cr1}
={2\over\sqrt3}$$

$$b_3^\prime=b_3-\langle b_3,\hat b_1\rangle\hat b_1-\langle b_3,\hat b_2\rangle\hat b_2
=\pmatrix{1\cr0\cr1\cr1}-\pmatrix{0\cr0\cr1\cr0}-{2\over3}\pmatrix{1\cr1\cr0\cr1}
=\pmatrix{1/3\cr-2/3\cr0\cr1/3}$$

$$|b_3^\prime|=\sqrt{(1/3)^2+(-2/3)^2+(1/3)^2}=\sqrt{6/9}=\sqrt{2/3}$$

$$\hat b_3=|b_3^\prime|^{-1}b_3^\prime=\sqrt{3/2}\pmatrix{1/3\cr-2/3\cr0\cr1/3}
={1\over\sqrt6}\pmatrix{1\cr-2\cr0\cr1}$$

Answer...

$$
\hat b_1=\pmatrix{0\cr0\cr1\cr0},\quad
\hat b_2={1\over\sqrt3}\pmatrix{1\cr1\cr0\cr1},\quad
\hat b_3={1\over\sqrt6}\pmatrix{1\cr-2\cr0\cr1}
$$

\end