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Erschienen in:
2014 | OriginalPaper | Buchkapitel
11. Solutions
verfasst von : João Pedro Morais, Svetlin Georgiev, Wolfgang Sprößig
Erschienen in: Real Quaternionic Calculus Handbook
Verlag: Springer Basel
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Excerpt
1.
(a) p + q = 3 + i + 5i + j − 2k = 3 + 6i + j − 2k; (b) 3 − 4i − j + 2k; (c) 6 − 33i − 7j + 14k; (d) 9 + 23i + 4j − 8k.
2.
(a) − 1; (b) 1; (c) 1; (d) − 1; (e) − 1; (f) 1.
3.
(b) 3 + 3i − 3j + k; (c) − 1 + 4i + 7j + 12k; (d) 14 − 2i − 8j + 24k; (e) − 2 + 2i − 2j − 2k; (f) − 162 + 12i + 6j + 24k; (g) 14 − 22i + 24j + 68k; (h) − 12 + 24i − 26j − 70k; (i) 1 + 4i − j + 2k; (j) − 147 − 9i + 29j + 91k.
$$\displaystyle\begin{array}{rcl} (a)\quad pq& =& (1 + i - j - k)(3 + 2i + j + 4k) {}\\ & =& 3 + 2i + j + 4k + 3i - 2 + k - 4j - 3j + 2k + 1 {}\\ & -& 4i - 3k - 2j + i + 4 = 6 + 2i - 8j + 4k; {}\\ \end{array}$$
4.
(a) First we consider \(p(1 + i - j) = 1_{\mathbb{H}}\). We have
From here we obtain the following system for a, b, c and d:
$$\displaystyle\begin{array}{rcl} 1_{\mathbb{H}}& =& (a + bi + cj + dk)(1 + i - j) {}\\ & =& (a - b + c) + (a + b + d)i + (-a + c + d)j + (-b - c + d)k. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left \{\begin{array}{l} a - b + c = 1, \\ a + b + d = 0, \\ - a + c + d = 0, \\ - b - c + d = 0.\end{array} \right.& & {}\\ \end{array}$$
Its solution is
Consequently,
It is easy to check that
Now we consider
We have
From the last equality we get for a, b, c, d the system
Its solution is
Therefore,
It is easy to check that
(b) p = −i; (c) p = −j; (d) p = −k.
$$\displaystyle\begin{array}{rcl} a = \frac{1} {3},\;b = -\frac{1} {3},\;c = \frac{1} {3},\;d = 0.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p = \frac{1} {3} -\frac{1} {3}i + \frac{1} {3}j.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left (\frac{1} {3} -\frac{1} {3}i + \frac{1} {3}j\right )(1 + i - j) = 1_{\mathbb{H}}.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} (1 + i - j)p = 1_{\mathbb{H}}.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} 1_{\mathbb{H}}& =& (1 + i - j)(a + bi + cj + dk) {}\\ & =& a - b + c + (b + a - d)i + (c - a - d)j + (d + c + b)k. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left \{\begin{array}{l} a - b + c = 1, \\ a + b - d = 0, \\ c - a - d = 0, \\ d + c + b = 0.\end{array} \right.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} a = \frac{1} {3},\;b = -\frac{1} {3},\;c = \frac{1} {3},\;d = 0.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p = \frac{1} {3} -\frac{1} {3}\ i + \frac{1} {3}\ j.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} (1 + i - j)\left (\frac{1} {3} -\frac{1} {3}\ i + \frac{1} {3}\ j\right ) = 1_{\mathbb{H}}.& & {}\\ \end{array}$$
5.
(a), (b)
$$\displaystyle\begin{array}{rcl} p& =& \frac{a} {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}} - \frac{b} {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}i {}\\ & -& \frac{c} {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}j - \frac{d} {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}k. {}\\ \end{array}$$
6.
(a) We have
which implies
From here we obtain for a, b, c, d the following system
Consequently,
It is easy to check that p satisfies the given equation. (b) \(p = \frac{1\pm \sqrt{5}} {2}\); (c) \(p = a + \frac{1} {2a-1}i\), where a is a solution to the equation
(such solution exists: prove!), \(a\neq \frac{1} {2}\); (d) \(p = a + \frac{1} {2a-1}j\); (e) \(p = a + \frac{1} {2a-1}k\), where a is a solution to the Eq. (11.1); (f) \(p = \frac{1} {2} + \frac{1} {2}i + k\); (g) \(p = \frac{1} {2} + \frac{1} {2}i + j\); (h) \(p = -\frac{3} {7} + \frac{5} {7}i -\frac{5} {7}j + \frac{5} {7}k\); (i) p = bi + cj + dk so that \({b}^{2} + {c}^{2} + {d}^{2} = 1\); (j) \(p = \pm \frac{\sqrt{3}} {2} -\frac{1} {2}i\); (k) \(p = \pm \frac{1} {2} -\frac{1} {2}i \pm \frac{1} {2}j -\frac{1} {2}k\); (l) \(p = \pm \frac{1} {2} -\frac{1} {2}i + \frac{1} {2}j \pm \frac{1} {2}k\).
$$\displaystyle\begin{array}{rcl} 2p = 1 - i + j + 3k,& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} 2a + 2bi + 2cj + 2dk = 1 - i + j + 3k.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left \{\begin{array}{l} 2a = 1,\;2b = -1,\;2c = 1,\;2d = 3.\end{array} \right.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p = \frac{1} {2} -\frac{1} {2}\ i + \frac{1} {2}\ j + \frac{3} {2}\ k.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} 4{a}^{4} - 8{a}^{3} + 5{a}^{2} - a - 1 = 0& & {}\end{array}$$
(11.1)
7.
(a) 2 + i − j; (b) 1 − i − j − k; (c) 2 − 2i − 4j; (d) 2 − 2j − 4k; (e) 2 − 2i − 4j; (f) 2 − 2j − 4k.
8.
$$\displaystyle\begin{array}{rcl} \mathrm{(a)}& & \left (a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2}\right ) -\left (a_{1}b_{2} + b_{1}a_{2} + c_{1}d_{2} - d_{1}c_{2}\right )i {}\\ & -&\left (a_{1}c_{2} - b_{1}d_{2} + c_{1}a_{2} + d_{1}b_{2}\right )j -\left (a_{1}d_{2} + b_{1}c_{2} - c_{1}b_{2} + d_{1}a_{2}\right )k; {}\\ \mathrm{(b)}& & \left (a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2}\right ) + \left (-a_{1}b_{2} - b_{1}a_{2} + c_{1}d_{2} - d_{1}c_{2}\right )i {}\\ & +& \left (-a_{1}c_{2} - b_{1}d_{2} - c_{1}a_{2} + d_{1}b_{2}\right )j + \left (-a_{1}d_{2} + b_{1}c_{2} - c_{1}b_{2} - d_{1}a_{2}\right )k; {}\\ \mathrm{(c)}& & \left (a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2}\right ) -\left (a_{2}b_{1} + a_{1}b_{2} + c_{2}d_{1} - d_{2}c_{1}\right )i {}\\ & -&\left (a_{2}c_{1} - b_{2}d_{1} + c_{2}a_{1} + d_{2}b_{1}\right )j -\left (a_{2}d_{1} + b_{2}c_{1} - c_{2}b_{1} + d_{2}a_{1}\right )k; {}\\ \mathrm{(d)}& & \left (a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2}d_{1}d_{2}\right ) -\left (a_{2}b_{1} + b_{2}a_{1} + c_{2}d_{1} - d_{2}c_{1}\right )i {}\\ & -&\left (a_{2}c_{1} - b_{2}d_{1} + c_{2}a_{1} + d_{2}b_{1}\right )j -\left (a_{2}d_{1} + b_{2}c_{1} - c_{2}b_{1} + d_{2}a_{1}\right )k. {}\\ \end{array}$$
9.
Hint. Use the previous exercise.
10.
We have that \(p{p}^{-1} = 1_{\mathbb{H}}\) and \(\vert p{\vert }^{2} = p\overline{p}\). Consequently,
$$\displaystyle\begin{array}{rcl} \frac{\overline{p}} {\vert p{\vert }^{2}} = \frac{\overline{p}} {p\overline{p}} = {p}^{-1}.& & {}\\ \end{array}$$
11.
(a) Here
Then the first matrix representation is
$$\displaystyle\begin{array}{rcl} a = 1,\quad b = -2,\quad c = 3,\quad d = 4.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left (\begin{array}{rr} 1 - 4i&2 + 3i\\ - 2 + 3i &1 + 4i\end{array} \right );& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \mathrm{(b)}\;\left (\begin{array}{cc} 2 &1 + i\\ - 1 + i & 2\end{array} \right );\qquad \mathrm{(c)}\;\left (\begin{array}{cc} \frac{1} {2} - i & -\frac{1} {3} - 2i \\ \frac{1} {3} - 2i& \frac{1} {2} + i\end{array} \right ).& & {}\\ \end{array}$$
12.
(a) Let p = a + bi + cj + dk \((a,b,c,d \in \mathbb{R})\) be a quaternion that corresponds to the given matrix. Then
therefore we obtain the system
Consequently, a = 1, b = −2, c = 4, and d = 1. That is,
(b) p = 1 + 2i + j; (c) \(p = 1 - 2i + 8j -\frac{1} {2}k\).
$$\displaystyle\begin{array}{rcl} \left (\begin{array}{cl} \ 1 - i &2 + 4i\\ - 2 + 4i &1 + i\end{array} \right )\; =\; \left (\begin{array}{rr} a - di& - b + ci\\ b + ci & a + di\end{array} \right ),& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left \{\begin{array}{rcl} a - di& =&1 - i,\\ - b + ci & = &2 + 4i.\end{array} \right.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p = 1 - 2i + 4j + k;& & {}\\ \end{array}$$
13.
(a) Let us suppose that the given matrix corresponds to a quaternion
Then
whence
which is a contradiction. Therefore the given matrix does not correspond to any quaternion; (b) no solution; (c) no solution; (d) p = 1 − 2i − j + 2k.
$$\displaystyle\begin{array}{rcl} p = a + bi + cj + dk,\quad a,b,c,d \in \mathbb{R}.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left (\begin{array}{cc} 1 + i&1 - i\\ 1 + i &2 + i\end{array} \right )\; =\; \left (\begin{array}{cc} a - di& - b + ci\\ b + ci & a + di\end{array} \right ),& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left \{\begin{array}{rcl} a - di& =&1 + i, \\ - b + ci& =&1 - i, \\ b + ci& =&1 + i, \\ a + di& =&2 + i,\end{array} \right.& & {}\\ \end{array}$$
14.
We have that
Also
from here \(\vert z\vert = \sqrt{\det A}\).
$$\displaystyle\begin{array}{rcl} \vert z\vert = \sqrt{{a}^{2 } + {b}^{2 } + {c}^{2 } + {d}^{2}}.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \det A = \left \vert \begin{array}{rr} a - di& - b + ci\\ b + ci & a + di\end{array} \right \vert & =& (a - di)(a + di) - (-b + ci)(b + ci) {}\\ & =& {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}, {}\\ \end{array}$$
15.
(a) The first matrix representation of p = 1 − i + 2j − k is
Next,
is the first matrix representation of the quaternion q = −2i + 3j + k. Then
The last matrix corresponds to the quaternion
(b) pq = 3 + 3k.
$$\displaystyle\begin{array}{rcl} A_{1} = \left (\begin{array}{cl} \ 1 + i &1 + 2i\\ - 1 + 2i &1 - i\end{array} \right ).& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} A_{2} = \left (\begin{array}{cc} - i &2 + 3i\\ - 2 + 3i & i\end{array} \right )& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} A_{1}A_{2}& =& \left (\begin{array}{cc} \ 1 + i &\;\ 1 + 2i\\ - 1 + 2i & 1 - i\end{array} \right )\left (\begin{array}{cc} - i &2 + 3i\\ - 2 + 3i & i.\end{array} \right ) {}\\ & =& \left (\begin{array}{rl} - 7 - 2i& - 3 + 6i\\ 3 + 6i & - 7 + 2i\end{array} \right ). {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} pq = -7 + 3i + 6j + 2k;& & {}\\ \end{array}$$
16.
− 3 + 5i − 2j + 3k.
17.
(a) Here
Then the second matrix representation is
$$\displaystyle\begin{array}{rcl} a = 1,\quad b = -2,\quad c = \frac{1} {2},\quad d = 1.& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left (\begin{array}{rrrr} 1& 2& 1& -\frac{1} {2} \\ - 2& 1& -\frac{1} {2} & - 1 \\ - 1&\frac{1} {2} & 1& 2 \\ \frac{1} {2} & 1& - 2& 1\end{array} \right );& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} & & \mathrm{(b)}\;\left (\begin{array}{cccc} 0 & - 1&3& 1\\ 1 & 0 &1 & -3 \\ - 3& - 1&0& - 1\\ - 1 & 3 &1 & 0 \end{array} \right );\qquad \mathrm{(c)}\;\left (\begin{array}{cccc} 2 & - 4& - 1& 1\\ 4 & 2 & 1 & 1 \\ 1 & - 1& 2 & - 4\\ - 1 & - 1 & 4 & 2 \end{array} \right ); {}\\ & & \mathrm{(d)}\;\left (\begin{array}{cccc} 0 & - 1& 1 & - 1\\ 1 & 0 & - 1 & - 1 \\ - 1& 1 & 0 & - 1\\ 1 & 1 & 1 & 0\end{array} \right ). {}\\ \end{array}$$
18.
For p = 1 − i + j − k we have the representation
while for q = 2 − i − j − k we have
$$\displaystyle\begin{array}{rcl} A_{1} = \left (\begin{array}{rrrr} 1&1& 1& - 1\\ - 1 &1 & - 1 & - 1 \\ - 1&1& 1& 1\\ 1 &1 & - 1 & 1 \end{array} \right ),& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} A_{2} = \left (\begin{array}{rrrr} 2& 1& - 1&1\\ - 1 & 2 & 1 &1 \\ 1& - 1& 2&1\\ - 1 & - 1 & - 1 &2 \end{array} \right ).& & {}\\ \end{array}$$
Then
which yields
i.e. pq = 3 − 3i − j + 3k.
$$\displaystyle\begin{array}{rcl} A_{1}A_{2} = \left (\begin{array}{rrrr} 3& 3& 3& 1\\ - 3 & 3 & 1 & - 3 \\ - 3& - 1& 3& 3\\ - 1 & 3 & - 3 & 3 \end{array} \right )\ =\ \left (\begin{array}{rrrr} a& - b& d& - c \\ b& a& - c& - d \\ - d& c& a& - b \\ c& d& b& a\end{array} \right ),& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} a = 3,\quad b = -3,\quad c = -1,\quad d = 3,& & {}\\ \end{array}$$
19.
(a) Let p = a + bi + cj + dk, with \(a,b,c,d \in \mathbb{R}\) be the quaternion that corresponds to the given matrix. For its second matrix representation we have
It follows that
i.e.
(b) p = 7 + i + 9k; (c) p = 1 − i + j + k; (d) p = 2 − 2i + j − k; (e) \(p = 3 -\frac{1} {2}i + j + k\); (f) \(p = 1 -\frac{1} {3}i + \frac{2} {3}j + 4k\).
$$\displaystyle\begin{array}{rcl} \left (\begin{array}{rrrr} a& - b& d& - c \\ b& a& - c& - d \\ - d& c& a& - b \\ c& d& b& a\end{array} \right ).& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} a = \frac{1} {2},\quad b = -2,\quad c = -3,\quad d = -9,& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p = \frac{1} {2} - 2i - 3j - 9k;& & {}\\ \end{array}$$
20.
(a) no solution; (b) no solution; (c) has a solution; (d) no solution; (e) no solution; (f) no solution.
21.
2 − 5i + 5j + 5k.
22.
(a) Here p 0 = 0, p = (1, 0, 0), q 0 = 0, \(\mathbf{q} = (1,0,0)\). Then
(b) 1; (c) 1; (d) 0; (e) 0; (f) 0; (g) 0; (h) 0; (i) 0; (j) \(-\frac{2} {3}\); (k) − 3; (l) 1.
$$\displaystyle\begin{array}{rcl} p \cdot q = p_{0}q_{0} + \mathbf{p}\mathbf{q} = 0 + (1,0,0)(1,0,0) = 1;& & {}\\ \end{array}$$
23.
(a) We have
Therefore
(b) − 18; (c) − 10 − 2j − 2k.
$$\displaystyle\begin{array}{rcl} pq& =& (1 - i - j - k)(2 + i + j) = 4 - 2j - 2k, {}\\ p_{0}& =& 1,\quad q_{0} = 2, {}\\ \mathbf{p}& =& (-1,-1,-1), {}\\ \mathbf{q}& =& (1,1,0), {}\\ p \cdot q& =& p_{0}q_{0} + \mathbf{p}\mathbf{q} = 2 + (-1,-1,-1)(1,1,0) = 0, {}\\ p \cdot q + 2q& =& 4 + 2i + 2j, {}\\ p(p \cdot q + 2q)& =& (1 - i - j - k)(4 + 2i + 2j) = 8 - 4j - 4k, {}\\ p - q& =& -1 - 2i - 2j - k, {}\\ \mathrm{Sc}(p - q)& =& -1,\quad \mathrm{Vec}(p - q) = (-2,-2,-1), {}\\ q \cdot (p - q)& =& q_{0}\mathrm{Sc}(p - q) + \mathbf{q}\mathrm{Vec}(p - q) {}\\ & =& -2 + (1,1,0)(-2,-2,-1) = -6. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} pq - p(p \cdot q + 2q) + q \cdot (p - q)& =& 4 - 2j - 2k - (8 - 4j - 4k) - 6 {}\\ & =& -10 + 2j + 2k; {}\\ \end{array}$$
24.
We have
$$\displaystyle\begin{array}{rcl} p_{0}& =& a_{1},\quad q_{0} = a_{2}, {}\\ \mathbf{p}& =& (b_{1},c_{1},d_{1}),\quad \mathbf{q} = (b_{2},c_{2},d_{2}), {}\\ p \cdot q& =& p_{0}q_{0} + \mathbf{p}\mathbf{q} = a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2}. {}\\ \end{array}$$
25.
We have
and, moreover
From here and from the previous exercise we obtain our equality.
$$\displaystyle\begin{array}{rcl} \overline{p}& =& a_{1} - b_{1}i - c_{1}j - d_{1}k, {}\\ \overline{p}q& =& (a_{1} - b_{1}i - c_{1}j - d_{1}k)(a_{2} + b_{2}i + c_{2}j + d_{2}k) {}\\ & =& a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2} + (a_{1}b_{2} - b_{1}a_{2} - c_{1}d_{2} + d_{1}c_{2})i {}\\ & +& (a_{1}c_{2} + b_{1}d_{2} - c_{1}a_{2} - d_{1}b_{2})j + (a_{1}d_{2} - b_{1}c_{2} + c_{1}b_{2} - d_{1}a_{2})k, {}\\ \overline{q}& =& a_{2} - b_{2}i = c_{2}j - d_{2}k, {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \overline{q}p& =& (a_{2} - b_{2}i - c_{2}j - d_{2}k)(a_{1} + b_{1}i + c_{1}j + d_{1}k) {}\\ & =& a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2} + (a_{2}b_{1} - b_{2}a_{1} - c_{2}d_{1} + d_{2}c_{1})i {}\\ & +& (a_{2}c_{1} + b_{2}d_{1} - c_{2}a_{1} - d_{2}b_{1})j + (a_{2}d_{1} - b_{2}c_{1} + c_{2}b_{1} - d_{2}a_{1})k, {}\\ \frac{1} {2}\left (\overline{p}q + \overline{q}p\right )& =& a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2}. {}\\ \end{array}$$
26.
(a) We have
(b) 0; (c) − 2.
$$\displaystyle\begin{array}{rcl} \overline{p}& =& 1 + i - j, {}\\ \overline{p}q& =& (1 + i - j)(i - j - k)\, =\, -2 + 2i - k, {}\\ \overline{q}& =& -i + j + k, {}\\ \overline{q}p& =& (-i + j + k)(1 - i + j)\, =\, -2 - 2i + k, {}\\ p \cdot q& =& \frac{1} {2}\left (\overline{p}q + \overline{q}p\right ) = -2; {}\\ \end{array}$$
27.
It is easy to find
$$\displaystyle\begin{array}{rcl} \mathrm{Sc}(1_{\mathbb{H}})& =& 1,\quad p_{0} = a, {}\\ \mathrm{Vec}(1)& =& (0,0,0),\quad \mathbf{p} = (b,c,d), {}\\ 1_{\mathbb{H}} \cdot p& =& \mathrm{Sc}(1_{\mathbb{H}})p_{0} + (0,0,0)(b,c,d) = a = p_{0}. {}\\ \end{array}$$
28.
(a) Here p 0 = 1, q 0 = 0, p = (−1, 0, 0), q = (0, 1, 1). Then
i.e. (p, q) = 2k; (b) − 3i − j − k; (c) − i − j + k.
$$\displaystyle\begin{array}{rcl} (p,q)& =& p_{0}\mathbf{q} - q_{0}\mathbf{p} -\mathbf{p} \times \mathbf{q} {}\\ & =& (0,1,1) - (0,1,-1) = (0,0,2), {}\\ \end{array}$$
29.
(a) We have
$$\displaystyle\begin{array}{rcl} & & p_{0} = 1,\quad q_{0} = 1,\quad \mathbf{p} = (-1,-1,-1),\quad \mathbf{q} = (1,1,-1), {}\\ & & p \cdot q = p_{0}q_{0} + \mathbf{p}\mathbf{q}\, =\, 0, {}\\ & & -2p + q = -1 + 3i + 3j + k,\quad pq = 2 + 2i - 2j - 2k, {}\\ & & (p,q) = p_{0}\mathbf{q} - q_{0}\mathbf{p} -\mathbf{p} \times \mathbf{q} {}\\ & & =\; (1,1,-1) - (-1,-1,-1) - (2,-2,0) = (0,4,0)\, =\, 4j, {}\\ \end{array}$$
and also
Consequently
(b) 4 − 2i − 6j + 4k.
$$\displaystyle\begin{array}{rcl} & & \mathrm{Sc}(p,q) = 0,\quad \mathrm{Vec}(p,q) = (0,4,0), {}\\ & & q \cdot (p,q) = q_{0}\mathrm{Sc}(p,q) + \mathbf{q}\mathrm{Vec}(p,q) = 4. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p \cdot q - 2p + q - pq + q \cdot (p,q) = 1 + i + 5j + 3k;& & {}\\ \end{array}$$
30.
We have
i.e.
$$\displaystyle\begin{array}{rcl} & & p_{0} = a_{1},\quad q_{0} = a_{2}, {}\\ & & V ec(p) = (b_{1},c_{1},d_{1}),\quad \mathbf{q} = (b_{2},c_{2},d_{2}), {}\\ & & (p,q) = p_{0}\mathbf{q} - q\mathbf{p} -\mathbf{p} \times \mathbf{q} = {}\\ & & = a_{1}(b_{2},c_{2},d_{2}) - a_{2}(b_{1},c_{1},d_{1}) - (b_{1},c_{1},d_{1}) \times (b_{2},c_{2},d_{2}) {}\\ & & = (a_{1}b_{2},a_{1}c_{2},a_{1}d_{2}) - (a_{2}b_{1},a_{2}c_{1},a_{2}d_{1}) {}\\ & & -\;(c_{1}d_{2} - c_{2}d_{1},b_{2}d_{1} - b_{1}d_{2},b_{1}c_{2} - b_{2}c_{1}) {}\\ & & = (a_{1}b_{2} - a_{2}b_{1} - c_{1}d_{2} + c_{2}d_{1},a_{1}c_{2} - a_{2}c_{1} - b_{2}d_{1} + b_{1}d_{2}, {}\\ & & a_{1}d_{2} - a_{2}d_{1} - b_{1}c_{2} + b_{2}c_{1}), {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} & & \mathbf{p} = (b_{1},c_{1},d_{1}),\quad \mathbf{q} = (b_{2},c_{2},d_{2}), {}\\ & & (p,q) = (a_{1}b_{2} - a_{2}b_{1} - c_{1}d_{2} + c_{2}d_{1})i {}\\ & & +(a_{1}c_{2} - a_{2}c_{1} - b_{2}d_{1} + b_{1}d_{2})j + (a_{1}d_{2} - a_{2}d_{1} - b_{1}c_{2} + b_{2}c_{1})k. {}\\ \end{array}$$
31.
We have
From the previous exercise we obtain
$$\displaystyle\begin{array}{rcl} \overline{p}& =& a_{1} - b_{1}i - c_{1}j - d_{1}k, {}\\ \overline{p}q& =& a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2} + (a_{1}b_{2} - b_{1}a_{2} - c_{1}d_{2} + d_{1}c_{2})i + {}\\ & +& \!(a_{1}c_{2} + b_{1}d_{2} - c_{1}a_{2} - d_{1}b_{2})j + (a_{1}d_{2} - b_{1}c_{2} + c_{1}b_{2} - d_{1}a_{2})k, {}\\ \overline{q}& =& a_{2} - b_{2}i - c_{2}j - d_{2}k, {}\\ \overline{q}p& =& a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2} + (a_{2}b_{1} - b_{2}a_{1} - c_{2}d_{1} + d_{2}c_{1})i + {}\\ & +& (a_{2}c_{1} + b_{2}d_{1} - c_{2}a_{1} - d_{2}b_{1})j + (a_{2}d_{1} - b_{2}c_{1} + c_{2}b_{1} - d_{2}a_{1})k, {}\\ \frac{1} {2}\left (\overline{p}q -\overline{q}p\right )& =& (a_{1}b_{2} - b_{1}a_{2} - c_{1}d_{2} + c_{2}d_{1})i + {}\\ & +& (a_{1}c_{2} - a_{2}c_{1} + b_{1}d_{2} - b_{2}d_{1})j + (a_{1}d_{2} - a_{2}d_{1} - b_{1}c_{2} + c_{1}b_{2})k. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} (p,q) = \frac{\overline{p}q -\overline{q}p} {2}.& & {}\\ \end{array}$$
32.
(a) We have
(b) i + 6j + 4k; (c) − 3i − 10j + 8k; (d) 3i − j + 2k; (e) 2i + j + 4k; (f) 3j; (g) k; (h) j; (i) − i.
$$\displaystyle\begin{array}{rcl} \overline{p}& =& -i - j, {}\\ \overline{p}q& =& (-i - j)(i + j + k) = 2 - i + j, {}\\ \overline{q}& =& -i - j - k, {}\\ \overline{q}p& =& (-i - j - k)(i + j) = 2 + i - j, {}\\ (p,q)& =& \frac{1} {2}\left (\overline{p}q -\overline{q}p\right ) = -i + j; {}\\ \end{array}$$
33.
One has that
$$\displaystyle\begin{array}{rcl} (p,q)& =& \frac{1} {2}\left (\overline{p}q -\overline{q}p\right ), {}\\ (q,p)& =& \frac{1} {2}\left (\overline{q}p -\overline{p}q\right )\, =\, -\frac{1} {2}\left (\overline{p}q -\overline{q}p\right )\, =\, -(p,q). {}\\ \end{array}$$
34.
Let \(p = a_{1} + b_{1}i + c_{1}j + d_{1}k\) and \(q = a_{2} + b_{2}i + c_{2}j + d_{2}k\), where \(a_{i},b_{i},c_{i},d_{i} \in \mathbb{R}\) (i = 1, 2). Then we have
and
Therefore, \((p,q) =\mathrm{ Vec}(\overline{p}q)\). Also \((p,q) = -(q,p) = -\mathrm{Vec}(\overline{q}p)\).
$$\displaystyle\begin{array}{rcl} \overline{p}q& =& a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} + d_{1}d_{2} + (a_{1}b_{2} - b_{1}a_{2} - c_{1}d_{2} + d_{1}c_{2})i + {}\\ & +& (a_{1}c_{2} + b_{1}d_{2} - c_{1}a_{2} - d_{1}b_{2})j + (a_{1}d_{2} - b_{1}c_{2} + c_{1}b_{2} - d_{1}a_{2})k, {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \mathrm{Vec}(\overline{p},q)& =& (a_{1}b_{2} - b_{1}a_{2} - c_{1}d_{2} + d_{1}c_{2}, {}\\ & & a_{1}c_{2} + b_{1}d_{2} - c_{1}a_{2} - d_{1}b_{2},a_{1}d_{2} - b_{1}c_{2} + c_{1}b_{2} - d_{1}a_{2}). {}\\ \end{array}$$
35.
(a) We have
(b) 6 + 2j + 3k; (c) − 2 − i + 4j + 2k; (d) \(0_{\mathbb{H}}\); (e) \(0_{\mathbb{H}}\); (f) \(0_{\mathbb{H}}\).
$$\displaystyle\begin{array}{rcl} p_{0}& =& 2,\quad q_{0} = 1,\quad \mathbf{p} = (-1,1,-1),\quad \mathbf{q} = (1,1,1), {}\\ \left [p,q\right ]& =& p_{0}q_{0} -\mathbf{p}\mathbf{q} + p_{0}\mathbf{q} + q_{0}\mathbf{p} {}\\ & =& 2 - (-1,1,-1)(1,1,1) + 2(1,1,1) + (-1,1,-1) {}\\ & =& 3 + (1,3,1) = 3 + i + 3j + k; {}\\ \end{array}$$
36.
We immediately get
Therefore
$$\displaystyle\begin{array}{rcl} 2p& =& 2 + 2i + 2j - 2k,\quad 2p + q = 4 + i + j - k, {}\\ pq& =& (1 + i + j - k)(2 - i - j + k) = 5 + i + j - k, {}\\ p_{0}& =& 1,\quad q_{0} = 2,\quad \mathbf{p} = (1,1,-1),\quad \mathbf{q} = (-1,-1,1), {}\\ \left [p,q\right ]& =& p_{0}q_{0} -\mathbf{p}\mathbf{q} + p_{0}\mathbf{q} {}\\ & +& q_{0}\mathbf{p} = 2 - (1,1,-1)(-1,-1,1) + (-1,-1,1) {}\\ & +& 2(1,1,-1) = 5 + (1,1,-1) = 5 + i + j - k {}\\ \left [p,q\right ] + p& =& 6 + 2i + 2j - 2k, {}\\ \mathrm{Sc}(\left [p,q\right ] + p)& =& 6, {}\\ \mathrm{Vec}(\left [p,q\right ] + p)& =& (2,2,-2), {}\\ q \cdot (\left [p,q\right ] + p)& =& q_{0}\mathrm{Sc}(\left [p,q\right ] + p) + \mathbf{q}\mathrm{Vec}(\left [p,q\right ] + p) {}\\ & =& 12 + (-1,-1,1)(2,2,-2) = 6, {}\\ (p,q)& =& p_{0}q_{0} - q_{0}\mathbf{p} -\mathbf{p} \times \mathbf{q} {}\\ & =& (-1,-1,1) - 2(1,1,-1) - (1,1,-1) \times (-1,-1,1) {}\\ & =& (-3,-3,3)\, =\, -3i - 3j + 3k. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} & & 2p + q - pq + q \cdot (\left [p,q\right ] + p) + (p,q) {}\\ & =& 4 + i + j - k - (5 + i + j - k) + 6 - 3i - 3j + 3k {}\\ & =& 5 - 3i - 3j + 3k. {}\\ \end{array}$$
37.
We have p 0 = a 1, \(q_{0} = a_{2}\), \(\mathbf{p} = (b_{1},c_{1},d_{1})\), \(\mathbf{q} = (b_{2},c_{2},d_{2})\), and
$$\displaystyle\begin{array}{rcl} \left [p,q\right ]& =& p_{0}q_{0} -\mathbf{p}\mathbf{q} + p_{0}\mathbf{q} + q_{0}\mathbf{p} {}\\ & =& a_{1}a_{2} - (b_{1},c_{1},d_{1})(b_{2},c_{2},d_{2}) + a_{1}(b_{2},c_{2},d_{2}) + a_{2}(b_{1},c_{1},d_{1}) {}\\ & =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{1}b_{2},a_{1}c_{2},a_{1}d_{2}) + {}\\ & +& (a_{2}b_{1},a_{2}c_{1},a_{1}d_{1}) {}\\ & =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} {}\\ & +& (a_{1}b_{2} + a_{2}b_{1},a_{1}c_{2} + a_{2}c_{1},a_{1}d_{2} + a_{2}d_{1}), {}\\ & =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{1}b_{2} + a_{2}b_{1})i {}\\ & +& (a_{1}c_{2} + a_{2}c_{1})j + (a_{1}d_{2} + a_{2}d_{1})k. {}\\ \end{array}$$
38.
Let \(p = a_{1} + b_{1}i + c_{1}j + d_{1}k\) and \(q = a_{2} + b_{2}i + c_{2}j + d_{2}k\), where \(a_{i},b_{i},c_{i},d_{i} \in \mathbb{R}\), i = 1, 2. Then
Consequently,
$$\displaystyle\begin{array}{rcl} pq& =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{1}b_{2} + b_{1}a_{2} + c_{1}d_{2} - d_{1}c_{2})i + {}\\ & +& (a_{1}c_{2} - b_{1}d_{2} + c_{1}a_{2} + d_{1}b_{2})j + (a_{1}d_{2} + b_{1}c_{2} - c_{1}b_{2} + d_{1}a_{2})k, {}\\ qp& =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{2}b_{1} + b_{2}a_{1} + c_{2}d_{1} - d_{2}c_{1})i + {}\\ & +& (a_{2}c_{1} - b_{2}d_{1} + c_{2}a_{1} + d_{2}b_{1})j + (a_{2}d_{1} + b_{2}c_{1} - c_{2}b_{1} + d_{2}a_{1})k, {}\\ \frac{1} {2}\left (pq + qp\right )& =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{2}b_{1} + b_{2}a_{1})i + {}\\ & +& (a_{2}c_{1} + a_{1}c_{2})j + (a_{1}d_{2} + a_{2}d_{1})k. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left [p,q\right ] = \frac{pq + qp} {2}.& & {}\\ \end{array}$$
39.
(a) It is easy to see that
(b) − 1 + 2i; (c) 3 + i + j + k.
$$\displaystyle\begin{array}{rcl} pq = (i + j)(i - j) = -2k,\quad qp = (i - j)(i + j) = 2k,& & {}\\ \left [p,q\right ] = \frac{pq + qp} {2} = 0_{\mathbb{H}};& & {}\\ \end{array}$$
40.
We have
$$\displaystyle\begin{array}{rcl} \left [p,q\right ] = \frac{pq + qp} {2} = \frac{qp + pq} {2} = \left [q,p\right ].& & {}\\ \end{array}$$
41.
(a) Obviously, it holds that
(b) 3i − 3j; (c) \(-2i -\frac{3} {2}j + \frac{1} {2}k\); (d) k; (e) − j; (f) i.
$$\displaystyle\begin{array}{rcl} & & \mathbf{p} = (1,2,-1),\quad \mathbf{q} = (-1,1,1), {}\\ & & \mathbf{p} \times \mathbf{q} = (3,0,3) = 3i + 3k; {}\\ \end{array}$$
42.
We have
i.e.,
$$\displaystyle\begin{array}{rcl} & & \mathbf{p} = (b_{1},c_{1},d_{1}),\quad \mathbf{q} = (b_{2},c_{2},d_{2}), {}\\ & & \mathbf{p} \times \mathbf{q} = (b_{1},c_{1},d_{1}) \times (b_{2},c_{2},d_{2}) = (c_{1}d_{2} - c_{2}d_{1},b_{2}d_{1} - b_{1}d_{2},b_{1}c_{2} - b_{2}c_{1}), {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p \times q = (c_{1}d_{2} - c_{2}d_{1})i + (b_{2}d_{1} - b_{1}d_{2})j + (b_{1}c_{2} - b_{2}c_{1})k.& & {}\\ \end{array}$$
43.
Let \(p = a_{1} + b_{1}i + c_{1}j + d_{1}k\) and \(q = a_{2} + b_{2}i + c_{2}j + d_{2}k\), where \(a_{i},b_{i},c_{i},d_{i} \in \mathbb{R}\) (i = 1, 2). Then
Therefore,
whence
$$\displaystyle\begin{array}{rcl} pq& =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{1}b_{2} + b_{1}a_{2} + c_{1}d_{2} - c_{2}d_{1})i + {}\\ & +& (a_{1}c_{2} - b_{1}d_{2} + c_{1}a_{2} + d_{1}b_{2})j + (a_{1}d_{2} + b_{1}c_{2} - c_{1}b_{2} + d_{1}a_{2})k, {}\\ qp& =& a_{1}a_{2} - b_{1}b_{2} - c_{1}c_{2} - d_{1}d_{2} + (a_{2}b_{1} + b_{2}a_{1} + c_{2}d_{1} - d_{2}c_{1})i + {}\\ & +& (a_{2}c_{1} - b_{2}d_{1} + c_{2}a_{1} + d_{2}b_{1})j + (a_{2}d_{1} + b_{2}c_{1} - c_{2}b_{1} + d_{2}a_{1})k, {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \frac{1} {2}\left (pq - qp\right ) = (c_{1}d_{2} - c_{2}d_{1})i + (d_{1}b_{2} - d_{2}b_{1})j + (b_{1}c_{2} - c_{1}b_{2})k,& & {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p \times q = \frac{pq - qp} {2}.& & {}\\ \end{array}$$
44.
(a) We obtain
(b) − i − k, (c) i − j + k.
$$\displaystyle\begin{array}{rcl} pq& =& (i + j - k)(1 - i + j + k) = 1 + 3i + j + k, {}\\ qp& =& (1 - i + j + k)(i + j - k) = 1 - i + j - 3k, {}\\ \frac{1} {2}\left (pq - qp\right )& =& 2i + 2k; {}\\ \end{array}$$
45.
(a) We get
(b) 2 + i + 6j − 5k; (c) 6 − i − 2j + k.
$$\displaystyle\begin{array}{rcl} p_{0}& =& 1,\quad q_{0} = 2, {}\\ \mathbf{p}& =& (-1,1,1),\quad \mathbf{q} = (-1,0,-1), {}\\ p \cdot q& =& p_{0}q_{0} + \mathbf{p}\mathbf{q}\, =\, 2 + (-1,1,1)(-1,0,-1) = 2, {}\\ (p,q)& =& p_{0}\mathbf{q} - q_{0}\mathbf{p} -\mathbf{p} \times \mathbf{q} {}\\ & =& (-1,0,-1) - 2(-1,1,1) - (-1,1,1) \times (-1,0,-1) {}\\ & =& (2,0,-4)\, =\, 2i - 4k, {}\\ p \times q& =& \mathbf{p} \times \mathbf{q} = (-1,1,1) \times (-1,0,-1) = (-1,-2,1), {}\\ p \times q& =& -i - 2j + k, {}\\ p \cdot q - (p,q)& +& p \times q = 2 - 2i + 4k - i - 2j + k = 2 - 3i - 2j + 5k; {}\\ \end{array}$$
46.
(a) We have
(b) 1 + 2i + 2j + 2k; (c) 1 + i; (d) − 2 + 3i − 2k.
$$\displaystyle\begin{array}{rcl} 2p - r& =& 2i + 2j - (i + j + k) = i + j - k, {}\\ p + q& =& i + j + 1 - i - k = 1 + j - k, {}\\ \mathrm{Sc}(p + q) = 1,& & \mathrm{Vec}(p + q) = (0,1,-1), {}\\ r_{0} = 0,& & \mathbf{r} = (1,1,1), {}\\ r \cdot (p + q)& =& r_{0}\mathrm{Sc}(p + q) + \mathbf{r}\mathrm{Vec}(p + q)\,=\,(1,1,1)(0,1,-1)\,=\,0, {}\\ q + r& =& 1 + j, {}\\ p(q + r)& =& (i + j)(1 + j) = -1 + i + j + k, {}\\ 2p - r - r \cdot (p + q)& -& p(q + r)\,=\,i + j - k - (-1 + i + j + k)\,=\,1 - 2k; {}\\ \end{array}$$
47.
(a) We have
(b) \(\frac{\sqrt{3}} {3} -\frac{\sqrt{3}} {3} i + \frac{\sqrt{3}} {3} k\); (c) \(\frac{\sqrt{2}} {2} i -\frac{\sqrt{2}} {2} j\).
$$\displaystyle\begin{array}{rcl} \vert p\vert & =& \sqrt{9 + 1 + 1 + 1} = 2\sqrt{3}, {}\\ \mathrm{sgn}(p)& =& \frac{p} {\vert p\vert } = \frac{\sqrt{3}} {6} \ (3 + i - j + k)\, =\, \frac{\sqrt{3}} {2} + \frac{\sqrt{3}} {6} \ i -\frac{\sqrt{3}} {6} \ j + \frac{\sqrt{3}} {6} \ k; {}\\ \end{array}$$
48.
(a) After a straightforward calculation we get
(b) \((3 + \sqrt{2})i - j\); (c) \(-\sqrt{2}i -\sqrt{2}j - 3\sqrt{2}k\); (d) \(-2 -\frac{\sqrt{3}} {3} -\frac{2\sqrt{3}} {3} i -\frac{\sqrt{3}} {3} k\); (e) \(-2 -{\Bigl ( 1 + \frac{\sqrt{2}} {2} \Bigr )}i -{\Bigl ( 3 + \frac{\sqrt{2}} {2} \Bigr )}j - 2k\); (f) \(-3i + (\sqrt{2} - 3)j - 3k\).
$$\displaystyle\begin{array}{rcl} pq& =& (i + j)(1 - k) = 2j, {}\\ \vert r\vert & =& \sqrt{3}, {}\\ \mathrm{sgn}(r)& =& \frac{r} {\vert r\vert } = \frac{\sqrt{3}} {3} \ i + \frac{\sqrt{3}} {3} \ j + \frac{\sqrt{3}} {3} \ k, {}\\ pq - p + 2q - 3\mathrm{sgn}(r)& =& 2 - (1 + \sqrt{3})i + (1 -\sqrt{3})j - (2 + \sqrt{3})k; {}\\ \end{array}$$
49.
The following calculation solves our exercise:
$$\displaystyle\begin{array}{rcl} \vert p\vert & =& \sqrt{{a}^{2 } + {b}^{2 } + {c}^{2 } + {d}^{2}}, {}\\ \mathrm{sgn}(p)& =& \frac{a} {\sqrt{{a}^{2 } + {b}^{2 } + {c}^{2 } + {d}^{2}}} + \frac{b} {\sqrt{{a}^{2 } + {b}^{2 } + {c}^{2 } + {d}^{2}}}\ i {}\\ & +& \frac{c} {\sqrt{{a}^{2 } + {b}^{2 } + {c}^{2 } + {d}^{2}}}\ j + \frac{d} {\sqrt{{a}^{2 } + {b}^{2 } + {c}^{2 } + {d}^{2}}}\ k. {}\\ \end{array}$$
50.
(a) We have
(b) \(\arccos {\Bigl (-\frac{1} {2}\Bigr )}\); (c) \(\arccos \frac{2} {\sqrt{6}}\).
$$\displaystyle\begin{array}{rcl} p_{0} = 1,\quad \vert p\vert = \sqrt{6},\quad \mathrm{arg}(p) =\arccos \ \frac{\sqrt{6}} {6};& & {}\\ \end{array}$$
51.
(a) We compute
(b) \(-\frac{\pi }{2} - i + \frac{\sqrt{2}} {2} j +{\Bigl ( -4 + \frac{\sqrt{2}} {2} \Bigr )}k\); (c) \(3 -\frac{3\pi } {2} +{\Bigl ( 2 + \frac{\pi } {2}\Bigr )}i +{\Bigl ( 4 - \frac{\pi } {2}\Bigr )}j + 6k\).
$$\displaystyle\begin{array}{rcl} & & \vert q\vert = \sqrt{3},\quad q_{0} = 1,\quad \mathrm{arg}(q) =\arccos \ \frac{\sqrt{3}} {3}, {}\\ & & p - r = 2,\quad \mathbf{q} = (-1,0,-1),\quad r_{0} = 0,\quad \mathbf{r} = (0,1,1), {}\\ & & \left [q,r\right ] = q_{0}r_{0} -\mathbf{q}\mathbf{r} + q_{0}\mathbf{r} + r_{0}\mathbf{q} = 1 + (0,1,1), {}\\ & & \left [q,r\right ] = 1 + j + k, {}\\ & & \mathrm{arg}(q)(p - r) -\left [q,r\right ] = 2\arccos \ \frac{\sqrt{3}} {3} - 1 - j - k; {}\\ \end{array}$$
52.
(a) An easy computation yields
Consequently,
(b) \(\sqrt{2}{\Bigl (\cos \frac{\pi }{2} +{\Bigl ( \frac{\sqrt{2}} {2} i + \frac{\sqrt{2}} {2} j\Bigr )}\sin \frac{\pi } {2}\Bigr )}\); (c) \(\sqrt{2}{\Bigl (\cos \frac{\pi }{2} +{\Bigl ( \frac{\sqrt{2}} {2} i + \frac{\sqrt{2}} {2} k\Bigr )}\sin \frac{\pi } {2}\Bigr )}\); (d) \(\sqrt{2}{\Bigl (\cos \frac{\pi }{2} +{\Bigl ( \frac{\sqrt{2}} {2} j + \frac{\sqrt{2}} {2} k\Bigr )}\sin \frac{\pi } {2}\Bigr )}\); (e) \(\sqrt{2}{\Bigl (\cos \frac{\pi }{2} +{\Bigl ( \frac{\sqrt{2}} {2} i -\frac{\sqrt{2}} {2} j\Bigr )}\sin \frac{\pi } {2}\Bigr )}\); (f) \(\sqrt{2}{\Bigl (\cos \frac{\pi }{2} +{\Bigl ( \frac{\sqrt{2}} {2} i -\frac{\sqrt{2}} {2} k\Bigr )}\sin \frac{\pi } {2}\Bigr )}\); (g) \(\sqrt{2}{\Bigl (\cos \frac{\pi }{2} +{\Bigl ( \frac{\sqrt{2}} {2} j -\frac{\sqrt{2}} {2} k\Bigr )}\sin \frac{\pi } {2}\Bigr )}\).
$$\displaystyle\begin{array}{rcl} & & \vert p\vert = 2,\quad p_{0} = 1,\quad \mathrm{arg}(p) =\arccos \frac{1} {2} = \frac{\pi } {3}, {}\\ & & \mathbf{p} = (1,1,1),\quad \vert \mathbf{p}\vert = \sqrt{3}, \frac{\mathbf{p}} {\vert \mathbf{p}\vert } = \left (\frac{\sqrt{3}} {3}, \frac{\sqrt{3}} {3}, \frac{\sqrt{3}} {3} \right ). {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} p = 2\left [\cos \frac{\pi } {3} + \left (\frac{\sqrt{3}} {3} \ i + \frac{\sqrt{3}} {3} \ j + \frac{\sqrt{3}} {3} k\right )\sin \frac{\pi } {3}\right ];& & {}\\ \end{array}$$
53.
(a) p = 1 + i1 + 1j + i1j; (b) \(p = 2 + i{\Bigl ( -\frac{3} {2}\Bigr )} + (-4)j + i5j\); (c) \(p = \frac{1} {2} + i(-2) + 3j + i(-4)j\).
54.
(a) We have
(b) \(\frac{3} {2} + i{\Bigl ( -\frac{1} {2}\Bigr )} +{\Bigl ( -\frac{1} {2}\Bigr )}j + i{\Bigl ( -\frac{11} {2} \Bigr )}j\); (c) − 1 + i(6) + (−5)j + i5j.
$$\displaystyle\begin{array}{rcl} p& =& 1j + i1j, {}\\ q& =& 1 + i(-1) + i(-1)j, {}\\ pq& =& (1j + i1j)(1 + i(-1) + i(-1)j) = 1 + i(-1) + i2j; {}\\ \end{array}$$
55.
(a) 1 + i2 + 4j + i3j; (b) 1 + i1 + 1j + i1j; (c) 1 + i2 + 4j + i1j.
56.
(a) It is clear that
(b) \(p_{+} = \frac{3} {2} + i - j + \frac{3} {2}k\), \(p_{-} = -\frac{1} {2} + \frac{1} {2}k\); (c) \(p_{+} = -\frac{1} {2} - i + j -\frac{1} {2}k\), \(p_{-} = \frac{3} {2} -\frac{3} {2}k\).
$$\displaystyle\begin{array}{rcl} ipj& =& i\left (2 + \frac{3} {2}\ i -\frac{1} {2}\ j + k\right )j = 1 + \frac{1} {2}\ i -\frac{3} {2}\ j + 2k, {}\\ p + ipj& =& 3 + 2i - 2j + 3k, {}\\ p - ipj& =& 1 + i + j - k, {}\\ p_{+}& =& \frac{3} {2} + i - j + \frac{3} {2}\ k,\quad p_{-}\, =\, \frac{1} {2} + \frac{1} {2}\ i + \frac{1} {2}\ j -\frac{1} {2}\ k; {}\\ \end{array}$$
57.
We have
$$\displaystyle\begin{array}{rcl} ipj& =& d - ci - bj + ak, {}\\ p + ipj& =& a + d + (b - c)i - (b - c)j + (a + d)k, {}\\ p - ipj& =& a - d + (b + c)i + (b + c)j - (a - d)k, {}\\ p_{+}& =& \frac{a + d} {2} + \frac{b - c} {2} \ i -\frac{b - c} {2} \ j + \frac{a + d} {2} \ k, {}\\ p_{-}& =& \frac{a - d} {2} + \frac{b + c} {2} \ i + \frac{b + c} {2} \ j -\frac{a - d} {2} \ k. {}\\ \end{array}$$
58.
We have
where in the last equality we used the previous exercise. Also,
$$\displaystyle\begin{array}{rcl} \left (a + d + i(b - c)\right )\frac{1 + k} {2} & =& \frac{a + ak + d + dk + i(b - c) - j(b - c)} {2} {}\\ & =& \frac{a + d} {2} + \frac{b - c} {2} \ i -\frac{b - c} {2} \ j + \frac{a + d} {2} \ k {}\\ & =& p_{+}, {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \left (a - d + i(b + c)\right )\frac{1 - k} {2} & =& \frac{a - d - (a - d)k + i(b + c) + j(b + c)} {2} {}\\ & =& \frac{a - d} {2} + \frac{b + c} {2} \ i + \frac{b + c} {2} \ j -\frac{a - d} {2} \ k {}\\ & =& p_{-}. {}\\ \end{array}$$
59.
We have
$$\displaystyle\begin{array}{rcl} \frac{1 + k} {2} \left (a + d + j(c - b)\right )& =& \frac{a + d + j(c - b) + (a + d)k - i(c - b)} {2} {}\\ & =& \frac{a + d} {2} + \frac{b - c} {2} \ i -\frac{b - c} {2} \ j + \frac{a + d} {2} \ k {}\\ & =& p_{+}, {}\\ \frac{1 - k} {2} \left (a - d + j(b + c)\right )& =& \frac{(a - d) + j(b + c) - k(a - d) + i(b + c)} {2}{}\\ & = & p_{ -}. {}\\ \end{array}$$
60.
Let p = a + bi + cj + dk \((a,b,c,d \in \mathbb{R})\). Then
Consequently
$$\displaystyle\begin{array}{rcl} & & \vert p_{+}{\vert }^{2} = \frac{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} + 2ad - 2bc} {2}, {}\\ & & \vert p_{-}{\vert }^{2} = \frac{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} - 2ad + 2bc} {2}. {}\\ \end{array}$$
$$\displaystyle\begin{array}{rcl} \vert p_{+}{\vert }^{2} + \vert p_{ -}{\vert }^{2}\, =\, {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\, =\, \vert p{\vert }^{2}.& & {}\\ \end{array}$$
62.
(a) \(0_{\mathbb{H}}\); (b) 1 − i − j − k; (c) 4; (d) j + k; (e) 2 − 2k.
64.
Note that
$$\displaystyle\begin{array}{rcl} \mathrm{Sc}\left (p\mathbf{q}\overline{p}\right ) = \frac{p\mathbf{q}\overline{p} + \overline{p\mathbf{q}\overline{p}}} {2} = \frac{p} {2}\left (\mathbf{q} + \overline{\mathbf{q}}\right )\overline{p} = 0.& & {}\\ \end{array}$$
66.
The following computation leads to the solution:
$$\displaystyle\begin{array}{rcl} \left \vert q\mathbf{p}\overline{q}\right \vert = \vert q\vert \left \vert \mathbf{p}\right \vert \vert \overline{q}\vert = \left \vert \mathbf{p}\right \vert \vert \overline{q}\vert = \left \vert \mathbf{p}\right \vert.& & {}\\ \end{array}$$
68.
Indeed,
$$\displaystyle\begin{array}{rcl} q\left (\alpha \mathbf{p} + \mathbf{q}\right )\overline{q} = q\alpha \mathbf{p}\overline{q} + q\mathbf{q}\overline{q} =\alpha q\mathbf{p}\overline{q} + q\mathbf{q}\overline{q}.& & {}\\ \end{array}$$
70.
(a) \(- \frac{2} {\sqrt{3}}i + \frac{2} {\sqrt{3}}j - \frac{2} {\sqrt{3}}k\); (b) \(i\); (c) \(j\); (d) k; (e) \(101{\Bigl ( - \frac{i} {6} + \frac{2j} {\sqrt{6}} - \frac{k} {\sqrt{6}}\Bigr )}\); (f) \(2{\Bigl ( - \frac{i} {\sqrt{19}} + \frac{3j} {\sqrt{19}} - \frac{3k} {\sqrt{19}}\Bigr )}\).