Homework #1 Solutions
Calculate problems #1 and #2 manually and use
http://www.continuummechanics.org/interactivecalcs.html to double-check.
- Calculate the length of each vector and find the angle between them:
\({\bf a} = (12, 3, 4)\) and \({\bf b} = (16, 48, 12)\).
\[
\begin{eqnarray}
|{\bf a}| & = & \sqrt { 12^2 + 3^2 + 4^2 } = 13
\\
\\
|{\bf b}| & = & \sqrt { 16^2 + 48^2 + 12^2 } = 52
\\
\\
|{\bf a}| |{\bf b}| \cos(\theta) & = & {\bf a} \cdot {\bf b} = 12 * 16 + 3 * 48 + 4 * 12 = 384
\\
\\
\text{so } \cos(\theta) & = & 384 / (13 * 52) = 0.5680
\\
\\
\text{ and } \theta & = & \cos^{-1}(0.5680) = 55.4^\circ
\end{eqnarray}
\]
-
Find the area of a triangle whose edges are the two vectors in #1 above.
Ignore units.
\[
\begin{eqnarray}
{\bf a} \times {\bf b} & = & (12, 3, 4) \times (16, 48, 12)
\\
\\
& = & (3 * 12 - 4 * 48) {\bf i} + (4 * 16 - 12 * 12) {\bf j} + (12 * 48 - 3 * 16) {\bf k}
\\
\\
& = & -156 {\bf i} - 80 {\bf j} + 528 {\bf k}
\end{eqnarray}
\]
The length of the vector is 556.35
So the area of the triangle is
\[
\begin{eqnarray}
Area & = & {1 \over 2} | {\bf a} \times {\bf b} |
\\
\\
& = & {1 \over 2} ( 556.35)
\\
\\
& = & 278.17
\end{eqnarray}
\]
Don't do the remaining ones manually. It's too tedious. Just use the above webpage directly and write out the results.
-
Given \(
\quad {\bf A} = \left[
\matrix {
2 & 5 & 1 \\
4 & 8 & 2 \\
6 & 2 & 4 }
\right] \quad
\)
and
\(
\quad {\bf B} =
\left[ \matrix {
3 & 4 & 2 \\
1 & 7 & 5 \\
3 & 2 & 4 }
\right] \quad
\)
Demonstrate that \({\bf A} \cdot {\bf B} \ne {\bf B} \cdot {\bf A}\), but
\( ( {\bf A} \cdot {\bf B} )^T = {\bf B}^T \cdot {\bf A}^T \).
\[
{\bf A} \cdot {\bf B}
= \left[ \matrix{
14 & 45 & 33 \\
26 & 76 & 56 \\
32 & 46 & 38 }
\right]
\]
and
\[
{\bf B} \cdot {\bf A}
= \left[ \matrix{
34 & 51 & 19 \\
60 & 71 & 35 \\
38 & 39 & 23 }
\right]
\quad \ne \quad {\bf A} \cdot {\bf B}
\]
\[
({\bf A} \cdot {\bf B})^T
= \left[ \matrix{
14 & 26 & 32 \\
45 & 76 & 46 \\
33 & 56 & 38 }
\right]
\]
and
\[
{\bf B}^T \cdot {\bf A}^T
= \left[ \matrix{
14 & 26 & 32 \\
45 & 76 & 46 \\
33 & 56 & 38 }
\right]
\]
-
Calculate the double dot product of the two matrices, \({\bf A} : {\bf B}\).
\[
{\bf A} : {\bf B} = 136
\]
-
Calculate the inverse of \({\bf A}\) in #3 and confirm that
\({\bf A} \cdot {\bf A}^{-1} \; = \; {\bf A}^{-1} \cdot {\bf A} \; = \; {\bf I}\).
\[
{\bf A} = \left[
\matrix {
2 & 5 & 1 \\
4 & 8 & 2 \\
6 & 2 & 4 }
\right]
\qquad \qquad
{\bf A}^{-1} = \left[
\matrix {
-7 & 4.5 & -0.5 \\
1 & -0.5 & 0 \\
10 & -6.5 & 1 }
\right]
\]
\[
{\bf A} \cdot {\bf A}^{-1} \; = \;
\left[
\matrix {
2 & 5 & 1 \\
4 & 8 & 2 \\
6 & 2 & 4 }
\right]
\left[
\matrix {
-7 & 4.5 & -0.5 \\
1 & -0.5 & 0 \\
10 & -6.5 & 1 }
\right]
\; = \;
\left[
\matrix {
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 }
\right]
\]
\[
{\bf A}^{-1} \cdot {\bf A} \; = \;
\left[
\matrix {
-7 & 4.5 & -0.5 \\
1 & -0.5 & 0 \\
10 & -6.5 & 1 }
\right]
\left[
\matrix {
2 & 5 & 1 \\
4 & 8 & 2 \\
6 & 2 & 4 }
\right]
\; = \;
\left[
\matrix {
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 }
\right]
\]