Proof.

If \(\alpha = 0\), all the products are zero. If \(\alpha > 0\), \( \theta \) is an angle between \(\underline {a}\) and \(\underline {b}\) and \(\widehat {\textbf {n}}\) is a vector in the direction of \(\underline {a} \times \underline {b}\), then \begin {align*} \alpha (\underline {a} \times \underline {b}) & = \alpha \Big ( \left |a\right |\left |b\right | \sin \theta \hspace {0.1cm}\widehat {\textbf {n}}\Big )\\ & =\Big (\alpha \left |a\right |\left |b\right |\Big ) \sin \theta \hspace {0.1cm}\widehat {\textbf {n}}\\ & = \left |\alpha a\right |\left |b\right | \sin \theta \hspace {0.1cm}\widehat {\textbf {n}}=(\alpha \underline {a} ) \times \underline {b}\\ & = \left |a\right |\left |\alpha b\right | \sin \theta \hspace {0.1cm}\widehat {\textbf {n}}= \underline {a}\times (\alpha \underline {b}) \end {align*}

If \(\alpha < 0\), then \(\alpha (\underline {a} \times \underline {b})\) is a vector that is parallel to \(\underline {a} \times \underline {b}\) but in the opposite direction. On the hand, \(\alpha \underline {a}\) is parallel to \(\underline {a}\) but also points in the opposite direction. Thus \((\alpha \underline {a}) \times \underline {b}\) points in the opposite direction \((\underline {a} \times \underline {b})\).

In addition the length or the vectors \(\alpha (\underline {a} \times \underline {b})\) and \((\alpha \underline {a}) \times \underline {b}\) are equal. Therefore the vectors \(\alpha (\underline {a} \times \underline {b})\) and \((\alpha \underline {a})\times \underline {b}\) are equal. A similar argument shows that \(\alpha (\underline {a}\times \underline {b}) = \underline {a} \times (\alpha \underline {b})\). □

Proof. Expand the product using the distributive law, taking the scalar coefficients outside by the scalar law stated above. This gives nine terms, one for each pair of basis vectors: \begin {align*} \underline {a}\times \underline {b} &= \big (a_1\textbf {i} + a_2\textbf {j} + a_3\textbf {k}\big )\times \big (b_1\textbf {i} + b_2\textbf {j} + b_3\textbf {k}\big )\\ &= a_1b_1(\textbf {i}\times \textbf {i}) + a_1b_2(\textbf {i}\times \textbf {j}) + a_1b_3(\textbf {i}\times \textbf {k})\\ &\quad + a_2b_1(\textbf {j}\times \textbf {i}) + a_2b_2(\textbf {j}\times \textbf {j}) + a_2b_3(\textbf {j}\times \textbf {k})\\ &\quad + a_3b_1(\textbf {k}\times \textbf {i}) + a_3b_2(\textbf {k}\times \textbf {j}) + a_3b_3(\textbf {k}\times \textbf {k}) \end {align*}

Now evaluate the nine basis products from the definition of the cross product. For the three like pairs the angle between the vectors is \(0\), so \(\sin \theta = 0\) and \[\textbf {i}\times \textbf {i} = \textbf {j}\times \textbf {j} = \textbf {k}\times \textbf {k} = \underline {0}.\] For the unlike pairs the vectors are perpendicular unit vectors, so the magnitude is \(1\cdot 1\cdot \sin 90^\circ = 1\), and the right-hand rule fixes the direction: \[\textbf {i}\times \textbf {j} = \textbf {k},\qquad \textbf {j}\times \textbf {k} = \textbf {i},\qquad \textbf {k}\times \textbf {i} = \textbf {j},\] \[\textbf {j}\times \textbf {i} = -\textbf {k},\qquad \textbf {k}\times \textbf {j} = -\textbf {i},\qquad \textbf {i}\times \textbf {k} = -\textbf {j}.\]

Substituting these, the three like terms vanish and the remaining six become \begin {align*} \underline {a}\times \underline {b} &= a_1b_2\textbf {k} - a_1b_3\textbf {j} - a_2b_1\textbf {k} + a_2b_3\textbf {i} + a_3b_1\textbf {j} - a_3b_2\textbf {i}\\ &= \big (a_2b_3 - a_3b_2\big )\textbf {i} - \big (a_1b_3 - a_3b_1\big )\textbf {j} + \big (a_1b_2 - a_2b_1\big )\textbf {k} \end {align*}

Finally, expanding the determinant in the statement along its first row gives \[\begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end {vmatrix} = \textbf {i}\begin {vmatrix} a_2 & a_3\\ b_2 & b_3\end {vmatrix} - \textbf {j}\begin {vmatrix} a_1 & a_3\\ b_1 & b_3\end {vmatrix} + \textbf {k}\begin {vmatrix} a_1 & a_2\\ b_1 & b_2\end {vmatrix}\] \[= \big (a_2b_3 - a_3b_2\big )\textbf {i} - \big (a_1b_3 - a_3b_1\big )\textbf {j} + \big (a_1b_2 - a_2b_1\big )\textbf {k},\] which is the same expression. Hence the two agree. □

Solution.

Part 1

\begin {align*} \underline {a} \times \underline {b} & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ 2 & -1 & 1\\ 1 & 2 & -3\\ \end {vmatrix} = \begin {vmatrix} -1 & 1\\ 2 & -3\\ \end {vmatrix}\textbf {i} - \begin {vmatrix} 2 & 1\\ 1 & -3\\ \end {vmatrix}\textbf {j} + \begin {vmatrix} 2 & -1\\ 1 & 2\\ \end {vmatrix}\textbf {k}\\ & = \textbf {i} + 7\textbf {j} + 5\textbf {k} \end {align*}

\(\displaystyle { \underline {a} \times \underline {b} = \left |\underline {a}\right |\left |\underline {b}\right | \sin \theta \hspace {0.1cm} \left |\widehat {\textbf {n}}\right |}\)

\[\left |\underline {a}\times \underline {b}\right | = \left ||\underline {a}||\underline {b}|\sin \theta |\widehat {\textbf {n}}|\right |\hspace {0.1cm}, \hspace {0.2cm}\text {but}\hspace {0.2cm} |\widehat {\textbf {n}}| = 1\]

So that \(\left |\underline {a}\times \underline {b}\right |= \left |\underline {a}\right |\left |b\right |\sin \theta \)

\[\sin \theta = \frac {\left |\underline {a} \times \underline {b}\right | }{\left |\underline {a}\right |\left |\underline {b}\right | }\]

\[\left |\underline {a}\times \underline {b}\right |=\sqrt {75} = 5\sqrt {3}\hspace {0.5cm},\hspace {0.5cm} \left |\underline {a}\right |=\sqrt {6}\hspace {0.5cm},\hspace {0.5cm} \left |\underline {b}\right |=\sqrt {14}\]

\[\therefore \hspace {1cm} \sin \theta = \frac {5\sqrt {3}}{\sqrt {6}\sqrt {14}}= \frac {5}{2\sqrt {7}}\]

Part 2

Note that the direction vectors of the given lines are \(\underline {a} = \textbf {i} + 2\textbf {j} + 3\textbf {k}\) and \(\underline {b} = 5\textbf {i} + 2\textbf {j} - 3\textbf {k}\). The vector orthogonal to \(\underline {a}\) and \(\underline {b}\)

\[\underline {a} \times \underline {b} = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ 1 & 2 & 3\\ 5 & 2 & -3\ \end {vmatrix}= -12 \textbf {i} + 18\textbf {j} - 8\textbf {k}\] Thus, the direction vector of the line is \( -12 \textbf {i} + 18\textbf {j} - 8\textbf {k}\). So symmetric equations are \[\frac {x + 1}{-12} = \frac {y - 1}{18} = \frac {z - 2}{-8}\]

Part 3

\begin {align*} d & = \left |\overrightarrow {AP}\right |\sin \theta = \left |\overrightarrow {AP}\right |\cdot \frac {\left |\overrightarrow {AP}\times \overrightarrow {AB}\right | }{\left |\overrightarrow {AP}\right |\left |\overrightarrow {AB}\right | }\\ d & = \frac {\left |\overrightarrow {AP}\times \overrightarrow {AB}\right | }{\left |\overrightarrow {AB}\right | } = \frac {\sqrt {30}}{\sqrt {11}}\\ \end {align*}

Solution.

\(\overrightarrow {PQ} = -3\textbf {i} + \textbf {j} - 7\textbf {k}\hspace {0.2cm}, \hspace {0.2cm} \overrightarrow {PR} = -5\textbf {j} - 5\textbf {k}\)

\[\overrightarrow {PQ}\times \overrightarrow {PR} = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ -3 & 1 & -7\\ 0 & -5 & -5\\ \end {vmatrix} = -40\textbf {i} -15\textbf {j} + 15\textbf {k}\]

Area of triangle \(PQR= \dfrac {1}{2}\left |-40\textbf {i} -15\textbf {j} + 15\textbf {k}\right | = \dfrac {5}{2}\sqrt {82}\)