Parallel vectors, perpendicular vectors and area Problem set A Worked solutions

Find the area of the triangle with vertices \(A(1, 2, 3)\), \(B(3, 0, 4)\), \(C(2, 5, 1)\).

• \(\mathbf{a}=\overrightarrow{AB}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\), \(\mathbf{b}=\overrightarrow{AC}=\begin{pmatrix}1\\3\\-2\end{pmatrix}\)
• \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\times\begin{pmatrix}1\\3\\-2\end{pmatrix}=\begin{pmatrix}(-2)\times(-2)-1\times3\\ 1\times1-2\times(-2)\\ 2\times3-(-2)\times1\end{pmatrix}=\begin{pmatrix}1\\5\\8\end{pmatrix}\)
• \(|\mathbf{a}\times\mathbf{b}|=\sqrt{90}=9.487\)
• Area \(=\tfrac12|\mathbf{a}\times\mathbf{b}|=4.74\) square units

Find the area of each shape.

1. (a) The triangle with vertices at the points with position vectors \(\mathbf{a}=\mathbf{i} + 2\mathbf{k}\), \(\mathbf{b}=3\mathbf{i} + \mathbf{j}\) and \(\mathbf{c}=2\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\). Find its area.

   • \(\overrightarrow{AB}=\begin{pmatrix}2\\1\\-2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}1\\4\\1\end{pmatrix}\)
   • \(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}9\\-4\\7\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{146}=6.04\)

2. (b) The parallelogram with sides \(\mathbf{u}=2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(\mathbf{w}=\mathbf{i} + 4\mathbf{j}\). Find its area.

   • \(\mathbf{u}\times\mathbf{w}=\begin{pmatrix}2\\-1\\3\end{pmatrix}\times\begin{pmatrix}1\\4\\0\end{pmatrix}=\begin{pmatrix}-12\\3\\9\end{pmatrix}\)
   • Area \(=|\mathbf{u}\times\mathbf{w}|=\sqrt{234}=15.3\)

The points \(P(1, 2, 0)\), \(Q(3, 3, 2)\), \(R(7, 5, 6)\) and \(S(2, 6, 1)\) are given.

1. (a) Show that P, Q and R lie on a straight line, by finding \(\overrightarrow{PQ}\times\overrightarrow{PR}\).

   • \(\overrightarrow{PQ}=\begin{pmatrix}2\\1\\2\end{pmatrix}\), \(\overrightarrow{PR}=\begin{pmatrix}6\\3\\6\end{pmatrix}\)
   • \(\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{pmatrix}0\\0\\0\end{pmatrix}=\mathbf{0}\)
   • \(\overrightarrow{PR}=3\overrightarrow{PQ}\): parallel vectors from the same point, so P, Q, R are on one line

2. (b) Find the area of the triangle PQS.

   • \(\overrightarrow{PS}=\begin{pmatrix}1\\4\\1\end{pmatrix}\)
   • \(\overrightarrow{PQ}\times\overrightarrow{PS}=\begin{pmatrix}-7\\0\\7\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{98}=4.95\)

A solar roof panel is a triangle with corners \(P(0, 0, 2)\), \(Q(3, 1, 5)\), \(R(1, 4, 3)\), in metres. The panel costs \$64 per square metre.

1. (a) Find the area of the roof panel.

   • \(\overrightarrow{PQ}=\begin{pmatrix}3\\1\\3\end{pmatrix}\), \(\overrightarrow{PR}=\begin{pmatrix}1\\4\\1\end{pmatrix}\)
   • \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}3\\1\\3\end{pmatrix}\times\begin{pmatrix}1\\4\\1\end{pmatrix}=\begin{pmatrix}1\times1-3\times4\\ 3\times1-3\times1\\ 3\times4-1\times1\end{pmatrix}=\begin{pmatrix}-11\\0\\11\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{242}=7.78\ \text{m}^2\)

2. (b) Find the cost.

   • \(\tfrac12\sqrt{242}\times64=\$497.80\)

A triangular flower bed has its corners at \(A(1, 1, 0)\), \(B(6, 2, 1)\), \(C(3, 6, 4)\), in metres. A path is to run along the edge AB.

1. (a) Find the area of triangle ABC.

   • \(\overrightarrow{AB}=\begin{pmatrix}5\\1\\1\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}2\\5\\4\end{pmatrix}\)
   • \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}5\\1\\1\end{pmatrix}\times\begin{pmatrix}2\\5\\4\end{pmatrix}=\begin{pmatrix}1\times4-1\times5\\ 1\times2-5\times4\\ 5\times5-1\times2\end{pmatrix}=\begin{pmatrix}-1\\-18\\23\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{854}=14.61\)

2. (b) A gardener stands at C. Find the shortest distance from C to the path (the line AB).

   • \(|\overrightarrow{AB}|=\sqrt{27}=5.196\)
   • \(\tfrac12\times5.196\times h=14.61\)
   • \(h=5.62\)

The triangle has vertices \(A(1, 0, 0)\), \(B(0, 2, 0)\) and \(C(0, 0, k)\), where \(k>0\). The area of the triangle is 5 square units.

1. (a) Show that \(|\overrightarrow{AB}\times\overrightarrow{AC}|^2=5 k^{2} + 4\).

   • \(\overrightarrow{AB}=\begin{pmatrix}-1\\2\\0\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}-1\\0\\k\end{pmatrix}\)
   • \(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}2 k\\k\\2\end{pmatrix}\)
   • \(|\overrightarrow{AB}\times\overrightarrow{AC}|^2=5 k^{2} + 4\)

2. (b) Find the value of \(k\).

   • Area \(=\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|=5\), so \(|\overrightarrow{AB}\times\overrightarrow{AC}|^2=100\)
   • \(5 k^{2} + 4=100\)
   • From the calculator: \(k=4.38\) (positive solution)