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

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

• \(\mathbf{a}=\overrightarrow{AB}=\begin{pmatrix}2\\2\\-3\end{pmatrix}\), \(\mathbf{b}=\overrightarrow{AC}=\begin{pmatrix}5\\-1\\-2\end{pmatrix}\)
• \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}2\\2\\-3\end{pmatrix}\times\begin{pmatrix}5\\-1\\-2\end{pmatrix}=\begin{pmatrix}2\times(-2)-(-3)\times(-1)\\ (-3)\times5-2\times(-2)\\ 2\times(-1)-2\times5\end{pmatrix}=\begin{pmatrix}-7\\-11\\-12\end{pmatrix}\)
• \(|\mathbf{a}\times\mathbf{b}|=\sqrt{314}=17.72\)
• Area \(=\tfrac12|\mathbf{a}\times\mathbf{b}|=8.86\) square units

Find the area of each shape.

1. (a) The triangle with vertices at the points with position vectors \(\mathbf{a}=2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), \(\mathbf{b}=3\mathbf{j} + 4\mathbf{k}\) and \(\mathbf{c}=5\mathbf{i} + \mathbf{j} + 3\mathbf{k}\). Find its area.

   • \(\overrightarrow{AB}=\begin{pmatrix}-2\\1\\3\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}3\\-1\\2\end{pmatrix}\)
   • \(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}5\\13\\-1\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{195}=6.98\)

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

   • \(\mathbf{u}\times\mathbf{w}=\begin{pmatrix}3\\1\\-2\end{pmatrix}\times\begin{pmatrix}-1\\2\\4\end{pmatrix}=\begin{pmatrix}8\\-10\\7\end{pmatrix}\)
   • Area \(=|\mathbf{u}\times\mathbf{w}|=\sqrt{213}=14.6\)

The points \(P(0, 4, 1)\), \(Q(2, 5, 3)\), \(R(6, 7, 7)\) and \(S(1, 1, 3)\) 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.

   • \(\overrightarrow{PS}=\begin{pmatrix}1\\-3\\2\end{pmatrix}\)
   • \(\overrightarrow{PQ}\times\overrightarrow{PS}=\begin{pmatrix}8\\-2\\-7\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{117}=5.41\)

A sail for a dinghy is a triangle with corners \(P(2, 0, 0)\), \(Q(5, 3, 1)\), \(R(1, 4, 2)\), in metres. The sailcloth costs \$23 per square metre.

1. (a) Find the area of the sail.

   • \(\overrightarrow{PQ}=\begin{pmatrix}3\\3\\1\end{pmatrix}\), \(\overrightarrow{PR}=\begin{pmatrix}-1\\4\\2\end{pmatrix}\)
   • \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}3\\3\\1\end{pmatrix}\times\begin{pmatrix}-1\\4\\2\end{pmatrix}=\begin{pmatrix}3\times2-1\times4\\ 1\times(-1)-3\times2\\ 3\times4-3\times(-1)\end{pmatrix}=\begin{pmatrix}2\\-7\\15\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{278}=8.34\ \text{m}^2\)

2. (b) Find the cost.

   • \(\tfrac12\sqrt{278}\times23=\$191.74\)

A triangular field on a hillside has its corners at \(A(0, 2, 1)\), \(B(4, 5, 3)\), \(C(1, 7, 6)\), in metres. A fence is to be built along the edge AB.

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

   • \(\overrightarrow{AB}=\begin{pmatrix}4\\3\\2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}1\\5\\5\end{pmatrix}\)
   • \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}4\\3\\2\end{pmatrix}\times\begin{pmatrix}1\\5\\5\end{pmatrix}=\begin{pmatrix}3\times5-2\times5\\ 2\times1-4\times5\\ 4\times5-3\times1\end{pmatrix}=\begin{pmatrix}5\\-18\\17\end{pmatrix}\)
   • Area \(=\tfrac12\sqrt{638}=12.63\)

2. (b) Find the shortest distance from the corner C to the fence (the line AB).

   • \(|\overrightarrow{AB}|=\sqrt{29}=5.385\)
   • \(\tfrac12\times5.385\times h=12.63\)
   • \(h=4.69\)

The triangle has vertices \(A(2, 0, 0)\), \(B(0, 3, 0)\) and \(C(0, 0, k)\), where \(k>0\). The area of the triangle is 6 square units.

1. (a) Show that \(|\overrightarrow{AB}\times\overrightarrow{AC}|^2=13 k^{2} + 36\).

   • \(\overrightarrow{AB}=\begin{pmatrix}-2\\3\\0\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}-2\\0\\k\end{pmatrix}\)
   • \(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}3 k\\2 k\\6\end{pmatrix}\)
   • \(|\overrightarrow{AB}\times\overrightarrow{AC}|^2=13 k^{2} + 36\)

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

   • Area \(=\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|=6\), so \(|\overrightarrow{AB}\times\overrightarrow{AC}|^2=144\)
   • \(13 k^{2} + 36=144\)
   • From the calculator: \(k=2.88\) (positive solution)