Question

Determine the area of the parallelogram with vertices\(A( - 2,1),B(0,4),C(4,2)\) and\(D(2, - 1)\).

Short answer

The area of the parallelogram with vertices \(A( - 2,1),B(0,4),C(4,2)\) and \(D(2, - 1)\) is 16.

Step-by-step solution

Formula used

Consider the vectors\({\rm{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \)and\({\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \).

1. The area of the parallelogram with vectors\(a\)and\(b\)is\(|a \times b|\)

2. The cross product between\(a\)and\(b\)is\(a \times b = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|\).

Calculate the value of the vector \(\overrightarrow {AB} \)

Consider the vertices of the parallelogram are \(A( - 2,1),B(0,4),C(4,2)\), and \(D(2, - 1)\).

\(\begin{array}{l}\overrightarrow {AB} = B(0,4) - A( - 2,1)\\\overrightarrow {AB} = \langle (0 - ( - 2)),(4 - 1)\rangle \\\overrightarrow {AB} = \langle (0 + 2),(3)\rangle \\\overrightarrow {AB} = \langle 2,3\rangle \end{array}\)

Calculate the value of the vector \(\overrightarrow {AD} \)

\(\begin{array}{l}\overrightarrow {AD} = D(2, - 1) - A( - 2,1)\\\overrightarrow {AD} = \langle (2 - ( - 2)),( - 1 - 1)\rangle \\\overrightarrow {AD} = \langle (2 + 2), - 2\rangle \\\overrightarrow {AD} = \langle 4, - 2\rangle \end{array}\)

Consider the \(\overrightarrow {AB} \) and \(\overrightarrow {AD} \) as the three dimensional vectors are \(\langle 2,3,0\rangle \) and \(\langle 4, - 2,0\rangle \).

Calculate the cross product of \(\overrightarrow {AB} \) and\(\overrightarrow {AD} \)

\(\begin{array}{l}\overrightarrow {AB} \times \overrightarrow {AD} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\2&3&0\\4&{ - 2}&0\end{array}} \right|\\\overrightarrow {AB} \times \overrightarrow {AD} = \left| {\begin{array}{*{20}{c}}3&0\\{ - 2}&0\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}2&0\\4&0\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}2&3\\4&{ - 2}\end{array}} \right|{\rm{k}}\\\overrightarrow {AB} \times \overrightarrow {AD} = (0 - 0){\rm{i}} - (0 - 0){\rm{j}} + ( - 4 - 12){\rm{k}}\\\overrightarrow {AB} \times \overrightarrow {AD} = 0{\rm{i}} - 0{\rm{j}} - 16{\rm{k}}\end{array}\)

Calculate the area of the parallelogram

As given vertices as follows

Substitute the vector \(0{\rm{i}} - 0{\rm{j}} - 16{\rm{k}}\) for \(\overrightarrow {AB} \times \overrightarrow {AD} \)

\(\begin{array}{l}{\rm{Area of the parallelogram}} = |0{\rm{i}} - 0{\rm{j}} - 16{\rm{k}}|\\{\rm{Area of the parallelogram}} = \sqrt {{{(0)}^2} + {{(0)}^2} + {{( - 16)}^2}} \\{\rm{Area of the parallelogram}} = \sqrt {256} \\{\rm{Area of the parallelogram}} = 16\end{array}\)

Thus, the area of the parallelogram with vertices \(A( - 2,1),B(0,4),C(4,2)\) and \(D(2, - 1)\) is 16.