Question

Determine the area of the parallelogram with vertices\(K(1,2,3),L(1,3,6),M(3,8,6)\) and\(N(3,7,3)\).

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 \(\overrightarrow {KL} \)

Consider the vertices of the parallelogram are \(K(1,2,3),L(1,3,6),M(3,8,6)\) and \(N(3,7,3)\)

\(\begin{array}{l}\overrightarrow {KL} = L(1,3,6) - K(1,2,3)\\\overrightarrow {KL} = \langle (1 - 1),(3 - 2),(6 - 3)\rangle \\\overrightarrow {KL} = \langle 0,1,3\rangle \end{array}\)

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

\(\begin{array}{l}\overrightarrow {KN} = N(3,7,3) - K(1,2,3)\\\overrightarrow {KN} = \langle (3 - 1),(7 - 2),(3 - 3)\rangle \\\overrightarrow {KN} = \langle 2,5,0\rangle \end{array}\)

Hence, the three dimensional vectors \(\overrightarrow {KL} \) and \(\overrightarrow {KL} \) are \(\langle 0,1,3\rangle \) and\(\langle 2,5,0\rangle \).

Calculate the cross product of \(\overrightarrow {KL} \) and\(\overrightarrow {KN} \)

\(\begin{array}{l}\overrightarrow {KL} \times \overrightarrow {KN} = \left| {\begin{array}{*{20}{l}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\0&1&3\\2&5&0\end{array}} \right|\\\overrightarrow {KL} \times \overrightarrow {KN} = \left| {\begin{array}{*{20}{c}}1&3\\5&0\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}0&3\\2&0\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}0&1\\2&5\end{array}} \right|{\rm{k}}\\\overrightarrow {KL} \times \overrightarrow {KN} = (0 - 15){\rm{i}} - (0 - 6){\rm{j}} + (0 - 2){\rm{k}}\\\overrightarrow {KL} \times \overrightarrow {KN} = - 15{\rm{i}} + 6{\rm{j}} - 2{\rm{k}}\end{array}\)

Calculate the area of the parallelogram

As given vertices as follows

Substitute the vector \( - 15{\rm{i}} + 6{\rm{j}} - 2{\rm{k}}\) for \(\overrightarrow {KL} \times \overrightarrow {KN} \)

\(\begin{array}{l}{\rm{Area of the parallelogram}} = | - 15{\rm{i}} + 6{\rm{j}} - 2{\rm{k}}|\\{\rm{Area of the parallelogram}} = \sqrt {{{( - 15)}^2} + {{(6)}^2} + {{( - 2)}^2}} \\{\rm{Area of the parallelogram}} = \sqrt {225 + 36 + 4} \\{\rm{Area of the parallelogram}} = \sqrt {265} \end{array}\)

Thus, the area of the parallelogram with vertices \(K(1,2,3),L(1,3,6),M(3,8,6)\) and \(N(3,7,3)\) is \(\sqrt {265} \).