Question

How do you find the area of parallelogram with vertices \[k\left( {1,2,3} \right),l\left( {1,3,6} \right),m\left( {3,8,6} \right)\] and \[n\left( {3,7,3} \right)\]?

Answer 1

Hint: In the given question, we have been given a parallelogram with the coordinates of its vertices. We have to find the area of the given parallelogram. First, we are going to draw the figure of the parallelogram. Then we are going to find the area. To do that, we first express any two adjacent vectors. Then we find their cross product using matrix and find the answer.

Formula Used:
We are going to use the formula of area of parallelogram with two adjacent vectors \[\vec a\] and \[\vec b\],
Area, \[A = \vec a \times \vec b\]

Complete step-by-step answer:

The vertices of the parallelogram are \[k\left( {1,2,3} \right),l\left( {1,3,6} \right),m\left( {3,8,6} \right)\] and \[n\left( {3,7,3} \right)\].
Now, \[\overrightarrow {KL} = \left( {1,3,6} \right) - \left( {1,2,3} \right) = \left( {0,1,3} \right)\]
and, \[\overrightarrow {KN} = \left( {3,7,3} \right) - \left( {1,2,3} \right) = \left( {2,5,0} \right)\]
Now, we are going to use the formula of area of parallelogram with two adjacent vectors \[\vec a\] and \[\vec b\],
Area, \[A = \vec a \times \vec b\]
Hence, \[A = \left| {\left( {0,1,3} \right) \times \left( {2,5,0} \right)} \right|\]
\[A = \left| {\begin{array}{*{20}{c}}i&j&k\\0&1&3\\2&5&0\end{array}} \right|\]
Now applying the formula of matrices,
\[A = \left| {i\left| {\begin{array}{*{20}{c}}1&3\\5&0\end{array}} \right| - j\left| {\begin{array}{*{20}{c}}0&3\\2&0\end{array}} \right| + k\left| {\begin{array}{*{20}{c}}0&1\\2&5\end{array}} \right|} \right|\]
\[A = \left| {i\left( {0 - 15} \right) - j\left( {0 - 6} \right) + k\left( {0 - 2} \right)} \right|\]
Solving the expression,
$\Rightarrow$ \[A = \left| { - 15i + 6j - 2k} \right|\]
Now, finding the value of the area,
$\Rightarrow$ \[A = \sqrt {{{\left( { - 15} \right)}^2} + {{\left( 6 \right)}^2} + {{\left( { - 2} \right)}^2}} \]
Opening the squares and adding,
$\Rightarrow$ \[A = \sqrt {225 + 36 + 4} \]
Hence, the area of the parallelogram is \[\sqrt {265} \] square units.

Note: In the given question, we had to calculate the area of a parallelogram with given vertices. We found the area by converting the vertices into vector form, taking just two adjacent ones, finding their matrix product and calculating the answer. But it is very important that we know how anything of such form is done and how we can use the formulae to do that.