Chapter 11.3, Problem 39E

To determine

ToEvaluate:The area of the parallelogram with adjacent sides are $u=4i+3j+7k$ and $v=2i-4j+6k$

Answer to Problem 39E

The area of the parallelogram is $A=2\sqrt{579}\text{ sq units}$

Explanation of Solution

Given:

The adjacent sides of parallelogram $u=4i+3j+7k$ and $v=2i-4j+6k$

Concept Used:

For the two adjacent sides of parallelogram $u=\langle u_1,u_2,u_3\rangle$ and $v=\langle v_1,v_2,v_3\rangle$ , the areacan be calculated as $A=\Vert u\times v\Vert$ where $u\times v=\left|\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|$

Calculation:

Forthe adjacent sides of parallelogram $u=4i+3j+7k$ and $v=2i-4j+6k$

That is

  $u=\langle 4,3,7\rangle$ and $u=\langle 2,-4,6\rangle$

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 4& 3& 7\\ 2& -4& 6\end{array}\right|\\ =i\left|\begin{array}{cc}3& 7\\ -4& 6\end{array}\right|-j\left|\begin{array}{cc}4& 7\\ 2& 6\end{array}\right|+k\left|\begin{array}{cc}4& 3\\ 2& -4\end{array}\right|\\ =i\left(\left(3\right)\left(6\right)-\left(-4\right)\left(7\right)\right)-j\left(\left(4\right)\left(6\right)-\left(2\right)\left(7\right)\right)+k\left(\left(4\right)\left(-4\right)-\left(2\right)\left(3\right)\right)\\ =i\left(18+28\right)-j\left(24-14\right)+k\left(-16+6\right)\end{array}$

  $=46i-10j-10k$

The area is $A=\Vert u\times v\Vert$

  $\begin{array}{c}A=\sqrt{{\left(46\right)}^2+{\left(-10\right)}^2+{\left(-10\right)}^2}\\ =\sqrt{2116+100+100}\\ =\sqrt{2316}\\ =2\sqrt{579}\end{array}$

Therefore, the area of the parallelogram is $A=2\sqrt{579}\text{ sq units}$

Conclusion:

The area of the parallelogram is $A=2\sqrt{579}\text{ sq units}$