Chapter 11.3, Problem 36E

To determine

ToEvaluate:The area of the parallelogram with adjacent sides are $u=\langle 4,-3,2\rangle$ and $v=\langle 5,0,1\rangle$

Answer to Problem 36E

The area of the parallelogram is $A=3\sqrt{30}\text{ sq units}$

Explanation of Solution

Given:

The adjacent sides of parallelogram $u=\langle 4,-3,2\rangle$ and $v=\langle 5,0,1\rangle$

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 $u=\langle 4,-3,2\rangle$ and $v=\langle 5,0,1\rangle$

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 4& -3& 2\\ 5& 0& 1\end{array}\right|\\ =i\left|\begin{array}{cc}-3& 2\\ 0& 1\end{array}\right|-j\left|\begin{array}{cc}4& 2\\ 5& 1\end{array}\right|+k\left|\begin{array}{cc}4& -3\\ 5& 0\end{array}\right|\\ =i\left(\left(-3\right)\left(1\right)-\left(0\right)\left(2\right)\right)-j\left(\left(4\right)\left(1\right)-\left(5\right)\left(2\right)\right)+k\left(\left(4\right)\left(0\right)-\left(5\right)\left(-3\right)\right)\\ =i\left(-3-0\right)-j\left(4-10\right)+k\left(0+15\right)\end{array}$

  $=-3i+6j+15k$

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

  $\begin{array}{c}A=\sqrt{{\left(-3\right)}^2+{\left(6\right)}^2+{\left(15\right)}^2}\\ =\sqrt{9+36+225}\\ =\sqrt{270}\\ =3\sqrt{30}\end{array}$

Therefore, the area of the parallelogram is $A=3\sqrt{30}\text{ sq units}$

Conclusion:

The area of the parallelogram is $A=3\sqrt{30}\text{ sq units}$