Chapter 11.3, Problem 40E

To determine

ToEvaluate:The area of the parallelogram with adjacent sides are $u=-2i+3j+2k$ and $v=1i+2j+4k$

Answer to Problem 40E

The area of the parallelogram is $A=\sqrt{213}\text{ sq units}$

Explanation of Solution

Given:

The adjacent sides of parallelogram $u=-2i+3j+2k$ and $v=1i+2j+4k$

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 $u=-2i+3j+2k$ and $v=1i+2j+4k$

That is

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

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ -2& 3& 2\\ 1& 2& 4\end{array}\right|\\ =i\left|\begin{array}{cc}3& 2\\ 2& 4\end{array}\right|-j\left|\begin{array}{cc}-2& 2\\ 1& 4\end{array}\right|+k\left|\begin{array}{cc}-2& 3\\ 1& 2\end{array}\right|\\ =i\left(\left(3\right)\left(4\right)-\left(2\right)\left(2\right)\right)-j\left(\left(-2\right)\left(4\right)-\left(1\right)\left(2\right)\right)+k\left(\left(-2\right)\left(2\right)-\left(1\right)\left(3\right)\right)\\ =i\left(12-4\right)-j\left(-8-2\right)+k\left(-4-3\right)\end{array}$

  $=8i+10j-7k$

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

  $\begin{array}{c}A=\sqrt{{\left(8\right)}^2+{\left(10\right)}^2+{\left(-7\right)}^2}\\ =\sqrt{64+100+49}\\ =\sqrt{213}\end{array}$

Therefore, the area of the parallelogram is $A=\sqrt{213}\text{ sq units}$

Conclusion:

The area of the parallelogram is $A=\sqrt{213}\text{ sq units}$