Chapter 9, Problem 25CT

To determine

To calculate: The area of parallelogram that has $u$ and $v$ as adjacent sides, where $u=2i-3j+k$ and $v=-i+3j+2k$ .

Answer to Problem 25CT

Solution:

The area of parallelogram that has $u$ and $v$ as adjacent sides is approximately $10.724$ square

units.

Explanation of Solution

Given information:

The vectors, $u=2i-3j+k$ and $v=-i+3j+2k$

Formula used:

1) Suppose $u=a_{1}i+b_{1}j+c_{1}k$ and $v=a_{2}i+b_{2}j+c_{2}k$ are two vectors in space, the cross product

  $u\times v$ is defined as the vector $u\times v=\left|\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$ .

2) If $v=a_{1}i+a_{2}j+a_{3}k$ , then $\Vert v\Vert =\sqrt{a_1{}^2+a_2{}^2+a_3{}^2}$ .

3) The area of a parallelogram with $u$ and $v$ as adjacent sides is given by $\Vert u\times v\Vert$ .

Calculation:

Let $u=2i-3j+k$ and $v=-i+3j+2k$ .

By using the formula of $u\times v$,

  $u\times v=\left|\begin{array}{ccc}i& j& k\\ 2& -3& 1\\ -1& 3& 2\end{array}\right|$

  $=\left|\begin{array}{cc}-3& 1\\ 3& 2\end{array}\right|i-\left|\begin{array}{cc}2& 1\\ -1& 2\end{array}\right|j+\left|\begin{array}{cc}2& -3\\ -1& 3\end{array}\right|k$

  $=\left(-6-(3)\right)i-\left(4-\left(-1\right)\right)j+\left(6-(3)\right)k$

  $=-9i-5j+3k$

Therefore, $u\times v=-9i-5j+3k$ .

Form area of parallelogram,

  $Area=\Vert u\times v\Vert$

By using norm,

  $\text{Area}=\Vert u\times v\Vert =\sqrt{{\left(-9\right)}^2+{\left(-5\right)}^2+3^2}$

  $\Rightarrow \Vert u\times v\Vert =\sqrt{81+25+9}$

  $\Rightarrow \Vert u\times v\Vert =\sqrt{115}$

  $\Rightarrow \Vert u\times v\Vert \approx 10.72380529$

  $\Rightarrow \Vert u\times v\Vert \approx 10.724$

Therefore, the area of parallelogram that has $u$ and $v$ as adjacent sides is approximately $10.724$

square units.