Chapter 11.3, Problem 42E

To determine

The area of the parallelogram that has vectors $u=i+2j+2k$ and $v=i+k$ as adjacent sides.

Answer to Problem 42E

The area of parallelogram is $3$ .

Explanation of Solution

Given information:

The value of the vectors in component form $u=i+2j+2k$ and $v=i+k$ .

Formula used:

If $a$ and $b$ are two vectors that represents two adjacent sides of the parallelogram, where $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ , then the area of parallelogram with its sides as a and b is equal to the magnitude of their cross product $\Vert a\times b\Vert$ .

Let $a\times b$ represents the vector x such that $x=x_{1}i+x_{2}j+x_{3}k$ then the magnitude of the vector $x$ is given by $\Vert x\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

If there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Calculation:

Consider the given value of the vectors $u=i+2j+2k$ and $v=i+k$ .

Recall that if there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Apply it,

So, the cross product of $u$ and $v$ vectors will be calculated as,

  $\begin{array}{c}u\times v=\begin{array}{ccc}i& j& k\\ 1& 2& 2\\ 1& 0& 1\end{array}\\ \text{= }\left(\left[2\times 1\right]-\left[0\times 2\right]\right)i-\left(\left[1\times 1\right]-\left[1\times 2\right]\right)j+\left(\left[1\times 0\right]-\left[2\times 1\right]\right)k\\ \text{=}\left(2-0\right)i-\left(1-2\right)j+\left(0-2\right)k\\ \text{=2}i+1j-2k\end{array}$

Recall that if $a$ and $b$ are two vectors that represents two adjacent sides of the parallelogram, where $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the area of parallelogram with its sides as a and b is equal to the magnitude of their cross product $\Vert a\times b\Vert$ .

Let $a\times b$ represents the vector $x$ such that $x=x_{1}i+x_{2}j+x_{3}k$ then the magnitude of the vector $x$ is given by $\Vert x\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

So, area of parallelogram with its adjacent sides as $u=\langle 4,4,-6\rangle$ and $v=\langle 0,4,6\rangle$ will be evaluated when we will calculate the magnitude of the vector $u\text{×}v$ .

  $\begin{array}{c}\Vert u\text{×}v\Vert =\sqrt{2^2+1^2+{\left(-2\right)}^2}\\ =\sqrt{4+1+4}\\ =\sqrt{9}\\ =3\end{array}$

Hence, the area of parallelogram is $3$ .