Chapter 11.3, Problem 43E

To determine

The area of the parallelogram that has vectors $u=4i+3j+7k$ and $v=2i-4j+6k$ as adjacent sides.

Answer to Problem 43E

The area of parallelogram is $51.96$ .

Explanation of Solution

Given information:

The value of the vectors in component form $u=4i+3j+7k$ and $v=2i-4j+6k$ .

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 $\text{a}\times \text{b}$ 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, $\text{a}=\langle a_1,a_2,a_3\rangle$ and $\text{b}=\langle b_1,b_2,b_3\rangle$ , 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 $u=4i+3j+7k$ and $v=2i-4j+6k$ .

Recall that if there are two three-dimensional vectors in a space given in the form of their components, say, $\text{a}=\langle a_1,a_2,a_3\rangle$ and $\text{b}=\langle b_1,b_2,b_3\rangle$ , 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,

  $\begin{array}{c}u\times v=\begin{array}{ccc}i& j& k\\ 4& 3& 7\\ 2& -4& 6\end{array}\\ \text{= }\left(\left[6\times 3\right]-\left[7\times \left(-4\right)\right]\right)i-\left(\left[6\times 4\right]-\left[7\times 2\right]\right)j+\left(\left[4\times \left(-4\right)\right]-\left[2\times 3\right]\right)k\\ \text{=}\left(18-\left(-28\right)\right)i-\left(24-14\right)j+\left(-16-6\right)k\\ \text{=46}i-10j-22k\end{array}$

Recall that if a and b are two vectors that represents two adjacent sides of the parallelogram, where $\text{a}=\langle a_1,a_2,a_3\rangle$ and $\text{b}=\langle b_1,b_2,b_3\rangle$ , 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 $\Vert \text{x}\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

So, area of parallelogram with its adjacent sides as $u=4i+3j+7k$ and $v=2i-4j+6k$ will be evaluated when we will calculate the magnitude of the vector $\text{u×v}$ .

  $\begin{array}{c}\Vert u\text{×}v\Vert =\sqrt{{46}^2+{\left(-10\right)}^2+{\left(-22\right)}^2}\\ =\sqrt{2116+100+484}\\ =\sqrt{2700}\\ \approx 51.96\end{array}$

Hence, the area of parallelogram is $51.96$ .