Question

Find the area of a parallelogram whose adjacent sides are given by the vectors
$$\overrightarrow{a}=3\hat{i}+\hat{j}+4\hat{k}$$ and $$\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$$ .

Answer 1

Hint: The adjacent side vectors of the parallelogram ABCD are $$\overrightarrow{a}=3\hat{i}+\hat{j}+4\hat{k}$$ and $$\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$$ . We know the formula that if the vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ are the adjacent sides of a parallelogram then the area of the parallelogram is the magnitude vector product of the adjacent sides of the parallelogram i.e., The area of the parallelogram = $$|\overrightarrow{a}\times \overrightarrow{b}|$$ . Use this formula and calculate the area vector. Now, get the magnitude of the area vector using the formula that magnitude of a vector $$xi+yj+zk$$ is $$\sqrt{x^2+y^2+z^2}$$ .

Complete step by step answer:
According to the question, we are given the adjacent sides of a parallelogram in vector form and we are asked to find the area of the parallelogram.
The vector form of side AB of the parallelogram is $$\overrightarrow{a}=3\hat{i}+\hat{j}+4\hat{k}$$ ……………………………………..(1)
The vector form of side BC of the parallelogram is $$\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$$ ……………………………………..(2)
Here, we have to find the area of the parallelogram.
We know the formula that if the vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ are the adjacent sides of a parallelogram then the area of the parallelogram is the magnitude vector product of the adjacent sides of the parallelogram i.e., The area of the parallelogram = $$|\overrightarrow{a}\times \overrightarrow{b}|$$ …………………………………..(3)

Now, from equation (1), equation (2), and equation (3), we get
The area of the parallelogram = $$[(3\hat{i}+\hat{j}+4\hat{k})\times (\hat{i}-\hat{j}+\hat{k})]$$ ………………………………………..(4)
We know the property that $$\hat{i}\times \hat{i}=0$$ , $$\hat{j}\times \hat{j}=0$$ , $$\hat{k}\times \hat{k}=0$$ , $$\hat{i}\times \hat{j}=\hat{k}$$ , $$\hat{i}\times \hat{k}=-\hat{j}$$ , $$\hat{j}\times \hat{i}=-\hat{k}$$ ,
$$\hat{j}\times \hat{k}=\hat{i}$$ , $$\hat{k}\times \hat{i}=\hat{j}$$ , and $$\hat{k}\times \hat{j}=\widehat{-i}$$ ……………………………………..(5)
Now, from equation (4) and equation (5), we get
The area of the parallelogram ABCD = $$[(3\hat{i}+\hat{j}+4\hat{k})\times (\hat{i}-\hat{j}+\hat{k})]$$ = $$(5\hat{i}+\hat{j}-4\hat{k})$$ ……………………………………..(6)
We know the formula for the magnitude of a vector $$xi+yj+zk$$ , Magnitude = $$\sqrt{x^2+y^2+z^2}$$ …………………………………………………(7)
Now, from equation (6) and equation (7), we get
The area of the parallelogram = $$\sqrt{5^2+1^2+{(-4)}^2}=\sqrt{25+1+16}=\sqrt{42}$$ .
Therefore, the area of the parallelogram is $$\sqrt{42}$$ sq. units.

Note:
 We can also solve this question using the matrix method formula that is, the area of the parallelogram whose adjacent side vectors are $$x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$$ and $$x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$$ is given by $$\left|\begin{array}{rl}& \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\end{array}\\ & \begin{array}{ccc}{x}_1& y_1& z_1\end{array}\\ & \begin{array}{ccc}{x}_2& y_2& z_2\end{array}\end{array}\right|$$ .