Name: Cross Product
Uploaded: 2019-07-19
Duration: 4 min 16 s
Description: The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a particular direction and magnitude. Watch this vi...

Question

Find the area of the parallelogram (in terms of sq units) whose adjacent sides are represented by the vectors: $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \vec{b} = - 3 \hat{i} - 2 \hat{j} + \hat{k}$

Accepted Answer

The area of the parallelogram $= | \vec{a} \times \vec{b} |$ , where $\vec{a}$ and $\vec{b}$ are vectors of it’s adjacent sides.

$\vec{a} \times \vec{b} = (a_{2} b_{3} - b_{2} a_{3}) \hat{i} + (a_{3} b_{1} - b_{3} a_{1}) \hat{j} + (a_{1} b_{2} - b_{1} a_{2}) \hat{k} . . . (i)$
Here, We have $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} and \vec{b} = - 3 \hat{i} - 2 \hat{j} + \hat{k}$ $\Rightarrow a_{1} = 1, a_{2} = 2, a_{3} = 3 and b_{1} = - 3, b_{2} = - 2, b_{3} = 1$

Thus, substituting the values of $a_{1}, a_{2}, a_{3} and b_{1}, b_{2} and b_{3}$ , in equation $(i)$ we get

$\Rightarrow \vec{a} \times \vec{b} = (2 \times 1 - (- 2 \times 3)) \hat{i} + ((3 \times - 3) - 1 \times 1) \hat{j} + ((1 \times - 2) - (- 3 \times 2)) \hat{k}$

$\Rightarrow \vec{a} \times \vec{b} = (8) \hat{i} + (- 10) \hat{j} + (4) \hat{k}$

$\Rightarrow Area = | \vec{a} \times \vec{b} |$

$\Rightarrow | \vec{a} \times \vec{b} | = \sqrt{(8)^{2} + (- 10)^{2} + (4)^{2}}$

$\Rightarrow | \vec{a} \times \vec{b} | = \sqrt{180}$

Therefore, area equal to $6 \sqrt{5} sq units$

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