First slide

Scalars and Vectors

Question

The area of a parallelogram formed from the vectors $\vec{A} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ as adjacent sides is

Moderate

A

$8 \sqrt{3} u n i t s$
B

64 units
C

32 units
D

$4 \sqrt{6} u n i t s$

Solution

$\vec{A} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{B} = 3 \hat{i} - 2 \hat{j} + \hat{k}$
Area of parallelogram $= | \vec{A} \times \vec{B} |$

Now, $\vec{A} \times \vec{B} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 3 & - 2 & 1 \end{matrix} | = 4 \hat{i} + 8 \hat{j} + 4 \hat{k}$

$| \vec{A} \times \vec{B} | = \sqrt{4^{2} + 8^{2} + 4^{2}} = \sqrt{96} = 4 \sqrt{6} u n i t s$ .

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