The area of a parallelogram whose adjacent sides are given by the vectors $\widehat{\mathbf{i}}\mathbf{+}2\widehat{\mathbf{j}}\mathbf{+}3\widehat{\mathbf{k}}$ and $- 3\widehat{\mathbf{i}}\mathbf{-}2\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$ (in sq unit), is

Question

Dr. Abubacker · Accepted Answer

(a)
Adjacent sides of parallelogram are $\overrightarrow{\mathbf{a}} = \widehat{\mathbf{i}} + 2\widehat{\mathbf{j}} + 3\widehat{\mathbf{k}}\ and\ \overrightarrow{\mathbf{b}} = - 3\widehat{\mathbf{i}} - 2\widehat{\mathbf{j}} + \widehat{\mathbf{k}}$
Now, $\overrightarrow{\mathbf{a}}\mathbf{	imes}\overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \
1 & 2 & 3 \
 - 3 & - 2 & 1
\end{matrix} ight|$
$= \widehat{\mathbf{i}}(2 + 6) - \widehat{\mathbf{j}}(1 + 9) + \widehat{\mathbf{k}}( - 2 + 6)$
$= 8\widehat{\mathbf{i}} - 10\widehat{\mathbf{j}} + 4\widehat{\mathbf{k}}$
Therefore, area of parallelogram
$= \left| \overrightarrow{\mathbf{a}}\mathbf{	imes}\overrightarrow{\mathbf{b}} ight|$
$= \sqrt{(8)^{2} + ( - 10)^{2} + (4)^{2}}\$
$= \sqrt{64 + 100 + 16} = \sqrt{180}$ sq unit

Q. The area of a parallelogram whose adjacent sides are given by the vectors $\widehat{\mathbf{i}}\mathbf{+}2\widehat{\mathbf{j}}\mathbf{+}3\widehat{\mathbf{k}}$ and $- 3\widehat{\mathbf{i}}\mathbf{-}2\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$ (in sq unit), is

Explanation

Reviewed by Dr. Abubacker

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