Browse Questions

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\overrightarrow a = \hat i - \hat j + 3\hat k$ and $\overrightarrow b = 2\hat i - 7 \hat j + \hat k$

$\begin{array}{1 1} (A) \large\frac{15}{\sqrt 2} \;cubic units. \\ (B) \;15\sqrt 2 \;sq units. \\ (C) 30\sqrt 2 \;sq units. \\ (D) \;\large\frac{30}{\sqrt 2} \;cubic \;units. \end{array}$

Toolbox:
• Area of the parallelogram = $|\overrightarrow a \times \overrightarrow b |$ where $\overrightarrow a\: and \: \overrightarrow b$ are adjacent sides.
• $\overrightarrow a\times \overrightarrow b = \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Step 1:
Let $\overrightarrow a=\hat i-\hat j+3\hat k$
$\quad\;\overrightarrow b=2\hat i-7\hat j+\hat k$
Area of the parallelogram is the cross product of its adjacent sides.
Here $\overrightarrow{AB}=\overrightarrow a=\hat i-\hat j+3\hat k$
$\quad\;\;\overrightarrow{AD}=\overrightarrow b=2\hat i-7\hat j+\hat k$
Therefore area of the parallelogram is given by $|\overrightarrow a\times\overrightarrow b|$
Step 2:
Let us determine $\overrightarrow a\times \overrightarrow b$.
$\overrightarrow a\times \overrightarrow b = \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
$\overrightarrow a\times \overrightarrow b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$
$\qquad\;\;=\hat i(-1\times 1-3\times -7)-\hat j(1\times 1-3\times 2)+\hat k(1\times -7-2\times -1)$
$\qquad\;\;=\hat i(-1+21)-\hat j(1-6)+\hat k( -7+2)$
$\qquad\;\;=20\hat i+5\hat j-5\hat k$
Step 3:
Area of the parallelogram = $|\overrightarrow a \times \overrightarrow b |$
$|\overrightarrow a\times\overrightarrow b|=\sqrt{{20}^2+5^2+(-5)^2}$
$\qquad\qquad=\sqrt{400+25+25}$
$\qquad\qquad=\sqrt{450}$
$\qquad\qquad=15\sqrt 2$
Therefore area of the parallelogram is $15\sqrt 2$sq.units.