Question

Find the area of a parallelogram for which the vectors $2\hat{i}+\hat{j}$ and $3\hat{i}+\hat{j}+4\hat{k}$ are adjacent sides.

Answer 1

$\overrightarrow{a}$ = $2\hat{i}+\hat{j}$

$\overrightarrow{b}$ = $3\hat{i}+\hat{j}+4\hat{k}$

$\left|\overrightarrow{a}\right|$ = $\sqrt{2^2+1^2}=\sqrt{5}$

$\left|\overrightarrow{b}\right|$ = $\sqrt{3^2+1^2+4^2}=\sqrt{26}$

$\overrightarrow{a}$ $.$ $\overrightarrow{b}$ = $\left|\overrightarrow{a}\right|$ $\left|\overrightarrow{b}\right|$ $cos\theta$

$(2\hat{i}+\hat{j})$ $.$ $(3\hat{i}+\hat{j}+4\hat{k})$ = $\sqrt{5}\ast \sqrt{26}\cos \theta$

$2\times 3+1\times 1+0\times 4=\sqrt{130}\cos \theta$

$⟹\cos \theta =\frac{7}{\sqrt{130}}$

$\therefore \sin \theta =\frac{\sqrt{130-7^2}}{\sqrt{130}}=\frac{9}{\sqrt{130}}$

Area of parallelogram $=\left|\overrightarrow{a}\right|$ $\left|\overrightarrow{b}\right|$ $\sin \theta$ = $\sqrt{5}\times \sqrt{26}\times \frac{9}{\sqrt{130}}$

$=9$