Question

If $\overrightarrow{\alpha }=3i-k,|\overrightarrow{\beta }|=\sqrt{5}$ and $\overrightarrow{\alpha }\cdot \overrightarrow{\beta }=3$, then the area of the parallelogram for which $\alpha$ and $\beta$ are adjacent sides, is

• A. $\frac{\sqrt{17}}{2}$
• B. $\frac{\sqrt{14}}{2}$
• C. $\frac{\sqrt{7}}{2}$
• D. $\sqrt{41}$
• E. $\sqrt{50}$

Answer 1

Answer: (D)

$\sqrt{41}$

Given, $\alpha =3i-k.|\beta |=\sqrt{5}$
and $\alpha \cdot \beta =3$
Now, $|\alpha |=\sqrt{9+1}=\sqrt{10}$
$\because \alpha \cdot \beta =3$
$\Rightarrow |\alpha ||\beta |\cos \theta =3$
$\Rightarrow \sqrt{10}\sqrt{5}\cos \theta =3$
$\Rightarrow \cos \theta =\frac{3}{\sqrt{50}}$
Thehrefore, $\sin \theta =\sqrt{1-\frac{9}{50}}(\because \sin \theta =\sqrt{1-{\cos }^2\theta })$
$=\sqrt{\frac{41}{50}}$
Since, $\alpha$ and $\beta$ are adjacent sides of the parallelogram.
Area of parallelogram $=|\alpha \times \beta |$
$=|\alpha ||\beta |\sin \theta$
$=\sqrt{10}\sqrt{5}\times \frac{\sqrt{41}}{\sqrt{50}}$ .....$(\because |n|=1)$
$=\sqrt{41}$.