Question

Let $\vec{a},$ $\vec{b}$ & $\vec{c}$ be non-coplanar unit vectors equally inclined to one another at an acute angle $\theta$ . Then $\left|\left[\vec{a}\vec{b}\vec{c}\right]\right|$ in terms of $\theta$ is equal to

• A. $\left(1+cos\theta \right)\sqrt{cos2\theta }$
• B. $\left(1+cos\theta \right)\sqrt{1-2cos2\theta }$
• C. $\left(1-cos\theta \right)\sqrt{1+2cos\theta }$
• D. None of these

Answer 1

Answer: (C)

$\overrightarrow{OA}=\overrightarrow{a,}\overrightarrow{OB}=\vec{b}\&\overrightarrow{OC}=\vec{c}$ are unit vectors and equally inclined to each other at an acute angle $\theta$.
$\therefore$ ABC is an equilateral triangle
$andAB=\sqrt{O{A}^2+O{B}^2-2OA.OBcos\theta }$
$=\sqrt{2-2cos\theta }=\sqrt{2}\sqrt{1-cos\theta }$
$\therefore Areaof\Delta ABC,$
$=\frac{\sqrt{3}}{4}A{B}^2=\frac{\sqrt{3}}{4}\cdot 2\left(1-cos\theta \right)=\frac{\sqrt{3}}{2}\left(1-cos\theta \right)$
If G is the centroid of the $\Delta$ ABC, then
$OG=\frac{1}{3}\left|\vec{a}+\vec{b}+\vec{c}\right|$
$=\frac{1}{3}\sqrt{a^2+b^2+c^2+2\overrightarrow{a.}\vec{b}+2\vec{b}.\vec{c}+2\overrightarrow{c.}\vec{a}}$
$=\frac{1}{\sqrt{3}}\sqrt{1+2cos\theta }$
$\therefore \left[\vec{a}\vec{b}\vec{c}\right]=$ Volume of parallelopiped
$=OG\times 2ar\left(\Delta ABC\right)$
$2.\frac{1}{\sqrt{3}}\sqrt{1+2cos\theta }\times \frac{\sqrt{3}}{2}\left(1-cos\theta \right)$
$=\left(1-cos\theta \right)\sqrt{1+2cos\theta }$