Question

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

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

Answer 1

Answer: (C)

$|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
$\vec{a}\cdot \vec{b}=(1)(1)\cos \theta =\cos \theta$ and
$\vec{c}\cdot \vec{a}=\cos \theta ,\vec{b}\cdot \vec{c}=\cos \theta$
$\vec{a}\vec{b}{\vec{c}}^2=\left|\begin{array}{ccc}\vec{a}\cdot \vec{a}& \vec{a}\cdot \vec{b}& \vec{a}\cdot \vec{c}\\ \vec{b}\cdot \vec{a}& \vec{b}\cdot \vec{b}& \vec{b}\cdot \vec{c}\\ \vec{c}\cdot \vec{a}& \vec{c}\cdot \vec{b}& \vec{c}\cdot \vec{c}\end{array}\right|=\left|\begin{array}{ccc}1& \cos \theta & \cos \theta \\ \cos \theta & 1& \cos \theta \\ \cos \theta & \cos \theta & 1\end{array}\right|$
Operate $R_1\to R_1+R_2+R_3$
$=(1+2\cos \theta )\left|\begin{array}{ccc}1& 1& 1\\ \cos \theta & 1& \cos \theta \\ \cos \theta & \cos \theta & 1\end{array}\right|$
Operate $C_2\to C_2-C_1;C_3\to C_3-C_1$
$=(1+2\cos \theta )\left|\begin{array}{ccc}1& 0& 0\\ \cos \theta & 1-\cos \theta & 0\\ \cos \theta & 0& 1-\cos \theta \end{array}\right|$
$=(1+2\cos \theta )(1-\cos \theta {)}^2$
$\therefore [\vec{a}\vec{b}\vec{c}]=(1-\cos \theta )\sqrt{1+2\cos \theta }$