Question

If $\overrightarrow{a},\overrightarrow{b},$ and $\overrightarrow{c}$ are unit vectors such that $\overrightarrow{a}+2\overrightarrow{b}+2\overrightarrow{c}=\overrightarrow{0},$ then $|\overrightarrow{a}\times \overrightarrow{c}|$ is equal to :

• A. $\frac{\sqrt{15}}{4}$
• B. $\frac{1}{4}$
• C. $\frac{15}{16}$
• D. $\frac{\sqrt{15}}{16}$

Answer 1

The correct option is A $\frac{\sqrt{15}}{4}$

$\begin{array}{c}\text{Given that}\\ \begin{array}{c}\overrightarrow{a}+2\overrightarrow{b}+2\overrightarrow{c}=0\\ \text{and}|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=01\end{array}\\ \text{Now,}\overrightarrow{a}+2\overrightarrow{c}=-2\overrightarrow{b}\\ |\overrightarrow{a}+2\overrightarrow{c}{|}^2=2{\overrightarrow{b}}^2\\ {\overrightarrow{a}}^2+4{\overrightarrow{c}}^2+\Delta |\overrightarrow{a}||c|\cos \theta =2{\overrightarrow{b}}^2\\ 1+4+4\cos \theta =2\\ \Rightarrow \cos \theta =-\frac{1}{4}\end{array}$

$\begin{array}{c}\text{using}{\cos }^2\theta +{\sin }^2\theta =1\\ \Rightarrow \sin \theta =\frac{\sqrt{15}}{4}\\ \therefore |\overrightarrow{a}\times \overrightarrow{c}|=|\overrightarrow{a}||\overrightarrow{c}|\sin \theta \\ =\frac{\sqrt{15}}{4}\\ \text{hence,}\left(A\right)\text{is the courect option}\end{array}$