Question

If $\overrightarrow{X}+\overrightarrow{Y}+\overrightarrow{Z}=\overrightarrow{0}$ ,$\left|\overrightarrow{X}\right|=\left|\overrightarrow{Y}\right|=\left|\overrightarrow{Z}\right|=2$ where $\theta$ is the angle between the vectors $\overrightarrow{Y}$ and $\overrightarrow{Z}$ then the value of $2\cos e{c}^2\theta +3co{t}^2\theta$ is:

• A. $\frac{11}{3}$
• B. $\frac{8}{3}$
• C. $\frac{7}{3}$
• D. $1$

Answer 1

The correct option is A

$\frac{11}{3}$

Explanation for the correct answer:

$\overrightarrow{X}+\overrightarrow{Y}+\overrightarrow{Z}=\overrightarrow{0}$

$\Rightarrow$ $\overrightarrow{Y}+\overrightarrow{Z}=-\overrightarrow{X}$

Taking modulus on both sides we get

$\Rightarrow$$\left|\overrightarrow{Y}+\overrightarrow{Z}\right|=\left|\overrightarrow{X}\right|$

Squaring both sides we get

$\Rightarrow$ ${\left|\overrightarrow{Y}+\overrightarrow{Z}\right|}^2={\left|\overrightarrow{X}\right|}^2$

$\Rightarrow$${\left|\overrightarrow{Y}\right|}^2+{\left|\overrightarrow{Z}\right|}^2+2\left|\overrightarrow{Y}\right|\left|\overrightarrow{Z}\right|\cos \theta ={\left|\overrightarrow{X}\right|}^2$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \left|A+B\right|={\left|A\right|}^2+{\left|B\right|}^2+\left|A\right|\left|B\right|cos\alpha \right]$ ,where $\alpha$ is angle between $A$and $B$

Substituting value of $\left|\overrightarrow{X}\right|=\left|\overrightarrow{Y}\right|=\left|\overrightarrow{Z}\right|=2$ we get

$\Rightarrow$ $8+8\cos \theta =4$

$\Rightarrow$ $\cos \theta =\frac{-1}{2}$

$\Rightarrow$ ${\cos }^2\theta =\frac{1}{4}$$...\left(i\right)$

$\Rightarrow$ ${\sin }^2\theta =1-\frac{1}{4}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}$

$\Rightarrow$ ${\sin }^2\theta =\frac{3}{4}$

To find the value of the required expression we substitute the values from $\left(i\right),\left(ii\right)$

$2\cos e{c}^2\theta +3co{t}^2\theta =\frac{2}{{\sin }^2\theta }+\frac{3{\cos }^2\theta }{{\sin }^2\theta }$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because sin\theta =\frac{1}{cosec\theta },cot\theta =\frac{cos\theta }{sin\theta }\right]$

$\Rightarrow$ $=\frac{2}{{\displaystyle \frac{3}{4}}}+\frac{3\times {\displaystyle \frac{1}{4}}}{{\displaystyle \frac{3}{4}}}$

$\Rightarrow$ $2\cos e{c}^2\theta +3co{t}^2\theta =\frac{11}{3}$

Hence, the value of $2\cos e{c}^2\theta +3co{t}^2\theta$ is $\frac{11}{3}$ .

Hence, option(A) is the correct answer