Question

If $x+y+z=0,\left|x\right|=\left|y\right|=\left|z\right|=2$ and $\theta$ is angle between $y\mathrm{and}z$, then the value of $2\cos e{c}^2\theta +3co{t}^2\theta$ is:

• A. $\frac{11}{3}$
• B. $\frac{8}{3}$
• C. $\frac{5}{3}$
• D. $1$

Answer 1

The correct option is A

$\frac{11}{3}$

Explanation for the correct option.

Step 1: Find the value of $\overrightarrow{y}·\overrightarrow{z}$.

Given $x+y+z=0$, so it can be written as $\overrightarrow{x}+\overrightarrow{y}+\overrightarrow{z}=0....\left(1\right)$

Multiplying $\left(1\right)$ by $\overrightarrow{x}$, we get

$$\begin{gathered} \overrightarrow{x}·\overrightarrow{x}+\overrightarrow{y}·\overrightarrow{x}+\overrightarrow{z}·\overrightarrow{x}=0·\overrightarrow{x} \\ \Rightarrow \left|x^2\right|+\overrightarrow{y}·\overrightarrow{x}+\overrightarrow{z}·\overrightarrow{x}=0....\left(2\right) \end{gathered}$$

Similarly, by multiplying $\left(1\right)$ by $\overrightarrow{y},\overrightarrow{z}$, we get

$$\begin{gathered} \left|y^2\right|+\overrightarrow{x}·\overrightarrow{y}+\overrightarrow{z}·\overrightarrow{y}=0....\left(3\right) \\ \left|z^2\right|+\overrightarrow{x}·\overrightarrow{z}+\overrightarrow{y}·\overrightarrow{z}=0....\left(4\right) \end{gathered}$$

Now, $\left(3\right)+\left(4\right)$ will be

$\begin{array}{rcl}\left|y^2\right|+\left|z^2\right|+\overrightarrow{x}·\overrightarrow{y}+\overrightarrow{x}·\overrightarrow{z}+2\overrightarrow{y}·\overrightarrow{z}& =& 0\\ \Rightarrow -\left(\left|y^2\right|+\left|z^2\right|+2\overrightarrow{y}·\overrightarrow{z}\right)& =& \overrightarrow{x}·\overrightarrow{y}+\overrightarrow{x}·\overrightarrow{z}\end{array}$

By substituting the value of $\begin{array}{rcl}& & \overrightarrow{x}·\overrightarrow{y}+\overrightarrow{x}·\overrightarrow{z}\end{array}$ in $\left(2\right)$, we get

$\begin{array}{rcl}\left|x^2\right|-\left|y^2\right|-\left|z^2\right|-2\overrightarrow{y}·\overrightarrow{z}& =& 0\\ \Rightarrow 4-4-4-2\overrightarrow{y}·\overrightarrow{z}& =& 0\left[\mathrm{given}\left|x\right|=\left|y\right|=\left|z\right|=2\right]\\ \Rightarrow 2\overrightarrow{y}·\overrightarrow{z}& =& -4\\ \Rightarrow \overrightarrow{y}·\overrightarrow{z}& =& -2\end{array}$

Step 2: Find the value of $\cos \theta$

$\overrightarrow{y}·\overrightarrow{z}$ can be written as

$\begin{array}{rcl}\left|y\right|\left|z\right|\cos \theta & =& -2\left[\mathrm{given},\mathrm{angle}\mathrm{bwteen}y\mathrm{and}z\mathrm{is}\theta \right]\\ & \Rightarrow & \left(2\times 2\right)\cos \theta =-2\\ \Rightarrow \cos \theta & =& -\frac{1}{2}\\ \Rightarrow {\cos }^2\theta & =& \frac{1}{4}\\ \Rightarrow {\sin }^2\theta & =& \frac{3}{4}\left[\text{by}{\cos }^2\theta +{\sin }^2\theta =1\right]\end{array}$

Step 3: Find the value of $2\cos e{c}^2\theta +3co{t}^2\theta$

$$\begin{gathered} 2\cos e{c}^2\theta +3co{t}^2\theta =2\left(\frac{1}{{\sin }^2\theta }\right)+3\left(\frac{{\cos }^2\theta }{{\sin }^2\theta }\right) \\ =2\left(\frac{4}{3}\right)+3\left(\frac{1}{3}\right) \\ =\frac{8}{3}+1 \\ =\frac{11}{3} \end{gathered}$$

Hence, option A is correct.