Question

If $\vec{x}+\vec{y}+\vec{z}=\vec{0},âˆ\pounds \vec{x}âˆ\pounds =âˆ\pounds \vec{y}âˆ\pounds =âˆ\pounds \vec{z}âˆ\pounds =2$ and $θ$ is angle between $\vec{y}$ and $\vec{z},$ then the value of $\text{cose}{\text{c}}^2θ+{\cot }^2θ$ is equal to

• A. $4/3$
• B. $5/3$
• C. $1/3$
• D. $1$

Answer 1

Answer: (B)

Given, $\vec{x}+\vec{y}+\vec{z}=\vec{0}$ and
$âˆ\pounds \vec{x}âˆ\pounds =âˆ\pounds \vec{y}âˆ\pounds =âˆ\pounds \vec{z}âˆ\pounds =2$
$∴$ $\vec{x}=−\vec{y}−\vec{z}$
$⇒$ ${\vec{x}}^2={(\vec{y}+\vec{z})}^2$ $⇒$
$âˆ\pounds \vec{x}{âˆ\pounds }^2=âˆ\pounds \vec{y}{âˆ\pounds }^2+âˆ\pounds \vec{z}{âˆ\pounds }^2+2âˆ\pounds yâˆ\pounds âˆ\pounds zâˆ\pounds \cos θ$
$⇒$ $4=4+4+2×2×2\cos θ$
$⇒$ $\cos θ=−\frac{4}{8}=−\frac{1}{2}=\cos {120}^o$
$⇒$ $θ={120}^o$
Now, $\text{cose}{\text{c}}^2θ+{\cot }^2θ=\text{cose}{\text{c}}^2{120}^o+{\cot }^2{120}^o$
$={\left(\frac{2}{\sqrt{3}}\right)}^2+{\left(−\frac{1}{\sqrt{3}}\right)}^2$
$=\frac{4}{3}+\frac{1}{3}=\frac{5}{3}$