Question

The components of a vector along the $x$- and $y$-directions are $(n+1)$ and $1$, respectively. If the coordinate system is rotated by an angle $\theta ={60}^o$, then the components change to $n$ and $3$. The value of $n$ is

• A. $2$
• B. $1+\sqrt{3}$
• C. $1-\sqrt{3}$
  
• D. $1\pm \sqrt{3}$
  

Answer 1

The correct option is D $1\pm \sqrt{3}$

Let the vector is A having magnitude a and making an angle x with x axis.so, the components are

$a\cos x=n+1$

$a\sin x=1$

When coordinates are rotated by 60⁰, the angle can be $x+60orx-60$ ,the new coordinates are

$a\cos (x\pm 60)=n$

$a\sin (x\pm 60)=3$

Now, solving four equations we get

$a\cos (x\pm 60)=a(\mathrm{cosx}\cos 60\pm \mathrm{sinx}\sin 60)$

$n=a(\frac{1}{2}\mathrm{cosx}\pm \frac{\sqrt{3}}{2}sinx)$

$put,a\cos x=n+1anda\sin x=1$

$so,n=1\pm \sqrt{3}$

n depends on direction of rotation.