Question

A vector $\bar{a} 7$ has components $2 p$ and $1$ with respect to a two dimensional rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter - clockwise direction. If $\bar{a}$ has components $p+1$ and 1 with respect to the new system, then

• A. $p=1$ or $p=-\frac{1}{3}$
• B. $p=-1$ or $p=\frac{1}{3}$
• C. $p=1$ or $p=-1$
• D. $p=0$ or $p=\frac{1}{2}$

Answer 1

Answer: (A)

Let point $A(2 p, 1)$
A is rotated about origin in the anti-clock wise direction then new coordinate is $(p+1,1)$
$\because 2 p=(p+1) \cos \theta-\sin \theta \ldots (i) $
and $1=(p+1) \sin \theta+\cos \theta \ldots$ (ii)
Squaring and adding Eqs. (i)and (ii), we get
$4 p^{2}+1=(p+1)^{2}+1 $
$\Rightarrow 4 p^{2}=p^{2}+2 p+1 $
$\Rightarrow 3 p^{2}-2 p-1=0 $
$\Rightarrow (p-1)(3 p+1)=0$
$ \Rightarrow p=1, \frac{-1}{3}$