Question

A vector $\bar{a}7$ has components $2p$ 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 $A(2p,1)$
A is rotated about origin in the anti-clock wise direction then new coordinate is $(p+1,1)$
$\because 2p=(p+1)\cos \theta -\sin \theta \dots (i)$
and $1=(p+1)\sin \theta +\cos \theta \dots$ (ii)
Squaring and adding Eqs. (i)and (ii), we get
$4p^2+1=(p+1{)}^2+1$
$\Rightarrow 4p^2=p^2+2p+1$
$\Rightarrow 3p^2-2p-1=0$
$\Rightarrow (p-1)(3p+1)=0$
$\Rightarrow p=1,\frac{-1}{3}$