Question

A vector $\overrightarrow{a}$ has components $2p$ & $1$ with respect to a rectangular cartesian system. The system is rotated through a certain angle about the origin in the counterclockwise sense. If with respect to the new system, $\overrightarrow{a}$ has components $p+1$ & $1$ then:

• A. $p=0$
• B. $p=1$ or $p=-1/3$
• C. $p=-1$ or $p=-1/3$
• D. $p=1$ or $p=-1$

Answer 1

The correct option is B $p=1$ or $p=-1/3$

Magnitude of A remains same,
Thus, $\sqrt{(2p{)}^2+1^2}=\sqrt{{(p+1)}^2+1^2}4p^2+1=p^2+2p+23p^2-2p-1=0$

Solving the quadratic, we get:
$p=\frac{2\pm \sqrt{{(-2)}^2-\mathrm{4.3.}(-1)}}{2.3}=\frac{2\pm 4}{6}p=1orp=\frac{-1}{3}$