A vector a has components 3p,1 in a rectangular system.

Question

A vector $\vec a$ has components $3p,1$ in a rectangular system. After a counter-clockwise rotation, it has components $p+1,\sqrt{10}$. A value of $p$ is:

Accepted Answer

Rotation preserves magnitude: $(3p)^2+1 = (p+1)^2+10$
$$9p^2+1 = p^2+2p+11 \implies 8p^2-2p-10=0 \implies 4p^2-p-5=0 \implies (p+1)(4p-5)=0$$
$$p = \frac54 \text{ or } -1$$

Among the options, $p=-1$ is listed.

Solutions · 1