Aspire Faculty ID #16318 · Topic: NIMCET 2010 · Just now
NIMCET 2010

A vector $\vec{a}$ has components $2p$ and $1$. After rotation, it becomes $(p+1,\ 1)$. Find $p$.

Solution

Length of vector does not change under rotation: $\sqrt{(2p)^2 + 1^2} = \sqrt{(p+1)^2 + 1^2}$ Square both sides: $4p^2 + 1 = (p+1)^2 + 1$ $4p^2 + 1 = p^2 + 2p + 2$ $3p^2 - 2p - 1 = 0$ Solve: $p = 1$ or $p = -\frac{1}{3}$ But the option lists $\frac13$ (positive), so valid match is:

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