Question

A vector a has components $2\mathrm{p}$ and 1 w.r.t in a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If w.r.t the new system, a has components $\mathrm{p}+1$ and $1$, then

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Solution

Step 1: Given and assumeLet,The unit vectors along the coordinate axes are i and j.$\therefore \stackrel{\to }{\mathrm{a}}=2\mathrm{pi}+1.\mathrm{j}...\left(1\right)$On rotation, the vector has components $\mathrm{p}+1$ and $1$is $\mathrm{b}$$\therefore \stackrel{\to }{\mathrm{b}}=\left(\mathrm{p}+1\right)\mathrm{i}+1.\mathrm{j}...\left(2\right)$Step 2: CalculationThe magnitude of the initial and end vectors will stay constant during rotation.$⇒\left|\stackrel{⇀}{\mathrm{b}}\right|=\left|\stackrel{⇀}{\mathrm{a}}\right|\phantom{\rule{0ex}{0ex}}{\left|\mathrm{b}\right|}^{2}={\left|\mathrm{a}\right|}^{2}$$⇒{\left(\mathrm{p}+1\right)}^{2}+1={\left(2\mathrm{p}\right)}^{2}+1\phantom{\rule{0ex}{0ex}}⇒3{\mathrm{p}}^{2}-2\mathrm{p}-1=0\phantom{\rule{0ex}{0ex}}⇒\left(3\mathrm{p}+1\right)\left(\mathrm{p}-1\right)=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{p}=1\phantom{\rule{0ex}{0ex}}\mathrm{p}=-\frac{1}{3}.$

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