Question

The sum of magnitude of two vectors is $36$ units and resultant vector is perpendicular to the smaller vector. If magnitude of resultant vector is $24$ unit, then magnitude of given vectors are

• A. $26$ unit and $10$ unit
• B. $20$ unit and $16$ unit
• C. $24$ unit and $12$ unit
• D. $25$ unit and $11$ unit

Answer 1

The correct option is A $26$ unit and $10$ unit

$\begin{array}{c}A+B=\mathrm{36............}\left(1\right)\\ R=24\\ \sqrt{A^2+B^2+2AB\cos \theta }=24\\ A^2+B^2+2AB\cos \theta =\mathrm{576..........}\left(2\right)\\ \alpha ={90}^0\\ \tan \alpha =\tan {90}^0\\ \Rightarrow \infty =\frac{1}{0}\\ But\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }\\ \frac{1}{0}=\frac{B\sin \theta }{A+B\cos \theta }\\ A+B\cos \theta =\mathrm{0.......}\left(3\right)\\ \cos \theta =\frac{-A}{B},putin\left(2\right)\\ A^2+B^2+2AB\times \frac{-A}{B}=576\\ A^2+B^2-2A^2=576\\ B^2-A^2=\mathrm{576..............}\left(4\right)\\ \left(B+A\right)\left(B-A\right)=576\\ \Rightarrow 36\left(B-A\right)=576\\ B-A=\frac{576}{36}=16\\ +A+B=36\\ \underset{–}{-A+B=16}\\ 2B=52\\ \therefore B=26\\ A=10,B=26\end{array}$