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  4. The resultant of two vectors is perpendicular to first vector of magnitude 6 N . If the resultant has magnitude 6√3 text N , then magnitude of second vector is

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Q. The resultant of two vectors is perpendicular to first vector of magnitude $6 \, N$ . If the resultant has magnitude $6\sqrt{3}\text{ N}$ , then magnitude of second vector is

NTA AbhyasNTA Abhyas 2020

Solution:

Solution
$B Cos \, \theta =6$
$B Sin \, \theta =6\sqrt{3}$
$\text{B}=\sqrt{6^{2} + \left(6 \sqrt{3}\right)^{2}}=12$