The simple sum of two co-initial vectors is 10 units. Their vector sum is 8 units. The resultant of the vectors is perpendicular to the smaller vector. The magnitudes of the two vectors are

Question

Dr. Abubacker · Accepted Answer

(c) $P + Q = 16$ (i)
$P^{2} + Q^{2} + 2PQcos	heta = 64$ (ii)
tan$90{^\circ} = \frac{Qsin	heta}{P + Qcos	heta}$
$\infty = \frac{Qsin	heta}{P + Qcos	heta}$
$\Rightarrow \ \ \ \ P + Qcos	heta = 0\ $ or $Qcos	heta = - P$
From Eq. (ii)
$P^{2} + Q^{2} + 2P( - P) = 64$ or $Q^{2} - P^{2} = 64$
or $(Q - P)(Q + P) = 64$
or $Q - P = \frac{64}{16} = 4$ (iii)
Adding Eq. (i) and (iii), we get
$2Q = 20$ or $Q = 10$ units
From (i), $P + 10 = 16$ or $P = 6$ units

Q. The simple sum of two co-initial vectors is 10 units. Their vector sum is 8 units. The resultant of the vectors is perpendicular to the smaller vector. The magnitudes of the two vectors are

Explanation

Reviewed by Dr. Abubacker

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