The algebraic sum of two co-initial vectors is 16 units . Their vector sum is 8 units and the resultant of…

1. $2 \, \text{unit} \, \& \, 14 \, \text{unit}$
2. $4 \, \text{unit} \, \& \, 12 \, \text{unit}$
3. $6 \, \text{unit} \, \& \, 10 \, \text{unit}$
4. $8 \, \text{unit} \, \& \, 8 \, \text{unit}$

Solution

$P+Q=16$                             (i)
$P^2+Q^2+2PQ\cos \text{θ}=64$       (ii)
tan${\text{90}}^{\text{ο}}=\frac{Q\; sin\text{θ}}{P+Q\; cos\text{θ}}$

$\Rightarrow P+Qcos\text{θ}=0$
     $cos\text{θ}=-\frac{\text{P}}{\text{Q}}$
From Eq. (ii)
$P^2+Q^2+2\mathit{PQ}\cos \text{θ}=\text{64}$
$P^2+Q^2+\mathit{\text{2.P.Q}}\left(-\frac{P}{Q}\right)=\text{64}$
$-P^2+Q^2=\text{64}$
$\left(Q+P\right)\left(Q-P\right)=\text{64}$
$asP+Q=16$
      $Q-P=4$
Solving we get
$2Q=\text{20}$
$Q=\text{10}units$
$P=6units$

Asked in: MHT CET Full Test 11

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