Question

The algebraic sum of two co-initial vectors is $16 \, units$ . Their vector sum is $8 \, units$ and the resultant of the vectors are perpendicular to the smaller vector. Then magnitudes of the two vectors are -

• A. $2unit\&14unit$
• B. $4unit\&12unit$
• C. $6unit\&10unit$
• D. $8unit\&8unit$

Answer 1

Answer: (C)

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

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