Question

The algebraic sum of two co-initial vectors is $16units$ . Their vector sum is $8units$ 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)
$P^2+Q^2+2PQcos\theta =64$ (ii)
tan ${\text{90}}^{\text{ο}}=\frac{Qsin\theta }{P+Qcos\theta }$

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