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JEE Main
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Physics
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Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

The algebraic sum of two co-initial vectors is 16units . Their...

The algebraic sum of two co-initial vectors is 16units16units. Their vector sum is 8units8units and the resultant of the vectors are perpendicular to the smaller vector. Then magnitudes of the two vectors are -The algebraic sum of two co-initial vectors is 16units . Their vector sum is 8units and the resultant of the vectors

Options

  1. A2unit & 14unit
  2. B4unit & 12unit
  3. C6unit & 10unit
  4. D8unit & 8unit

Correct answer

C. 6unit & 10unit

Step-by-step solution

P+Q=16P + Q = 16 (i)

P2+Q2+2PQcos\theta=64P^{2} + Q^{2} + 2PQcos\text{\theta} = 64 (ii)

tan90ο=Qsin\thetaP+Qcos\theta{90}^{ο} = \dfrac{Q sin \text{\theta}}{P + Q cos \text{\theta}}

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

⇒P+Qcos\theta=0⇒ P + Q cos \text{\theta} = 0

cos\theta=−PQcos \text{\theta} = - \dfrac{\text{P}}{\text{Q}}

From Eq. (ii)

P2+Q2+2PQcos\theta=64P^{2} + Q^{2} + 2 PQ cos \text{\theta} = \text{64}

P2+Q2+2.P.Q(−PQ)=64P^{2} + Q^{2} + \text{2.P.Q} \left(- \dfrac{P}{Q}\right) = \text{64}

−P2+Q2=64- P^{2} + Q^{2} = \text{64}

\left(Q + P\right) \left(Q - P\right) = \text{64}<br/><br/><br/><br/>as P + Q = 16<br/><br/><br/><br/>Q - P = 4<br/><br/>Solvingweget<br/><br/><br/><br/>Solving we get<br/><br/>2 Q = \text{20}<br/><br/><br/><br/>Q = \text{10} units<br/><br/><br/><br/>P = 6 units
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