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QJEEAdv17Ph05f8: JEE Advanced 2017 Physics - Class 11 JEEAdv17Ph05f8 Solution: Three vectors and are shown in the figure. Let be any point on…

Three vectors P→,Q→\overrightarrow P ,\overrightarrow Q and R→\overrightarrow R are shown in the figure. Let SS be any point on the vector R→.\overrightarrow R . The distance between the points PP and SS is b∣R→∣.b\left| {\overrightarrow R } \right|. The general relation among vectors P→,Q→\overrightarrow P ,\overrightarrow Q and S→\overrightarrow S is :

JEE Advanced 2017 Paper 2 Offline Physics - Motion Question 6 English

Multiple Choice Options

  • A S→=(1−b)P→+bQ→\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + b\overrightarrow Q
  • B S→=(b−1)P→+bQ→\overrightarrow S = \left( {b - 1} \right)\overrightarrow P + b\overrightarrow Q
  • C S→=(1−b2)P→+bQ→\overrightarrow S = \left( {1 - {b^2}} \right)\overrightarrow P + b\overrightarrow Q
  • D S→=(1−b)P→+b2Q→\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + {b^2}\overrightarrow Q

Step-by-Step Verified Solution

From figure we can write

P→=bR→=S→\overrightarrow P = b\overrightarrow R = \overrightarrow S

JEE Advanced 2017 Paper 2 Offline Physics - Motion Question 6 English Explanation

It is given that R→=Q→−P→\overrightarrow R = \overrightarrow Q - \overrightarrow P . Put in above equation we get

P→+b(Q→−P→)=S→⇒P→+bQ→−bP→=S→\overrightarrow P + b(\overrightarrow Q - \overrightarrow P ) = \overrightarrow S \Rightarrow \overrightarrow P + b\overrightarrow Q - b\overrightarrow P = \overrightarrow S

⇒S→=P→(1−b→)+bQ→\Rightarrow \overrightarrow S = \overrightarrow P (1 - \overrightarrow b ) + b\overrightarrow Q

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