JEE Advance - Physics (2017 - Paper 2 Offline - No. 7)
Three vectors $$\overrightarrow P ,\overrightarrow Q $$ and $$\overrightarrow R $$ are shown in the figure. Let $$S$$ be any point on the vector $$\overrightarrow R .$$ The distance between the points $$P$$ and $$S$$ is $$b\left| {\overrightarrow R } \right|.$$ The general relation among vectors $$\overrightarrow P ,\overrightarrow Q $$ and $$\overrightarrow S $$ is :
$$\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + b\overrightarrow Q $$
$$\overrightarrow S = \left( {b - 1} \right)\overrightarrow P + b\overrightarrow Q $$
$$\overrightarrow S = \left( {1 - {b^2}} \right)\overrightarrow P + b\overrightarrow Q $$
$$\overrightarrow S = \left( {1 - b} \right)\overrightarrow P + {b^2}\overrightarrow Q $$
Explanation
From figure we can write
$$\overrightarrow P = b\overrightarrow R = \overrightarrow S $$
It is given that $$\overrightarrow R = \overrightarrow Q - \overrightarrow P $$. Put in above equation we get
$$\overrightarrow P + b(\overrightarrow Q - \overrightarrow P ) = \overrightarrow S \Rightarrow \overrightarrow P + b\overrightarrow Q - b\overrightarrow P = \overrightarrow S $$
$$ \Rightarrow \overrightarrow S = \overrightarrow P (1 - \overrightarrow b ) + b\overrightarrow Q $$
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