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Let $$\overrightarrow A $$ = $$\left( {\widehat i + \widehat j} \right)$$ and, $$\overrightarrow B = \left( {2\widehat i - \widehat j} \right).$$ The magnitude of a coplanar vector $$\overrightarrow C $$ such that $$\overrightarrow A .\overrightarrow C = \overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B ,$$ is given by :

JEE · Physics · previous-year question

  1. A.$$\sqrt {{{10} \over 9}} $$
  2. B.$$\sqrt {{{5} \over 9}} $$correct
  3. C.$$\sqrt {{{20} \over 9}} $$
  4. D.$$\sqrt {{{9} \over 12}} $$

Answer

B. $$\sqrt {{{5} \over 9}} $$

Explanation

Let $$\overrightarrow C $$ = a$$\widehat i$$ + b$$\widehat j$$ Given, $$\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B $$ $$ \Rightarrow $$ $$\,\,\,\,$$ a + b = 2 $$-$$ 1 $$ \Rightarrow $$ $$\,\,\,\,$$ a + b = 1 . . . . .(1) also given $$\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B $$ $$ \Rightarrow $$ $$\,\,\,\,$$ 2a $$-$$ b = 1 . . . . (2) Solving (1) and (2), we get, a = $${1 \over 3}$$ and b = $${2 \over 3}$$ $$\therefore\,\,\,\,$$ $$\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j$$ $$\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $$ = $$\sqrt {{5 \over 9}} $$

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