JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 20)
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 :
$$\sqrt {{{10} \over 9}} $$
$$\sqrt {{{5} \over 9}} $$
$$\sqrt {{{20} \over 9}} $$
$$\sqrt {{{9} \over 12}} $$
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 = $${2 \over 3}$$
$$\therefore\,\,\,\,$$ $$\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j$$
$$\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $$
= $$\sqrt {{5 \over 9}} $$
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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