Verified Solution • 2025–26 Session moderate

QJEEM18Ph57fd: JEE Main 2018 Physics - Class 11 JEEM18Ph57fd Solution: Let = and, The magnitude of a coplanar vector such that is…

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

Multiple Choice Options

  • A 109\sqrt {{{10} \over 9}}
  • B 59\sqrt {{{5} \over 9}}
  • C 209\sqrt {{{20} \over 9}}
  • D 912\sqrt {{{9} \over 12}}

Step-by-Step Verified Solution

Let C→\overrightarrow C = a i^\widehat i + b j^\widehat j

Given, A→.C→=A→.B→\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B

⇒\Rightarrow     \,\,\,\, a + b = 2 −- 1

⇒\Rightarrow     \,\,\,\, a + b = 1 . . . . .(1)

also given

B→.C→=A→.B→\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B

⇒\Rightarrow     \,\,\,\, 2a −- b = 1 . . . . (2)

Solving (1) and (2), we get,

a = 13{1 \over 3} and b = 23{2 \over 3}

∴    \therefore\,\,\,\, C→=13i^+23j^\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j

∣C→∣=(13)2+(23)2\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}}

= 59\sqrt {{5 \over 9}}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring