Verified Solution • 2025–26 Session moderate

QJEEAdv08Ma6301: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ma6301 Solution: Let two non-collinear unit vectors and form an acute angle. A…

Let two non-collinear unit vectors a^\widehat a and b^\widehat b form an acute angle. A point PP moves so that at any time tt the position vector OP→\overrightarrow {OP} (where OO is the origin) is given by a^cos⁡t+b^sin⁡t.\widehat a\cos t + \widehat b\sin t. When PP is farthest from origin O,O, let MM be the length of OP→\overrightarrow {OP} and u^\widehat u be the unit vector along OP→\overrightarrow {OP} . Then :

Multiple Choice Options

  • A u^=a^+b^∣a^+b^∣  and  M=(1+a^. b^)1/2\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}
  • B u^=a^−b^∣a^−b^∣  and  M=(1+a^. b^)1/2\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}
  • C u^=a^+b^∣a^+b^∣  and  M=(1+2a^. b^)1/2\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}
  • D u^=a^−b^∣a^−b^∣  and  M=(1+2a^. b^)1/2\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}

Step-by-Step Verified Solution

∣OP→∣=∣a^cos⁡t+b^sin⁡t∣|\overrightarrow {OP} | = |\widehat a\cos t + \widehat b\sin t|

∣OP→∣2=(a^cos⁡t+b^sin⁡t) . (a^cos⁡t+b^sin⁡t)|\overrightarrow {OP} {|^2} = (\widehat a\cos t + \widehat b\sin t)\,.\,(\widehat a\cos t + \widehat b\sin t)

=∣a^∣2cos⁡2t+2a^ . b^cos⁡tsin⁡t+∣b^∣2sin⁡2t= |\widehat a{|^2}{\cos ^2}t + 2\widehat a\,.\,\widehat b\cos t\sin t + |\widehat b{|^2}{\sin ^2}t

=cos⁡2t+a^ . b^sin⁡2t+sin⁡2t= {\cos ^2}t + \widehat a\,.\,\widehat b\sin 2t + {\sin ^2}t

=1+a^ . b^sin⁡2t= 1 + \widehat a\,.\,\widehat b\sin 2t

The greatest value of ∣OP→∣|\overrightarrow {OP} | is reached when sin⁡2t=1\sin 2t = 1

i.e., t=π4t = {\pi \over 4}

M=∣OP→∣M = |\overrightarrow {OP} | greatest =1+a^ . b^= \sqrt {1 + \widehat a\,.\,\widehat b}

The direction of OP→\overrightarrow {OP} is given by

u^=a^+b^2 . ∣a^+b^∣2=a^+b^∣a^+b^∣\widehat u = {{\widehat a + \widehat b} \over {\sqrt 2 \,.\,{{|\widehat a + \widehat b|} \over {\sqrt 2 }}}} = {{\widehat a + \widehat b} \over {|\widehat a + \widehat b|}}

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