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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    The points \[O,\,A,\,B,\,C,\,D\] are such that \[\overrightarrow{OA}=\mathbf{a},\] \[\overrightarrow{OB}=\mathbf{b},\,\] \[\overrightarrow{OC}=2\mathbf{a}+3\mathbf{b}\] and \[\overrightarrow{OD}=\mathbf{a}-2\mathbf{b}.\] If \[|\mathbf{a}|\,=3\,|\mathbf{b}|,\] then the angle between \[\overrightarrow{BD}\] and \[\overrightarrow{AC}\] is

    A) \[\frac{\pi }{3}\]

    B) \[\frac{\pi }{4}\]

    C) \[\frac{\pi }{6}\]

    D) None of these

    Correct Answer: D

    Solution :

    • We have \[\overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}=\mathbf{a}-2\mathbf{b}-\mathbf{b}=\mathbf{a}-3\mathbf{b}\]  and \[\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=2\mathbf{a}+3\mathbf{b}-\mathbf{a}=\mathbf{a}+3\mathbf{b}.\]                   
    • Let \[\theta \] be the angle between \[\overrightarrow{BD}\] and \[\overrightarrow{AC}.\]                   
    • Then \[\cos \theta =\frac{\overrightarrow{BD}\,.\,\overrightarrow{AC}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}=\frac{|\mathbf{a}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}\]                                      
    • \[=\frac{9|\mathbf{b}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}\],  
    • \[(\because \,\,\,\left| \,\mathbf{a}\, \right|=3|\mathbf{b}|)\]                   
    • \[\Rightarrow \cos \theta =0{}^\circ \Rightarrow \theta =\frac{\pi }{2}.\]


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