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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    If a of magnitude 50 is collinear with the vector \[\mathbf{b}=6\,\mathbf{i}-8\,\mathbf{j}-\frac{15\,\mathbf{k}}{2},\] and makes an acute angle with the positive direction of z-axis, then the vector a is equal to [Pb. CET 2004]

    A) \[24\,\mathbf{i}-32\,\mathbf{j}+30\,\mathbf{k}\]

    B) \[-24\,\mathbf{i}+32\,\mathbf{j}+30\,\mathbf{k}\]

    C) \[16\,\mathbf{i}-16\,\mathbf{j}-15\,\mathbf{k}\]

    D) \[-12\,\mathbf{i}+16\,\mathbf{j}-30\,\mathbf{k}\]

    Correct Answer: B

    Solution :

    • Let \[\mathbf{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\]                     
    • \[|\mathbf{a}|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=50\],  \[\mathbf{b}=6\mathbf{i}-8\mathbf{j}-\frac{15}{2}\mathbf{k}\]                   
    • Since \[\mathbf{a}\] and \[\mathbf{b}\] are collinear, so \[\mathbf{a}=k\,\mathbf{b}\] and \[\frac{x}{6}=\frac{y}{-8}=\frac{2z}{-15}=k\],  (constant)                   
    • Þ   \[2500={{k}^{2}}\left[ \frac{144+256+225}{4} \right]\]                   
    • Þ   \[k=\pm \sqrt{\frac{2500\times 4}{625}}=\pm 4\]                   
    • Since \[\mathbf{a}\]makes an acute angle with the direction of        
    • z-axis, Hence, its z-component must be positive. This is possible only when \[k=-4\].                   
    • \ \[\mathbf{a}=k\,\left[ 6\mathbf{i}-8\mathbf{j}-\frac{15}{2}\mathbf{k} \right]\],  \[[\because \mathbf{a}=k\mathbf{b}]\]                   
    • Hence, \[\mathbf{a}=-24\mathbf{i}+32\mathbf{j}+30\mathbf{k}\].


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