JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    Let \[\mathbf{a}=2\mathbf{i}-\mathbf{j}+\mathbf{k},\,\,\mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k}\] and \[\mathbf{c}=\mathbf{i}+\mathbf{j}-2\mathbf{k}\] be three vectors. A vector in the plane of b and c whose projection on a is of magnitude \[\sqrt{2/3}\] is [IIT 1993; Pb. CET 2004]

    A) \[2\mathbf{i}+3\mathbf{j}-3\mathbf{k}\]

    B) \[2\mathbf{i}+3\mathbf{j}+3\mathbf{k}\]

    C) \[-\,2\mathbf{i}-\mathbf{j}+5\mathbf{k}\]

    D) \[2\mathbf{i}+\mathbf{j}+5\mathbf{k}\]

    Correct Answer: A

    Solution :

    • Any vector \[\mathbf{r}\] in the plane of \[\mathbf{b}\] and \[\mathbf{c}\] is \[\mathbf{r}=\mathbf{b}+t\mathbf{c}\] or \[\mathbf{r}=(1+t)\mathbf{i}+(2+t)\mathbf{j}-(1+2t)\mathbf{k}\]                    ......(i)                   
    • Projection of \[\mathbf{r}\] on \[\mathbf{a}\] is \[\sqrt{\left( \frac{2}{3} \right)}\Rightarrow \frac{\mathbf{r}\,.\,\mathbf{a}}{|\mathbf{a}|}=\sqrt{\left( \frac{2}{3} \right)}\]                   
    • or  \[\frac{2(1+t)-(2+t)-(1+2t)}{\sqrt{6}}=\pm \sqrt{\left( \frac{2}{3} \right)}\]                   
    • \ \[\,-t-1=\pm 2\Rightarrow t=-3,\,\,1\]                   
    • Projection in (i),we get                   
    • \[\therefore \,\mathbf{r}=-2\mathbf{i}-\mathbf{j}+5\mathbf{k}\] or \[\mathbf{r}=2\mathbf{i}+3\mathbf{j}-3\mathbf{k}\].

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