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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    The straight lines whose direction cosines are given by \[al+bm+cn=0,fmn+gnl+hlm=0\]are perpendicular, if

    A) \[\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0\]

    B) \[\sqrt{\frac{a}{f}}+\sqrt{\frac{b}{g}}+\sqrt{\frac{c}{h}}=0\]

    C) \[\sqrt{af}=\sqrt{bg}=\sqrt{ch}\]

    D) \[\sqrt{\frac{a}{f}}=\sqrt{\frac{b}{g}}=\sqrt{\frac{c}{h}}\]

    Correct Answer: A

    Solution :

    • From the first relation, \[n=-\left( \frac{al+bm}{c} \right)\] Put the value of n in second relation,           
    • \[fm\,\left( -\frac{(al+bm)}{c} \right)+gl\,\left( -\frac{(al+bm)}{c} \right)+hlm=0\]           
    • or   \[afml+bf{{m}^{2}}+ag{{l}^{2}}+bglm-chlm=0\]                   
    • \[ag\frac{{{l}^{2}}}{{{m}^{2}}}+\frac{l}{m}(af+bg-ch)+bf=0\]    .....(i)           
    • Now if \[{{l}_{1}},\,\,{{m}_{1}},\,\,{{n}_{1}}\] and \[{{l}_{2}},\,\,{{m}_{2}},\,\,{{n}_{2}}\] be direction cosines of two lines, then from (i)                    \[\frac{{{l}_{1}}{{l}_{2}}}{{{m}_{1}}{{m}_{2}}}=\frac{bf}{ag}\],   \[\left[ \text{Since roots of (i) are }\frac{{{l}_{1}}}{{{m}_{1}}},\frac{{{l}_{2}}}{{{m}_{2}}} \right]\]  or        \[\frac{{{l}_{1}}{{l}_{2}}}{f/a}=\frac{{{m}_{1}}{{m}_{2}}}{g/b}\]           
    • Similarly, elimination of l will yield \[\frac{{{m}_{1}}{{m}_{2}}}{g/b}=\frac{{{n}_{1}}{{n}_{2}}}{h/c}\]           
    • \[\therefore \,\,\,\frac{{{l}_{1}}{{l}_{2}}}{f/a}=\frac{{{m}_{1}}{{m}_{2}}}{g/b}=\frac{{{n}_{1}}{{n}_{2}}}{h/c}=q\]           
    • (Say)            We know that the lines are perpendicular, if \[{{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0\]                   
    • i.e., \[\left( \frac{f}{a} \right)\,q+\left( \frac{g}{b} \right)\,q+\left( \frac{h}{c} \right)\,q=0\] or \[\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0\].                   
    • Note: Student should remember this question as a fact.


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