BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    The angle between the line \[\frac{x+1}{2!}=\frac{y}{3}=\frac{z-3}{6}\]and the plane \[10x+2y-11z=3\] is

    A)  \[\frac{\pi }{2}\]                             

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

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

    D)         \[{{\sin }^{-1}}\left( \frac{8}{21} \right)\]

    Correct Answer: D

    Solution :

    Let \[\theta \,\]be the angle between the line and the plane, we have \[a=2,\,b=3,\,c=6\]                 and        \[{{a}_{1}}=10,\,{{b}_{2}}=2,\,{{c}_{1}}=-11\]                 \[\therefore \]  \[\cos (90-\theta )=sin\theta \]                 \[=\left| \frac{a{{a}_{1}}+b{{b}_{1}}+c{{c}_{1}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}} \right|\]                 \[=\left| \frac{(2)(10)+(3)(2)+(6)(-11)}{\sqrt{{{2}^{2}}+{{3}^{2}}+{{6}^{2}}}\sqrt{{{10}^{2}}+{{2}^{2}}+{{(-11)}^{2}}}} \right|\] \[=\left| \frac{20+6-66}{\sqrt{49.}\sqrt{225}} \right|=\left| \frac{-40}{7.15} \right|\] \[\theta ={{\sin }^{-1}}\left( \frac{8}{21} \right)\]


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