JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • question_answer
    Distance of the point \[P(\vec{p})\] from the line \[\vec{r}=\vec{a}+\lambda \vec{b}\] is

    A) \[\left| (\vec{a}-\vec{p})+\frac{((\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{\left| {\vec{b}} \right|}^{2}}} \right|\]

    B) \[\left| (\vec{b}-\vec{p})+\frac{((\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{\left| {\vec{b}} \right|}^{2}}} \right|\]

    C) \[\left| (\vec{a}-\vec{p})+\frac{((\vec{p}-\vec{b}).\vec{b})\vec{b}}{{{\left| {\vec{b}} \right|}^{2}}} \right|\]

    D) None of these

    Correct Answer: B

    Solution :

    [c] Let Q \[(\vec{q})\] be the foot of altitude drawn from \[P(\vec{p})\] to the line \[\vec{r}=\vec{a}+\lambda \vec{b},\] \[\Rightarrow (\vec{q}-\vec{p}).\vec{b}=0\] and \[\vec{q}=\vec{a}+\lambda \vec{b}\] \[\Rightarrow (\vec{a}+\lambda \vec{b}-\vec{p}).\vec{b}=0\] or \[\left| (\vec{a}-\vec{p}) \right|.\vec{b}+\lambda {{\left| {\vec{b}} \right|}^{2}}=0\] or \[\lambda =\frac{(\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{\left| {\vec{b}} \right|}^{2}}}-\vec{P}\] \[\Rightarrow \left| \vec{q}-\vec{p} \right|=\left| (\vec{a}-\vec{p})+\frac{(\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{\left| {\vec{b}} \right|}^{2}}} \right|\]


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