CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The coordinates of the foot of the perpendicular drawn from the point A (1, 0, 3) to the join of the points B (4, 7, 1) and C (3, 5, 3) are

    A)  \[\left( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right)\]

    B)  \[(5,7,17)\]

    C)  \[\left( \frac{5}{7},-\frac{7}{3},\frac{17}{3} \right)\]

    D)  \[\left( -\frac{5}{3},\frac{7}{3},-\frac{17}{3} \right)\]

    E)  None of these

    Correct Answer: A

    Solution :

    Let D be the foot of the perpendicular and let it divide BC in the ratio\[\lambda :1\]. Then, the coordinates of Dare \[\left( \frac{3\lambda +4}{\lambda +1},\frac{5\lambda +7}{\lambda +1},\frac{3\lambda +1}{\lambda +1} \right)\]. Now, \[\overrightarrow{AD}\bot \overrightarrow{BC}\Rightarrow \overrightarrow{AD}.\overrightarrow{BC}=0\] \[\Rightarrow \]\[\left[ \frac{(2\lambda +3)}{\lambda +1}\hat{i}+\frac{(5\lambda +7)}{\lambda +1}\hat{j}-\frac{2}{\lambda +1}\hat{k} \right]\]                                                 \[(-\hat{i}-2\hat{j}+2\hat{k})\] \[\Rightarrow \] \[-(2\lambda +3)-2(5\lambda +7)-4=0\] \[\Rightarrow \]               \[\lambda =-\frac{7}{4}\] Hence, coordinates of D are\[\left( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right)\].


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