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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    The distance of the point \[-\hat{i}+2\hat{j}+6\hat{k}\]from the straight line that passes through the point \[2\hat{i}+3\hat{j}-4\hat{k}\] and is parallel to the vector \[6\hat{i}+3\hat{j}-4\hat{k}\] is   JEE Main Online Paper (Held On 26-May-2012)  

    A) 9                             

    B)                        8

    C)                        7     

    D)                        10

    Correct Answer: C

    Solution :

                    Point is (-1,2, 6) Line passes through the point (2, 3, -4)parallel to vector whose direction ratios is 6, 3,-4. Equation is \[\frac{x-2}{6}=\frac{y-3}{3}=\frac{z+4}{-4}=\lambda \] Any point on this line is given by \[x=6\] \[\lambda =2,y=3\lambda +3,z=-4\lambda -4\] Now, d. Rs of line passing through(-1,.2, 6) and \[\bot \] to this line is\[\{(x+1),(y-2),(z-6)\}\] So, 6 (x + 1) + 3 (y - 2) - 4 (z - 6) = 06x + 3y - 4z + 24 =0 Now, \[6(6\lambda +2)+3(3\lambda +3)+4(4\lambda +4)+24=0\] \[\Rightarrow \]\[61\lambda +61=0\Rightarrow \lambda =-1\] So,\[x=-4,y=0,z=0\] Now, distance between (-1, 2, 6) and(- 4, 0, 0) is\[\sqrt{9+4+36}=\sqrt{49}=7\]                                


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