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The coordinates of the foot perpendicular from the point (1,0,0) to the line \[\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\]are     JEE Main Online Paper (Held On 12 May 2012)

A)                 (2.-3.8)                                

B)                        (1,-1,-10)

C)                        (5,-8,-4)                               

D)                        (3,-4,-2)

Correct Answer: D

Solution :

                Let the equation of AB is \[\frac{x-1}{2}=\frac{y-(-1)}{-3}=\frac{z-(-10)}{8}=k\] Let L be the foot of the perpendicular drawn forma P(1, 0,0). \[\therefore \]\[L=(2k+1,-3k-1,8k-10).\] Now, direction ratio of PL = (2k, - 3k-1, 8k -10) and direction ratio of AB = (2, - 3,8) Since, PL is perpendicular to AB \[\therefore \]\[2(2k)-3(-3k-1)+8(8k-10)=0\] Now,\[k=\frac{2(1-1)+(-3)(0+1)+8(0+10)}{{{(2)}^{2}}+{{(-3)}^{2}}+{{(8)}^{2}}}\] \[=\frac{0-3+80}{4+9+64}=\frac{77}{77}=1\] \[\therefore \]Required co-ordinate \[=L=(2+1,-3-1,8-10)=(3,-4,-2).\]