question_answer

The length of the perpendicular distance of the point \[(-1,\,4,\,0)\] from the line \[\frac{x}{1}=\frac{y}{3}=\frac{z}{1}\] is           equal to

A)  \[\sqrt{6}\]   

B)  \[\sqrt{5}\]  

C)  \[2\] 

D)  \[1\]

Correct Answer: A

Solution :

Let L be the foot of the perpendicular drawn from the point \[P(-1,4,0)\] to the given line. The coordinated of a general point on \[\frac{x-0}{1}=\frac{y-0}{3}=\frac{z-0}{1}\] are given by \[\frac{x}{1}=\frac{y}{3}=\frac{z}{1}=r\] i.e.,  \[x=r,\,\,y=3r,\,\,z=r\] Let the coordinate of L be \[(r,\,3r,r)\] ?.. (i) Direction ratios of PL are \[=r+1,\,3r-4,\,r\] Direction ratios of the given line are \[1,\,\,3,\,\,1\] Since, PL perpendicular to the given, line, \[\therefore \] \[(r+1)+(3r-4).2+r.1=0\] \[\Rightarrow \] \[r+1.1+(3r-4).2+r.1=0\] \[\Rightarrow \] \[11r=11\] \[\Rightarrow \] \[r=1\] So, the coordinate of L is \[(1,3,1)\] \[\therefore \] \[PL=\sqrt{{{(1+1)}^{2}}+{{(3-4)}^{2}}+{{(1-0)}^{2}}}\] \[=\sqrt{4+1+1}=\sqrt{6}\]