A) \[\frac{1}{7}\sqrt{35}\]
B) \[\frac{4}{7}\sqrt{35}\]
C) \[\frac{2}{7}\sqrt{35}\]
D) \[\frac{3}{7}\sqrt{35}\]
Correct Answer: D
Solution :
Given, point \[P(2,3,4)\] and line \[\frac{x-1}{-1}=\frac{y-0}{2}=\frac{z+1}{3}=r\] (say) ?.(i) Then, coordinates of any point N on the line (i) are \[(-r+1,\,2r,\,3r-1)\] ?..(ii) Let N be the foot of the perpendicular to line (i). \[\therefore \] Direction ratios of PN are \[(-r+1-2,\,2r-3,\,3r-1-4)\] \[=(r-1,2r-3,3r-5)\] ?(iii) \[\because \] PN is perpendicular to line (i). \[\therefore \] Using the condition, \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] \[\Rightarrow \] \[-1(-r-1)+2(2r-3)+3(3r-5)=0\] \[\Rightarrow \] \[r+1+4r-6+9r-15=0\] \[\Rightarrow \] \[14r=20\,\,\Rightarrow \,\,r=\frac{20}{14}=\frac{10}{7}\] Then, from Eq. (ii), we get \[N=\left( -\frac{10}{7}+1,\frac{20}{7},\frac{30}{7}-1 \right)=\left( \frac{3}{7},\frac{20}{7},\frac{23}{7} \right)\] Now, perpendicular distance \[PN=\sqrt{{{\left( -\frac{3}{7}-2 \right)}^{2}}+{{\left( \frac{20}{7}-3 \right)}^{2}}+{{\left( \frac{23}{7}-4 \right)}^{2}}}\] \[=\frac{1}{7}\sqrt{289+1+25}=\frac{\sqrt{315}}{7}=\frac{3}{7}\sqrt{35}\]
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