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Q.
The distance of the point $ (2,3,4) $ from the line $ 1-x=\frac{y}{2}=\frac{1}{3}(1+z) $ is
J & K CETJ & K CET 2005
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} $