A) \[\sqrt{13}\]
B) \[13\]
C) \[2\sqrt{3}\]
D) none of these
Correct Answer: A
Solution :
Key Idea: The perpendicular distance from\[(\alpha ,\,\,\beta ,\,\,\gamma )\] to the line \[\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\] ? (i) is given by \[\sqrt{\begin{align} & {{(\alpha -{{x}_{1}})}^{2}}+{{(\beta -{{y}_{1}})}^{2}}+{{(\gamma -{{z}_{1}})}^{2}}- \\ & {{[l(\alpha -{{x}_{1}})+m(\beta -{{y}_{1}})+n(\gamma -{{z}_{1}})]}^{2}} \\ \end{align}}\] ? (ii) Given line\[\overset{\to }{\mathop{\mathbf{r}}}\,=(\widehat{\mathbf{j}}+2\widehat{\mathbf{k}})+\lambda (\widehat{\mathbf{i}}+2\widehat{\mathbf{j}}+3\widehat{\mathbf{k}})\]and point\[(1,\,\,6,\,\,3)\]. \[\therefore \]Line in cartesian co-ordinates is \[\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}\] ? (ii) \[\therefore \]\[l=\frac{1}{\sqrt{14}},\,\,m=\frac{2}{\sqrt{14}},\,\,n=\frac{3}{\sqrt{14}}\] are direction ratio's of line. Also, by comparing (i) and (iii), we get \[{{x}_{1}}=0,\,\,{{y}_{1}}=1,\,\,{{z}_{1}}=2\] and \[\alpha =1,\,\,\beta =6,\,\,\gamma =3\] \[\therefore \]From (ii) Required distance \[\sqrt{\begin{align} & {{(1-0)}^{2}}+{{(6-1)}^{2}}+{{(3-2)}^{2}} \\ & -\left[ \frac{1}{\sqrt{14}}\{1+2\times 5)+3{{(1)}^{2}}\} \right] \\ \end{align}}\]
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