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  • The distance of the point ( 1,1,9) from the point of intersection of the line

The distance of the point ( 1,1,9) from the point of intersection of the line \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}and the plane x+y+z=17is:  
Option: 1 2\sqrt{19}
 
Option: 2 \sqrt{38}
Option: 3 38  
Option: 4 19\sqrt{2}

Answers (1)

best_answer

 

\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \\ \\\Rightarrow x=\lambda+3, y=2 \lambda+4, z=2 \lambda+5

Which lies on given plane hence

\\\Rightarrow \lambda+3+2 \lambda+4+2 \lambda+5=17\\\\\Rightarrow \lambda=\frac{5}{5}=1

\begin{aligned} &\text { Hence, point of intersection is } \mathrm{Q}(4,6,7)\\ &\therefore \text { Required distance }=\mathrm{PQ} \end{aligned}

PQ=\sqrt{9+25+4} =\sqrt{38}

Posted by

Suraj Bhandari

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