question_answer

The vertices B and C of a \[\Delta ABC\]lie on the line, \[\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}\]such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A(1, -1, 2), is :-             [JEE Main 9-4-2019 Afternoon]

A) \[2\sqrt{34}\]              

B) \[\sqrt{34}\]

C) \[6\]                             

D) \[5\sqrt{17}\]

Correct Answer: D

Solution :

          \[.(3\hat{i}+4\hat{k})=0\] \[3(3\lambda -3)+0+4(4\lambda -2)=0\] \[(9\lambda -9)+(16\lambda -8)=0\] \[25\lambda =17\Rightarrow \lambda =\frac{17}{25}\] \[\therefore \]\[\overrightarrow{AD}=\left( \frac{51}{25}-3 \right)\hat{i}+2\hat{j}+\left( \frac{68}{25}-2 \right)\hat{k}\] \[=\frac{24}{25}\hat{i}+2\hat{j}+\frac{18}{25}\hat{k}\] \[\left| \overrightarrow{AD} \right|=\sqrt{\frac{576}{625}+4+\frac{324}{625}}\] \[=\sqrt{\frac{900}{625}+4}=\frac{3400}{625}\] \[=\sqrt{34}.\frac{10}{25}=\frac{2}{5}\sqrt{34}\] Area of\[\Delta =\frac{1}{2}\times 5\times \frac{2\sqrt{34}}{5}=\sqrt{34}\]