Let A be the point of intersection of the lines L_1:(x-7)/(1)=(y-5)/(0)=(z-3)/(-1) and L_2…

Question

Let A be the point of intersection of the lines ${L}_{1}:\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and ${L}_{2}:\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines ${L}_{1}$ and ${L}_{2}$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $ABC$ is :

Accepted Answer

$L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1} ; L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$

$$\begin{aligned}
& \cos \theta=\left|\frac{3+0-5}{\sqrt{2} \times \sqrt{50}}\right| \
& =\frac{2}{10}=\frac{1}{5} \
& \therefore \sin \theta=\frac{2 \sqrt{6}}{5} \
& \text { Area }=\frac{1}{2} a b \sin \theta \
& =\frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2 \sqrt{6}}{5} \
& =3 \sqrt{6} \
& (\text { Area })^2=9 \times 6=54
\end{aligned}$$

Solutions · 1