Let the area of the triangle formed by the lines x+2=y-1=z, (x-3)/(5)=(y)/(-1)=(z-1)/(1) a…

Question

Let the area of the triangle formed by the lines $x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 56

Accepted Answer

$\begin{gathered} {L}_{1}:x+2=y-1=z=l \ {L}_{2}:\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=m \ {L}_{3}:\frac{x}{-3}=\frac{y-3}{5}=\frac{z-2}{1}=n \end{gathered}$

$\begin{gathered} {L}_{1}{L}_{2} \ \begin{gathered} l-2=5m+3 \ l+1=-m \ l=m+1 \end{gathered}l=0,m=-1\, A(-2,1,0) \end{gathered}$

Point of intersection of ${L}_{2}$ and ${L}_{3}$

$\begin{gathered} 5m+3=-3n \ -m=3n+3 \ m+1=n+2 \end{gathered}m=0,n=-1,B(3,0,1)$

Point of intersection ${L}_{3}$ and ${L}_{4}$

$\begin{gathered} -3n=l-2 \ 3n+3=l+1 \ n+2=l \end{gathered}l=2,n=0,C(0,3,2)$

$\begin{gathered} Ar(	riangleABC)=|\frac{1}{2}|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ -5 \quad  1 \quad  -1 \ -3 \quad  3 \quad  1\end{array}|| \ A=\frac{1}{2}|\hat{i}(4)-\hat{j}(-8)+\hat{k}(-12)| \ A=\frac{1}{2}\sqrt{16+64+144}=\sqrt{56} \ {A}^{2}=56 \end{gathered}$

Solutions · 1