Let the area of the triangle formed by the lines L_1: x+2=y-1=z, L_2: x-35=y-1=z-11, and L…

Question

Let the area of the triangle formed by the lines $L_1:\ x+2=y-1=z$, $L_2:\ \dfrac{x-3}{5}=\dfrac{y}{-1}=\dfrac{z-1}{1}$, and $L_3:\ \dfrac{x}{-3}=\dfrac{y-3}{5}=\dfrac{z-2}{1}$ be $A$. Then $A^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 56

Accepted Answer

Intersection of $L_1,L_2$: with $\ell=0,\ m=-1$: $A(-2,1,0)$.
Intersection of $L_2,L_3$: with $m=0,\ n=-1$: $B(3,0,1)$.
Intersection of $L_3,L_1$: with $\ell=2,\ n=0$: $C(0,3,2)$.

$$\vec{AB}=(5,-1,1),\qquad \vec{AC}=(2,2,2).$$
$$\vec{AB}\times\vec{AC}=\hat i(4)-\hat j(-8)+\hat k(-12)\ \ (\text{computed from the vertex vectors used in the source}).$$
$$\text{Area}=\frac12\sqrt{16+64+144}=\frac12\sqrt{224}\ \Rightarrow\ A^2 = \frac14\times224 = 56.$$

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