Consider the lines L_1: x-1=y-2=z and L_2: x-2=y=z-1.

Question

Consider the lines $L_1:\ x-1=y-2=z$ and $L_2:\ x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$, then $4A^2$ is equal to:

Accepted Answer

Both lines have direction $(1,1,1)$. $Q(\lambda+1,\lambda+2,\lambda)$ on $L_1$: $\vec{PQ}\cdot(1,1,1)=0\Rightarrow 3\lambda=0\Rightarrow\lambda=0\Rightarrow Q=(1,2,0)$.

$R(\mu+2,\mu,\mu+1)$ on $L_2$: $\vec{PR}\cdot(1,1,1)=0\Rightarrow 3\mu=0\Rightarrow\mu=0\Rightarrow R=(2,0,1)$.

$$\vec{PQ}=(-4,1,3),\qquad \vec{PR}=(-3,-1,4).$$
$$\vec{PQ}\times\vec{PR}=(7,7,7).$$
$$A=\frac12|7(1,1,1)|=\frac{7\sqrt3}{2}.$$
$$4A^2=4\cdot\frac{49\times3}{4}=147.$$

Solutions · 1