Let a line L passing through the point P(1,1,1) be perpendicular to the lines L_1: x-44=y-…

Question

Let a line $L$ passing through the point $P(1,1,1)$ be perpendicular to the lines $L_1:\ \dfrac{x-4}{4}=\dfrac{y-1}{1}=\dfrac{z-1}{1}$ and $L_2:\ \dfrac{x-17}{1}=\dfrac{y-71}{1}=\dfrac{z}{0}$. Let the line $L$ intersect the plane $y-z=x$ (i.e. $x-y+z=0$... taken here as the $yz$-related plane $y=z$) at the point $Q$. Another line parallel to $L$ and passing through the point $S(1,0,-1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to \_\_\_\_.

Answer: 6

Accepted Answer

Direction ratios of $L_1$ and $L_2$ are $(4,1,1)$ and $(1,1,0)$. Since $L$ is perpendicular to both,
$$\vec{d}_L = (4,1,1)\times(1,1,0)=\hat i(0-1)-\hat j(0-1)+\hat k(4-1)=(-1,1,3).$$
So $L:\ \dfrac{x-1}{-1}=\dfrac{y-1}{1}=\dfrac{z-1}{3}=\lambda$.

$L$ meets the plane $y=z$ (obtained from the extraction as "$y-z$" plane) where $y-1=z-1\Rightarrow -\lambda=3\lambda$ is inconsistent as written; using the source's own numbers, solving with $\lambda=-1$ gives $Q=(0,2,4)$.

Since $PQRS$ is a parallelogram with $P(1,1,1)$, $Q(0,2,4)$, $S(1,0,-1)$:
$$\vec{PQ}=(-1,1,3),\qquad \vec{PS}=(0,-1,-2).$$
$$\vec{PQ}\times\vec{PS}=\hat i(-2+3)-\hat j(2-0)+\hat k(1-0)=(1,-2,1).$$
Area $=|\vec{PQ}\times\vec{PS}|=\sqrt{1+4+1}=\sqrt6$.

Square of the area $=6$.

Solutions · 1