Let L be the line x+12=y+13=z+36 and let S be the set of all points (a,b,c) on L whose dis…

Question

Let $L$ be the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$ and let $S$ be the set of all points $(a,b,c)$ on $L$ whose distance from the line $L_2:\ \dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$, measured along $L$, is $7$. Then $\displaystyle\sum_{(a,b,c)\in S}(a+b+c)$ is equal to:

Accepted Answer

$L_2$ has direction $(2,3,0)$ and constant $z=9$. Points on $L$: $(2\lambda-1,\ 3\lambda-1,\ 6\lambda-3)$. Since $L_2$'s $z$-value is fixed at $9$, the intersection $M$ of $L$ and $L_2$ occurs at $6\lambda-3=9\Rightarrow\lambda=2$, giving
$M=(3,5,9).$
Let $P=(2K-1,\ 3K-1,\ 6K-3)$ be a point on $L$ with $|PM|=7$:
$$\vec{PM}=(2K-4,\ 3K-6,\ 6K-12)=(K-2)(2,3,6),\qquad |\vec{PM}|=|K-2|\sqrt{49}=7|K-2|.$$
Setting $7|K-2|=7\Rightarrow K=1,3.$

$K=1: (1,2,3)$; $K=3: (5,8,15)$.

Sum of all coordinates $=1+2+3+5+8+15=34.$

Solutions · 1