A line l passing through the origin is perpendicular to the lines [l_1: r=(3+t) i+(-1+2t) …

Question

A line $l$ passing through the origin is perpendicular to the lines
$$l_1:\ \vec r=(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k,\qquad l_2:\ \vec r=(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k.$$
If the coordinates of the point in the first octant on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$ are $(a,b,c)$, then $18(a+b+c)$ is equal to:

Answer: 44

Accepted Answer

Direction of $l$: $(1,2,2)\times(2,2,1)=(-2,3,-2)$.

Intersection of $l$ (through origin) with $l_1$: solving gives $M=(2,-3,2)$.

Point on $l_2$: $(3+2s,3+2s,2+s)$; distance$^2$ from $M$:
$$(1+2s)^2+(6+2s)^2+s^2=17 \implies 9s^2+28s+20=0 \implies s=-2\ \text{or}\ -\frac{10}{9}.$$
$s=-2$ gives $(-1,-1,0)$ (not first octant). $s=-\tfrac{10}{9}$ gives
$$(a,b,c)=\left(\frac79,\frac79,\frac89\right)\ (\text{first octant}).$$
$$18(a+b+c)=18\times\frac{22}{9}=44.$$

Solutions · 1