Question

A line 'l' passing through origin is perpendicular to the lines
$l_1: \vec{r} = (3+t)\hat{i} + (-1+2t)\hat{j} + (4+2t)\hat{k}$
$l_2: \vec{r} = (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}$
If the co-ordinates 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 ________ .

Hint:

This multi-step 3D geometry problem requires careful execution of standard procedures: find a perpendicular vector using cross product, solve for the intersection of two lines by equating their component equations, and use the distance formula between two points in space.

Answer 1

Correct Answer: 44

Direction vectors: \[ \vec d_1=(1,2,2),\quad \vec d_2=(2,2,1) \] \[ \vec d=\vec d_1\times\vec d_2=(-2,3,-2) \] Intersection with \(l_1\): \[ I=(2,-3,2) \] Point on \(l_2\): \[ P(3+2s,3+2s,2+s) \] Distance: \[ |IP|^2=17 \Rightarrow s=-\frac{10}{9} \] \[ (a,b,c)=\left(\frac79,\frac79,\frac89\right) \] \[ 18(a+b+c)=44 \] \[ \boxed{44} \]