The lines r_1=( i+ j- k)+(a i- j), a0, and r_2=(4 i- k)+(2 i+a k) intersect.

Question

The lines $\vec r_1=(\hat i+\hat j-\hat k)+\lambda(a\hat i-\hat j)$, $a
eq0$, and $\vec r_2=(4\hat i-\hat k)+\mu(2\hat i+a\hat k)$ intersect. The square of the distance of their point of intersection from the origin is:

Accepted Answer

$\vec r_1=(1+a\lambda)\hat i+(1-\lambda)\hat j-\hat k$; $\vec r_2=(4+2\mu)\hat i+0\hat j+(a\mu-1)\hat k$.

Equating $\hat j$: $1-\lambda=0\Rightarrow\lambda=1$.
Equating $\hat k$: $-1=a\mu-1\Rightarrow a\mu=0$; since $a\neq0$, $\mu=0$.
Equating $\hat i$: $1+a(1)=4+2(0)\Rightarrow a=3$.

Point of intersection (from $\vec r_2$ with $\mu=0$): $(4,0,-1)$.
$$\text{Square of distance from origin}=16+0+1=17.$$

Solutions · 1