If the shortest distance between the lines x-1=y+1-1= z1 (-1) and x+y+z+1=0=2x-y+z+3 is 1√…

Question

If the shortest distance between the lines $\dfrac{x-1}\alpha=\dfrac{y+1}{-1}=\dfrac z1$ ($\alpha
eq-1$) and $x+y+z+1=0=2x-y+z+3$ is $\dfrac1{\sqrt3}$, then a value of $\alpha$ is:

Accepted Answer

Family through the second line: $(x+y+z+1)+\lambda(2x-y+z+3)=0$. Parallel to the first line (direction $(\alpha,-1,1)$):
$$\alpha(2\lambda+1)+2\lambda=0\implies \alpha=\frac{-2\lambda}{2\lambda+1}.$$
The distance from $(1,-1,0)$ (on line 1) to this plane equals $\dfrac1{\sqrt3}$:
$$\frac{|6\lambda+1|}{\sqrt{6\lambda^2+4\lambda+3}}=\frac1{\sqrt3}\implies 3(6\lambda+1)^2=6\lambda^2+4\lambda+3\implies102\lambda^2+32\lambda=0.$$
So $\lambda=0$ (giving $\alpha=0$) or $\lambda=-\tfrac{16}{51}$ (giving $\alpha=\tfrac{32}{19}$).

Solutions · 1