Given ell_1: x=a, y=b and ell_2: x+λ y+z=0, λ x-y-z=0 : 1^°…

Given $\ell_1:\ x=a,\ y=b$ and $\ell_2:\ x+\lambda y+z=0,\ \lambda x-y-z=0$: $1^\circ$ find the common normal; $2^\circ$ find the locus of these normals as $\lambda$ varies.

Step 1: Directions

$\vec d_1=\{0,0,1\}$. $\ell_2$ passes through $O$, and

\[ \vec d_2=\{1,\lambda,1\}\times\{\lambda,-1,-1\}=\{1-\lambda,\ 1+\lambda,\ -(1+\lambda^2)\}. \]

\[ \vec n=\vec d_1\times\vec d_2=\{-(1+\lambda),\ 1-\lambda,\ 0\}\parallel\{1+\lambda,\ \lambda-1,\ 0\}. \]

The common normal is horizontal.

Step 2: $1^\circ$ The common normal

Let $B=s\vec d_2\in\ell_2$. The foot on $\ell_1$ is $A=(a,b,z_B)$, at the same height since the normal is horizontal. The condition $(A-B)\parallel\vec n$ gives

\[ \big(a-s(1-\lambda)\big)(\lambda-1)-\big(b-s(1+\lambda)\big)(1+\lambda)=0\ \Longrightarrow\ s=\frac{a(1-\lambda)+b(1+\lambda)}{2(1+\lambda^2)} . \]

Hence $z_B=-s(1+\lambda^2)$, and

\[\frac{x-a}{1+\lambda}=\frac{y-b}{\lambda-1},\qquad z=-\frac{a(1-\lambda)+b(1+\lambda)}{2}\]

Step 3: $2^\circ$ Locus

With $u=x-a$, $v=y-b$: from the first equation $\lambda=\tfrac{u+v}{u-v}$. Substituting into $2z=-(a+b)-(b-a)\lambda$: $2z(u-v)=-(a+b)(u-v)-(b-a)(u+v)=-2bu+2av$.

\[z\,(x-y-a+b)=ay-bx\]

This is a hyperbolic paraboloid: the quadratic part is $z(x-y)$, a ruled surface. Special case $a=b$: $(x-y)(z+a)=0$, two planes.