3-D geometry: line intersecting 2 lines and parallel to plane

Question

Find a surface generated by line intersecting lines $$y=a=z$$ and $$x+3z=a=y+z$$ and parallel to plane $$x+y=0$$

I tried to form a line equation which intersects the given two lines i.e. $(y-a)+k1(z-a)=0$ and $(x+3z-a)+k2(y+z-a)=0$. But don't know how to use the other (plane) condition.

Answer 1

Reordering the equation of the line $$\alpha: y+k_1 z-a k_1-a=0;\;\beta:x+k_2 y+(k_2+3) z -ak_2-a=0$$ The normal vector of plane $\alpha$ is $n_1=(0,1,k_1)$ and the normal vector of $\beta$ is $n_2=(1,k_2,k_2+3)$

the direction vector of the line is the cross product $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$

The line $(y-a)+k_1(z-a)=0;\;(x+3z-a)+k_2(y+z-a)=0$ is parallel to the plane $x+y=0$ if its direction vector is perpendicular to the normal vector of the plane $x+y=0$ which is $n=(1,1,0)$

$m\cdot n=0\to (3+k_2-k_1 k_2,k_1,-1)\cdot (1,1,0)=3 + k_1 + k_2 - k_1 k_2$

the condition is that $3 + k_1 + k_2 - k_1 k_2=0$

$k_2=\dfrac{k_1+3}{k_1-1}$

$\alpha:y+k_1 z-a-a k_1=0$

$\beta:(-1+k_1) x+(3+k_1) y+(3+3 (-1+k_1)+k_1) z-a (-1+k_1)-a (3+k_1)=0$

The direction of the line is $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$ where $k_2=\dfrac{k_1+3}{k_1-1}$

$m=(k_1^2-k_1,k_1-k_1^2,k_1-1)$

a point on the line is $P\left(-6 a,0,\dfrac{3 a}{2}\right)$

a parametric equation of the line is

$(x,y,x)=P+tm$ that is

$x=-k_1 t+k_1^2 t-6a,y=k_1 t-k_1^2 t,z=\dfrac{3 a}{2}-t+k_1 t$

Note that from the first two equations $x=-y-6a$

Furthermore from the three equation we get

$k_1= -\dfrac{4 y}{x+y+4 z},\;t= -\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}$

substitute in the third equation $z=\dfrac{3 a}{2}-t+k_1 t$

$z=\dfrac{3 a}{2}+\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}+\dfrac{y (x+y+4 z)}{x+5 y+4 z}$

simplify and get the result.

The requested surface is the plane $x + y+6a=0$

Answer 2

Equation of the line L1 in symmetric form is $$\frac{x}{1} = \frac{y - a}{0} = \frac{z - a}{0} \tag1$$ Equation of the Line L2 in symmetric form is $$\frac{x -a}{-3} = \frac{y - a}{-1} = \frac{z}{1} \tag2$$ Let the required line 'L' intersect L1 at P and L2 at Q.

Coordinates of P are $$(r,a,a) \tag3$$ Coordinates of Q are $$(-3s+a, -s+a, s) \tag4$$ Equation of L is $$\frac{x -r}{-3s-r+a} = \frac{y - a}{-s} = \frac{z-a}{s-a} \tag5$$ Given that this line L is parallel to the plane $x+y=0 \tag6$ Hence, its normal will be perpendicular to L So $$(-3s-r+a)+(-s)=0 \tag7$$ or $$r=a-4s \tag8$$ From equation 5 we have $$\frac{y-a}{z-a} = \frac{-s}{s-a}$$ Solving this for s, we will have $$s=\frac{a(y-a)}{y+z-2a} \tag9$$ From equation 5, we also have $$\frac{x -r}{-3s-r+a} = \frac{y - a}{-s} \tag10$$ Substituting the values of r and s obtained from equations 8 and 9, we get the required surface as $$y^2 + xy + yz + zx - 2ax -2az = 0$$