Hi I would like to find the parametric equation of the line L passing by P0 (the intersection of D1 and D2) and perpendicular to the plan having D1 and D2.
D1 and D2 are others lines.
I had to find the parametric equation of D1:\begin{cases} x =x+y=2 \\[2ex] 3x+y+z=5 \end{cases} So i have found: D1:\begin{cases} x =\frac{-(t-3)}{2} \\[2ex] y=\frac{t+1}{2}\\[2ex] z=t \end{cases}
D2:\begin{cases} x =1-2t \\[2ex] y=1+2t\\[2ex] z=1-t \end{cases}
I have found $P0(5/3, 1/3, 4/3)$
I have tried to do the vector product to find a perpendicular vector. I have replaced t by 0 in each equations to find a new point so i have found : AP0=[5/3-3/2, 1/3-1/3, 4/3-3/2] (a vector on D1) and BP0=[5/3-1, 1/3-1, 4/3-1] (a vector on D2). So I have found [-1/9,-1/6,-1/9] as the vector product of these 2 vectors. After I have found this equation of L:\begin{cases} x =5/3 -1/9t \\[2ex] y=1/3-1/6t\\[2ex] z=4/3 \end{cases}
but it dont pass through P0
could you help me to find my error. Thank you
Hint:
First note that from $t=1-t$ we have $t=\frac{1}{2}$ so $z=\frac{1}{2}$ for $P_0$.
To solve your problem note that the orienting vectors of the two lines are: $$ \vec v_1=(-\frac{1}{2},\frac{1}{2},1)^T \qquad \vec v_2=(-2,2,1)^T $$ so the vector $\vec v_3=\vec v_1 \times \vec v_2$ is orthogonal to the plane that contain the two lines.
So the line that you want is the line oriented by such vector $\vec v_3$ and passing thorough the point $P_0$.