# parametric equation of line 1 perpendicular to a plan

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

• what are $D_1$ and $D_2$? – Dr. Sonnhard Graubner Sep 28 '16 at 18:52
• it's two lines in the space – Pierre-Luc Bolduc Sep 28 '16 at 18:56
• but they have the same parameters, this can not be! – Dr. Sonnhard Graubner Sep 28 '16 at 18:57
• It seems that your $z$ for $P_0$ is wrong. – Emilio Novati Sep 28 '16 at 18:57
• And: what vectors have you used in the vector product? – Emilio Novati Sep 28 '16 at 19:02

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 $P_0$.