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The parametric equations are

$$x = 2 + 3t$$ $$y = 3-t$$ $$z= 1+t$$

Let $\alpha: A_1x + B_1y + C_1z + D_1 = 0$ and $\beta: A_2x + B_2y + C_2z + D_2 = 0$ be the equations of two intersecting planes.

To make it clear, I'm asked to go from to $(x,y,z) \to \alpha$ and $\beta$.

It can be seen that the vector lying on the of intersection of the planes has coordinates $(3,-1,1)$. Since this vector is perpendicular to both normal vectors to the planes we have that

$$B_1C_2-B_2C_1=3$$ $$C_1A_2 - C_2A_1 =-1$$ $$A_1B_2- A_2B_1 = 1$$

And here I'm stuck. I don't this is the path, since I got 3 equations in 6 unknowns. The other thing I was thinking was to plug the parametric equations in the equations of the planes, in order to get something in terms of $t$ but it doesn't look that helpful either.

asd
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    It should come as no surprise that your system of equations is underdetermined since there is an infinite number of planes that contain this line, any two of which would serve as a solution. – amd Nov 03 '16 at 20:21

1 Answers1

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The parametric equations of the line $\;D\;$ are

$x=2+3t$

$y=3-t$

$z=1+t$.

from the second, we get $\;\;t=3-y\;\;$ which we replace in the others to obtain the planes equations

$$P_1\;:\;x+3y=11$$ and $$P_2\;:\;y+z=4.$$

we check that $D\in P_1$ and $D\in P_2$.

hamam_Abdallah
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