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What is the equation for plane $P$ that passes through $D_1:( x = 1+2t, ~y=t,~ z=t )$ and its parallel with $D_2 : ( x = -t,~ y=t, ~z = 2+3t )$?

I'm just learning this. I think that if $P$ plane passes through $D_1$, then the director vector of $D_1$ is the normal vector for my plane, $N(2,1,1)$. I'm pretty confused on how to solve this problem.Can you help me figure it out?

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The directions of D1 and D2 are $(2,1,1)$ and $(-1,1,3)$. Now the normal to the plane must be perpendicular to both these directions. So if we take the cross product of these two directions we have the normal.

$$\begin{vmatrix} \mathbf{i}&\mathbf{j}& \mathbf{k}\\ 2&1&1\\ -1&1&3\\ \end{vmatrix}=2\mathbf{i}-7\mathbf{j}+3 \mathbf{k}\\$$

so the equation of the plane is $$2x-7y+3z=a$$ and to find $a$ we use one point, say $(1,0,0)$ to get $$2x-7y+3z=2$$ it can be checked that this does indeed pass through D1 and is parallel to D2.

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  • $\begingroup$ Is the same principle applicable to the homogeneous coordinates case? $\endgroup$ – Nikola Jun 9 '18 at 15:39
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By "$P$ passes through $D_1$" I assume you mean that $D_1$ lies in $P$, in which case $D_1$ is certainly not the normal vector for the plane. In fact, $P$ contains both $D_1$ and $D_2$, right? So it contains both of those direction vectors. Can you use that to compute its normal vector?

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