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I have to find the Equation of a plane $A$

Data I do have : Two points on that plane $A$, Equations of two planes, whose line of intersection is parallel to plane $A$

  1. I found a vector $v_1$ with the two points on the plane $A$ (So that vector lies on the plane $A$).

  2. From two given planes, I calculated their cross product. It gives the vector $v_2$, which is the line of intersection.

  3. Now I know the vector $v_1$ is parallel to the $v_2$. So, the normal of vector $v_1$ is parallel to normal of vector $v_2$.

With all these information how can I find the equation of plane $A$ ?

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  • $\begingroup$ Hello and welcome! The people who try to help on this site are greatly interested in what correspondents have done toward answering their own questions prior to publishing them here. Again, welcome! $\endgroup$ – Senex Ægypti Parvi Jan 3 '17 at 5:33
  • $\begingroup$ 1. I found a vector V1 with the two points on the plane A (So that vector lies on the plane A). 2. From two given planes, I calculated their cross product. It gives the vector V2, which is the line of intersection 3. Now I know the vector v1 is parallel to the v2. So, the normal of vector v1 is parallel to normal of vector v2. With all these information how can I find the equation of plane A ? $\endgroup$ – LetmeTry_Math Jan 3 '17 at 5:50
  • $\begingroup$ \ With your permission, I shall put your summarizing comment into your question. $\endgroup$ – Senex Ægypti Parvi Jan 3 '17 at 9:48
  • $\begingroup$ Thank you. I am new here. So, I will correct myself in the future. $\endgroup$ – LetmeTry_Math Jan 3 '17 at 18:10
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If the two planes have normals $\vec n_1$ and $\vec n_2$, then $\vec n_1 \times \vec n_2$ will give a vector that is parallel to the line of intersection of the planes and thus parallel to the plane of interest. Call $\vec n_1 \times \vec n_2=\vec v$. You have two points so you can find another vector that lies in the plane call this $\vec w$. Taking the cross product of $\vec w$ and $\vec v$ will give you a normal to the plane of interest.

This assumes of course that your two points adds information about the plane. In other words they do not all lie on the line. Otherwise, there are an infinite amount of planes that fit the criteria.

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  • $\begingroup$ Thanks for your answer. So according to your answer, v and w are parallel vectors, I understand that. But, they aren't on the same plane, still their cross product will give their normal vector? Thanks $\endgroup$ – LetmeTry_Math Jan 3 '17 at 6:15
  • $\begingroup$ Yes, we may move the vectors so that they are on the same plane. I'm not talking position vectors @LetmeTry_Math $\endgroup$ – Ahmed S. Attaalla Jan 3 '17 at 6:17
  • $\begingroup$ Also $v$ and $w$ are not (necessarily) "parallel vectors", I hope what you mean they are parallel to the plane. That is true. $\endgroup$ – Ahmed S. Attaalla Jan 3 '17 at 6:19
  • $\begingroup$ If v vector is parallel to the line of intersection, w vector is lying on the plane of interest, then v and w vectors must be parallel (because that is one of the constraint), right? $\endgroup$ – LetmeTry_Math Jan 3 '17 at 6:25
  • $\begingroup$ No, $v$ is one vector on the plane $A$ (if we move it so it is) $w$ is another. No one said they had to be parallel. The line of intersection of the two planes are parallel to $A$ is what is given. That is what I use to conclude that $n_1 \times n_2$ is parallel to the plane. $\endgroup$ – Ahmed S. Attaalla Jan 3 '17 at 6:28
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Hint: use the fact that since the new plane passes through the intersection of both the planes , it will satisfy the equations of both the planes. Therefore Plane 1 + (c)*Pane 2 =0 where C is a constant. Use the given point and the direction ratioS

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  • $\begingroup$ Thanks for your answer. The new plane may or may not pass through the intersection. All we know is the new plane is parallel to the line of intersection of plane1 and plane2. $\endgroup$ – LetmeTry_Math Jan 3 '17 at 5:40
  • $\begingroup$ Yes off course but if you get the equation of the plane passing through the intersection and the new plane is parallel to this plane of intersection , we know the direction cosines of the new plane and we know the points through which it passes . $\endgroup$ – Shashaank Jan 3 '17 at 6:22
  • $\begingroup$ I am sorry, I am not quiet understanding. I guess, my mind isn't working right. Thanks for your answer still. $\endgroup$ – LetmeTry_Math Jan 3 '17 at 6:26
  • $\begingroup$ No problem. What I am saying is that you can find out the equation of intersecting planes and your new plane has to to be parallel to it , so you know the direction cosines of the new plane ....further you have 2 points which lie on it. It's just about direction ratios and algebraic manipulation. $\endgroup$ – Shashaank Jan 3 '17 at 6:37
  • $\begingroup$ Thank you. I guess I am understanding it better now. $\endgroup$ – LetmeTry_Math Jan 3 '17 at 6:39

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