# Find equation of plane

by skyturnred
Tags: equation, plane
 P: 116 1. The problem statement, all variables and given/known data The plane that passes through the point (−3, 2, 1) and contains the line of intersection of the planes x + y − z = 3 and 4x − y + 5z = 5 2. Relevant equations 3. The attempt at a solution So I plugged the planes into a matrix and found the line of intersection with one parameter. I took the coefficients as the direction vector of the line. Plugging that into an equation of the plane, I get the following: (5/6)X-(19/5)Y+(56/5)Z but it says I'm wrong. I only have 1 attempt left at the right solution.. Please help! Thanks in advance!
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 Quote by skyturnred 1. The problem statement, all variables and given/known data The plane that passes through the point (−3, 2, 1) and contains the line of intersection of the planes x + y − z = 3 and 4x − y + 5z = 5 2. Relevant equations 3. The attempt at a solution So I plugged the planes into a matrix and found the line of intersection with one parameter. I took the coefficients as the direction vector of the line. Plugging that into an equation of the plane, I get the following: (5/6)X-(19/5)Y+(56/5)Z but it says I'm wrong. I only have 1 attempt left at the right solution.. Please help! Thanks in advance!
The plane you're looking for is the plane containing the given point and the line of intersection.

What did you get for the parametrization of the line of intersection?
 P: 116 I did the work a couple days ago and lost it.. but now that I re-do it, I don't remember what it was. But I DO remember not have any clue what I was doing. Anyways, the equation of the line of intersection is x=(-4/5)t y=(9/5)t z=t so I have to include this line.. in other words, the norm of the plane will be orthogonal to it right? But finding a direction vector that is orthogonal to that is a plane on it's own.. and I can't quite think of what to do. All I know is that it probably has to do with some of the following: I know that the dot product between two orthogonal vectors is 0 I know that the cross product between two vectors produces a vector that is orthogonal to both I realize that my answer that I showed above was actually an expression rather than an equation... but regardless, I'm pretty sure it's wrong anyways. That's about all I know though..
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Find equation of plane

 Quote by skyturnred I did the work a couple days ago and lost it.. but now that I re-do it, I don't remember what it was. But I DO remember not have any clue what I was doing. Anyways, the equation of the line of intersection is x=(-4/5)t y=(9/5)t z=t ...
That parametrization goes through the point (0,0,0).

Neither plane contains such a point.

However, both planes contain the point, $\displaystyle \left(\frac{8}{5}\,,\ \frac{7}{5}\,,\ 0\right)\,,$ so the line of intersection passes through this point.

An interesting way to approach this problem is to realize that any linear combination of the equations for the two planes, results in another plane which intersects the two given planes along that same line. Find the linear combination which also contains the point, (-3, 2, 1) .
 P: 116 EDIT 2: I think I may have found the right answer. I don't know what I was finding before by solving the matrix of the coefficients of the two planes.. but I did it the following way: The line of intersect will be orthogonal to both of the normal vectors of the planes. So I did the cross product of both of their normal vectors and get (4,-9,-5) I found an arbitrary point on the line intersection by choosing z=0 and solving for x and y and I get the point: (8/5, 7/5, 0) The difference between this point and the given point in the question will be another direction vector in the unknown plane. So now that I know two directional vectors that are in the plane, I did the cross product to find the normal vector of the plane in question. I get the following: (6, -19, 39)=n And then using this and the point given in the question (and the equation of a plane) I get the following as the equation of the plane: 6(x+3)-19(y-2)+39(z-1)=0 Is this correct? I only have one attempt left at the right answer. Also, by solving the matrix made by the coefficients of the planes, what am I finding? I must be finding something useful.. 1 0 (4/5) (8/5) 0 1 (-9/5) (7/5) I get: z=t y=(9/5)t x=(-4/5)t
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 Quote by skyturnred EDIT 2: I think I may have found the right answer. I don't know what I was finding before by solving the matrix of the coefficients of the two planes.. but I did it the following way: The line of intersect will be orthogonal to both of the normal vectors of the planes. So I did the cross product of both of their normal vectors and get (4,-9,-5) I found an arbitrary point on the line intersection by choosing z=0 and solving for x and y and I get the point: (8/5, 7/5, 0) The difference between this point and the given point in the question will be another direction vector in the unknown plane. So now that I know two directional vectors that are in the plane, I did the cross product to find the normal vector of the plane in question. I get the following: (6, -19, 39)=n And then using this and the point given in the question (and the equation of a plane) I get the following as the equation of the plane: 6(x+3)-19(y-2)+39(z-1)=0
The method is correct.

It seems to check out.

It simplifies to 6x-19y+39z=-17.

 Is this correct? I only have one attempt left at the right answer. Also, by solving the matrix made by the coefficients of the planes, what am I finding? I must be finding something useful.. 1 0 (4/5) (8/5) 0 1 (-9/5) (7/5) I get: z=t y=(9/5)t x=(-4/5)t
If t=-5, you have (x,y,z) = (4, -9, -5) .

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