# Finding the equation of a plane

• Jun 4th 2010, 02:45 PM
Kakariki
Finding the equation of a plane
Hey, I was hoping you kind people here at mathhelpforum.com could help me out with a problem that I am stuck on.

Question
Find the equation of the plane that passes through the line of intersection of the planes 3x - y + z - 2 = 0 and x + 2y - 4z + 1 = 0, and that satisfies the given condition.
a) the plane passes through the point A(3,1,3)
b) the plane is parallel to the plane 5x + 3y - 7z - 6 = 0

Attempt at a solution
So taking the cross product of the 2 given planes gives the direction vector (2,13,7). Then setting a variable to 0, I chose x, I get the vector equation: $r = (0,\frac{-7}{2},\frac{3}{-2}) + t(2,13,7)$.

a)So, knowing that, how do I find the plane that goes through (3,1,3)? If I had the normal to the plane I could do it, so how do I find the normal to (2,13,7)?

b) The dot product of (5,3,-7) and (2,13,7) = 0. Which means I can sub in a point from the line of intersection in for x y and z, and solve for D. A point from the line of intersection is $(0,\frac{-7}{2},\frac{3}{-2})$. Solving for D gives me 21. So the equation of the plane that passes through $r = (0,\frac{-7}{2},\frac{3}{-2}) + t(2,13,7)$ and is parallel to [tex] 5x + 3y - 7z - 6 = 0 [tex] is $5x + 3y - 7z + 21$.

I am unsure if either of those are the correct way of doing it. If not, can someone put me on the right track?

Thank you so much for reading everyone! Hopefully you can help me out.
• Jun 4th 2010, 08:09 PM
Turiski
So, I took a slightly different approach.

$3x - y + z = 2$
$x + 2y -4z = 1$

I multiplied the top equation by two and added to get

$7x - 2z = 5~~~(1)$

What we want is two points, much more than a vector (right now). So I'll arbitrarily choose $(1,y_1,1)$ and $(-1,y_2,-6)$. Those clearly satisfy (1) and if you substitute, you get the points:

$(1,2,1)$ and $(-1,-11,-6)$

The reason you want two points is because for (a) now we have three points, which easily produce two vectors.

So the vector you pointed out: $(2,13,7)$, is from (1,2,1) to (-1,-11,-6). The other one we'll use is from (1,2,1) to (3,1,3) which is $(-2,-1,-2)$. Then we take the cross product.

$(2,13,7)$ x $(-2,-1,-2)= (19,10,-24)$ (or maybe negative that; as we both know it is irrelevant here). Barring any arithmetic errors, the equation for the plane is

$19x+10y-24z=15$

As for part (b), you have that $(2,13,7)$ vector from before, and you also know the given plane's vector. If you draw this out, you'll see the cross product of these two lie inside the plane that we want, and in fact form the best second vector for cross product with $(2,13,7)$. With less English, we want:

$[(2,13,7)$ x $(5,3,-7)]$ x $(2,13,7) = (-1110,-666,1554)$

Now we just use one of those two magic points:

$-1110x-666y+1554z=-888$. Divide by -222:

$5x+3y-7z=4$

(a) $19x+10y-24z=15$

(b) $5x+3y-7z=4$
• Jun 4th 2010, 08:42 PM
ANDS!
The point P(1,2,1) is not on the line of intersection of those two planes. It certainly works in the first equation, but it does not satisfy the second equation, thus that point is not on plane two, and it can not be in the line of intersection of the two points. It is a simple error that you have done, probably smack your head when you see it.

Quote:

Originally Posted by Kakariki
So taking the cross product of the 2 given planes gives the direction vector (2,13,7). Then setting a variable to 0, I chose x, I get the vector equation:

This is correct, although you do not need to get the vector equation, only the direction vector <2, 13, 7>. In part a), the want the plane to pass through the point A(3,1,3). Taking the point in the line of intersection that crosses the x-plane, Q(0,-7/2, -3/2), you can use point A and point Q to get the vector AQ (or QA). Now you have two vectors (and numerous points). You should be able here to find the normal vector to the plane (since you have two vectors), and use one of the points to define your plane.

Part B is correct.

Quote:

Originally Posted by Turiski
As for part (b)

Way too much work. If a plane is parallel to another plane, then they the normal vector of the first plane is also a normal to the second plane. Thus, using the same point from part a), and the normal, you can solve B in seconds, without actually taking any cross-products and what-nots.
• Jun 5th 2010, 05:23 AM
Kakariki
Quote:

Originally Posted by ANDS!
The point P(1,2,1) is not on the line of intersection of those two planes. It certainly works in the first equation, but it does not satisfy the second equation, thus that point is not on plane two, and it can not be in the line of intersection of the two points. It is a simple error that you have done, probably smack your head when you see it.

This is correct, although you do not need to get the vector equation, only the direction vector <2, 13, 7>. In part a), the want the plane to pass through the point A(3,1,3). Taking the point in the line of intersection that crosses the x-plane, Q(0,-7/2, -3/2), you can use point A and point Q to get the vector AQ (or QA). Now you have two vectors (and numerous points). You should be able here to find the normal vector to the plane (since you have two vectors), and use one of the points to define your plane.

Part B is correct.

Way too much work. If a plane is parallel to another plane, then they the normal vector of the first plane is also a normal to the second plane. Thus, using the same point from part a), and the normal, you can solve B in seconds, without actually taking any cross-products and what-nots.

Thank you very much for helping me! Seems you know your stuff very well, I am quite jealous.
Thanks again!!!
• Jun 5th 2010, 06:09 PM
Turiski
Ah! I knew I was missing something...