# Find an equation of the plane that passes through the....

• space-time
In summary: This will be the equation of the target plane.In summary, the problem is to find an equation of the plane that passes through the line of intersection of two given planes and is perpendicular to a third given plane. The solution involves finding the normal vectors of the first two planes, taking their cross product, finding a point on the line of intersection, finding the normal vector of the third plane, and using these to find the equation of the target plane. However, there may be different approaches to solving this problem, such as identifying points on the line of intersection and using those to find the equation of the plane.
space-time

## Homework Statement

Find an equation of the plane that passes through the line of intersection of the two planes 3x + 7y + 2z= 0 and -9x - 8y - 6z = 9, and is perpendicular to the plane -2x - 6y + 3z = -10.

## The Attempt at a Solution

Here is what I did:

I first took the two normal vectors (n1 and n2) of the first two planes mentioned.

n1 = (3, 7, 2)
n2 = (-9, -8, -6)

I then took the cross product of these two vectors:

n1 × n2 = (-26, 0, 39) We can call this vector V.

I then found the normal vector of the plane that is mentioned in the end of the problem (which I will denote as n3)

n3 = (-2, -6, 3)

Next I took the cross product between V and n3.

V × n3 = (234, 0, 156)

so then my normal vector for the plane that will be my solution is:

nfinal = (234, 0, 156)

I then found a point on the line of intersection in the plane by first setting z=0 and then solving a system of equations for x and y. Here is the system:

3x + 7y = 0
-9x - 8y = 9

solving for x and y yields:

x = -189/117 , y= 9/13 (and of course z = 0)

Now that I have my point and my direction vector, I can find the equation of the plane. I got:

(234, 0, 156) ⋅ ( x + 189/117 , y - 9/13 , z) = 0

which expands to:

(234x +378) + (0 - 0) + 156z = 0

which simplifies to

234x + 378 + 156z =0

or simply

234x + 156z = -378

That was the final answer that I plugged into the software for the equation of the desired plane. There were two answers that the problem wanted: They wanted the normal vector of the desired plane and the equation of the desired plane.

When I plugged in my normal vector of (234, 0, 156) and my equation 234x + 156z = -378, the software said that I got the normal vector right, but got the equation wrong.

Why?! I don't know what I did wrong here. Furthermore, I worked another example of this type of problem that simply had different numbers and equations with a friend earlier, and I got it right using the exact same process I used here! I've also looked up other examples of this online and every other example used the exact same process!

Can someone please help me with this (because I am really ticked off and don't know what is wrong)?

I don't understand why you are taking normal vectors to the first two planes, so I don't understand your approach.

Here's what I'd do:

1. Find the equation for the line L of intersection of the first two planes.

2. Identify two points P and Q on that line, eg by setting each of the two parameters to zero in turn.

3. Find the normal vector for the 3rd plane.

4. Add that normal vector to P to get a 3rd point R that is not on the line L. That plane must be in the target plane, since L is and the target plane is parallel to the normal vector.

5. Find the equation of the plane containing P, Q and R.

## 1. What is the equation of a plane?

The equation of a plane in three-dimensional space is typically written in the form ax + by + cz + d = 0, where a, b, and c are the coefficients of the variables x, y, and z, and d is a constant. This equation represents all the points (x, y, z) that lie on the plane.

## 2. How can I find the equation of a plane that passes through a specific point?

To find the equation of a plane that passes through a given point, you need to know the coordinates of the point and the normal vector of the plane. Then, you can use the point-normal form of the equation, which is (x - x0)n1 + (y - y0)n2 + (z - z0)n3 = 0, where (x0, y0, z0) are the coordinates of the given point and (n1, n2, n3) is the normal vector of the plane.

## 3. Can I find the equation of a plane using only two points?

No, you need at least three non-collinear points to uniquely define a plane. If you only have two points, there are infinitely many planes that can pass through them. However, if you have three points, you can find the equation of the plane using the point-point form, which is (x - x1)(y2 - y1) - (x2 - x1)(y - y1) = 0, where (x1, y1, z1) and (x2, y2, z2) are the coordinates of two points on the plane.

## 4. How do I find the equation of a plane if I know its normal vector and a point on the plane?

If you know the normal vector and a point on the plane, you can use the normal form of the equation, which is n1(x - x0) + n2(y - y0) + n3(z - z0) = 0, where (x0, y0, z0) is the point on the plane and (n1, n2, n3) is the normal vector.

## 5. Can I find the equation of a plane without knowing any specific points on the plane?

Yes, you can find the equation of a plane using only its normal vector. This is called the normal form of the equation, which is n1x + n2y + n3z + d = 0, where (n1, n2, n3) is the normal vector and d is a constant. However, keep in mind that this equation will represent all the planes with the given normal vector, so you may need to use additional information to narrow down the specific plane you are looking for.

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