# Finding the scalar equation for a plane

• silver1x
In summary, this student attempted to find the scalar equation for a plane containing two lines, but ran into problems. They found the equation for one line, and plugged in the information for the second line to find the equation for the plane.
silver1x

## Homework Statement

Find the scalar equation for the plane containing L1 and L2.
Consider the lines:
L1 : x = t + 1, y = 2t, z = 3t - 1
L2 : x = s - 1, y = 2s + 1, z = 3s - 1

## Homework Equations

Scalar equation for a plane:
a(x - x0) + b(y - y0) + c(z - z0) = 0

## The Attempt at a Solution

These two lines have the same vector <1, 2, 3>, so they are parallel. I let t = 0 and s = 0 so that I found a point (1, 0, -1) for L1 and a point (-1, 1, -1) for L2. Where do I go from here? Thanks and I appreciate it.

Find the vector which connects those points. It will not be parallel with the original lines.

ehild

Ok so I found the displacement vector between the two points to be (-1 - 1, 1 - 0, -1 -(-1)), which comes up to (-2, 1, 0). Now that I have the vector, does it matter which point I use to plug into the formula to find the scalar equation?

You have two vectors lying in the plane, but you need a normal vector of the plane to get the scalar equation. How do you get a vector that is normal to the vectors of the plane?

What does the scalar equation of the plane mean? What are the parameters a,b,c?

ehild

Okay, I took the cross product with the displacement vector <-2, 1, 0> and the other vector <1, 2, 3> which ended up as <-3, -6, 5>. So now I would use that vector and pick one of those points to come up with the scalar equation. I end up with -3x - 6y + 5z = -8

Now, it is easy to check if that is correct, isn't it? The line L1 is given by x = t + 1, y = 2t, z = 3t - 1. Putting that into your equation for the plane, -3(t+1)- 6(2t)+ 5(3t-1)= -3t- 3- 12t+ 15t- 5= (-3- 12+ 15)t+ (-3- 5)= -8 which is correct for all t. Every point of L1 is in that plane. The line L2 is given by x = s - 1, y = 2s + 1, z = 3s - 1. Putting that into your equation, -3(s- 1)- 6(2s+ 1)+ 5(3x- 1)= -3s+ 3- 12s- 6+ 15x- 5= (-3-12+15)s+ (3- 6- 5)= -8 which is correct for all s. Every point of L2 is in that plane.

Thanks a lot for the help! To follow up on this problem. If I wanted to find the distance between the lines, would I just use the distance formula between the two points?

The distance between two parallel lines is the length of the segment of line, perpendicular to them. You know already a vector which connects the two lines: find its normal component.

ehild

Using the displacement vector <-2, 1, 0>, the component would be ( (-2)^2 + 1^2 + 0^2 ) ^ (1/2), which ends up as square root of 5. That would then be the distance correct?

As the displacement vector happens to be perpendicular to the lines, the distance is correct. ehild

## 1. What is a scalar equation for a plane?

A scalar equation for a plane is a mathematical representation of a plane in three-dimensional space using a single equation. It is in the form of 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.

## 2. How do you find the scalar equation for a plane?

To find the scalar equation for a plane, you need to know the coordinates of three points that lie on the plane. Then, you can use the formula Ax + By + Cz + D = 0 and solve for the coefficients A, B, C, and D using the coordinates of the three points.

## 3. What is the significance of the coefficients in a scalar equation for a plane?

The coefficients A, B, and C determine the orientation of the plane in three-dimensional space. They also help to determine the distance of the plane from the origin. The constant D represents the distance of the plane from the origin in the direction perpendicular to the plane.

## 4. Can a scalar equation for a plane be written in different forms?

Yes, a scalar equation for a plane can be written in different forms. It can be rewritten in the form of ax + by + cz + d = 0, where a, b, and c are the direction numbers of the normal vector to the plane. It can also be written in the form of (x - x1)/a = (y - y1)/b = (z - z1)/c, where (x1,y1,z1) is a point on the plane.

## 5. What is the relationship between a scalar equation and a vector equation for a plane?

A scalar equation for a plane and a vector equation for a plane are two different representations of the same plane. The vector equation for a plane is in the form of r = r0 + ta + ub, where r0 is a point on the plane, a and b are two vectors parallel to the plane, and t and u are scalar parameters. By equating the scalar equation Ax + By + Cz + D = 0 to the vector equation, we can find the values of a and b and rewrite the vector equation in the form of a scalar equation.

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