Equation of Plane Passing Through (-1,2,1): 2x-3y+z-4=0

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In summary, the problem involves finding the equation of a plane passing through a given point and containing the line of intersection of two planes. To do this, we can take the normal vectors of the given planes, take the cross product to find a vector orthogonal to the desired plane, and then use this vector and the given point to find the equation of the plane. This involves finding the direction from the given point to a point on the line of intersection, finding a vector perpendicular to both directions, and using a known point on the plane to satisfy the scalar equations.
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Winzer
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Homework Statement


Find the equation of the plane that passes throughthe point (-1,2,1) and contains the line of intersection of the planes x+y-z=2, and 2x-y+3=1


Homework Equations


[tex]a(x-x_{o})+b(y-y_{o})+c(z-z_{o})=0[/tex]

The Attempt at a Solution


My reasoning is that we can take the normal vectors of the given planes, take the cross product, which will be orthoganol to the plane we want.
We then just plug the obtained normal vector and the point into the equation. Right?
 
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  • #2
that's correct
 
  • #3
Just wanted to check.
I get an anwer but it is wrong from the books, so it must be my doing.
 
  • #4
put up your calculations in tex and i'll follow em through
 
  • #5
Shouldn't you find the the line of intersection? And then the direction from the point (-1,2,1) to a point on the line? And then find a vector perpendicular to the two directions? And then pick any point that you know is going to be on your plane to satisfy your scalar equations?
 
  • #6
ZioX said:
Shouldn't you find the the line of intersection? And then the direction from the point (-1,2,1) to a point on the line? And then find a vector perpendicular to the two directions? And then pick any point that you know is going to be on your plane to satisfy your scalar equations?

yea that's right, i wasn't thinking straight. how do you find a vector that's perpindicular to the vector that points from the (-1,2,1) to the line of intersection? find a vector which when crossed with it = 0?
 
  • #7
ice109 said:
yea that's right, i wasn't thinking straight. how do you find a vector that's perpindicular to the vector that points from the (-1,2,1) to the line of intersection?

What is the angle between the cross product of two vectors and either of the two multiplicand vectors?
 
  • #8
D H said:
What is the angle between the cross product of two vectors and either of the two multiplicand vectors?

90deg , yea just cross the vector from (-1,2,1) and the line of intersection to the normal vector of the plane
 

1. What is the equation of a plane passing through the point (-1,2,1)?

The equation of a plane passing through the point (-1,2,1) is 2x-3y+z-4=0.

2. How do you determine if a point lies on the plane given by 2x-3y+z-4=0?

To determine if a point (x,y,z) lies on the plane given by 2x-3y+z-4=0, substitute the values of x, y, and z into the equation. If the resulting expression is equal to 0, then the point lies on the plane.

3. Can the equation of a plane passing through (-1,2,1) have different coefficients for x, y, and z?

Yes, the equation of a plane passing through a given point can have different coefficients for x, y, and z. For example, the equation 3x-2y+5z-7=0 also passes through the point (-1,2,1).

4. What is the normal vector of the plane given by 2x-3y+z-4=0?

The normal vector of a plane is a vector that is perpendicular to the plane. To find the normal vector of the plane given by 2x-3y+z-4=0, we can look at the coefficients of x, y, and z. In this case, the normal vector is (2,-3,1).

5. How do you find the distance from a point to the plane 2x-3y+z-4=0?

To find the distance from a point (x0,y0,z0) to the plane given by 2x-3y+z-4=0, we can use the formula d = |ax0 + by0 + cz0 + d| / √(a² + b² + c²), where a, b, and c are the coefficients of x, y, and z in the plane's equation. In this case, the distance from the point to the plane is |2x0 - 3y0 + z0 - 4| / √(2² + (-3)² + 1²).

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