Equation of Plane: Passes (5,-5,5), Perp to 3x-2z+1=0 & 4x+3y+7=0

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In summary, the equation of the plane that passes through (5,-5,5) and is perpendicular to the lines of intersection of the planes 3x-2z+1=0 and 4x+3y+7=0 can be found by using the cross product of the normals of the planes as the direction vector of the line of intersection. Then, by letting z=0 and solving for x and y using the given equations, a point on the line of intersection can be found. This point can then be used, along with the normal vector and the point (5,-5,5), to find the equation of the desired plane. The equation for the line of intersection, r=(-1/3,-17
  • #1
molly16
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Homework Statement


Determine the equation of the plane that passes through (5,-5,5) and is perpendicular to the lines of intersection of the planes 3x-2z+1=0 and 4x+3y+7=0

Homework Equations



Ax+By+Cz+D=0

The Attempt at a Solution



I found the cross product of the normals of the planes given. Then used that as the direction vector of the line of intersection. Then I let z=0 and solved for x and y using the equations of the planes given in order to find a point on the line of intersection. The equation for the line of intersection is r=(-1/3,-17/9,0) + s(6,8,9).

I'm not sure what to do now. Can someone please explain?
 
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  • #2
molly16 said:

Homework Statement


Determine the equation of the plane that passes through (5,-5,5) and is perpendicular to the lines of intersection of the planes 3x-2z+1=0 and 4x+3y+7=0


Homework Equations



Ax+By+Cz+D=0

The Attempt at a Solution



I found the cross product of the normals of the planes given. Then used that as the direction vector of the line of intersection. Then I let z=0 and solved for x and y using the equations of the planes given in order to find a point on the line of intersection.
Great! Now use the fact that if <A, B, C> is the normal to the plane and the plane contains the point [itex](x_0, y_0, z_0)[/itex] then the plane is given by [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex]. Of course, you use [itex](x_0, y_0, z_0)= (5, -5, 5).

The equation for the line of intersection is r=(-1/3,-17/9,0) + s(6,8,9).

I'm not sure what to do now. Can someone please explain?
The "equation of the line of intersection" is irrelevant. The point you are given is not near that line.
 
  • #3
Ohhh I see! Thanks for the help!
 

Related to Equation of Plane: Passes (5,-5,5), Perp to 3x-2z+1=0 & 4x+3y+7=0

What is the equation of the plane that passes through the point (5,-5,5), perpendicular to the planes 3x-2z+1=0 and 4x+3y+7=0?

The equation of the plane can be found by using the point-normal form of the equation of a plane. The normal vector can be found by taking the cross product of the normal vectors of the two given planes. From there, the equation can be written as (x-5)(3a-2c)+(y+5)(4c+3b)+(z-5)(3b+2a)=0, where a, b, and c are the coefficients of the normal vector.

What is the normal vector of the plane perpendicular to 3x-2z+1=0 and 4x+3y+7=0?

The normal vector can be found by taking the cross product of the normal vectors of the two given planes. In this case, the normal vector would be (3,-2,0) x (4,3,0) = (0,0,25).

How can I check if a point lies on the plane passing through (5,-5,5) and perpendicular to 3x-2z+1=0 and 4x+3y+7=0?

A point (x,y,z) lies on a plane if it satisfies the equation of the plane. Substituting the coordinates of the given point into the equation found in the first question, if the result is equal to 0, then the point lies on the plane.

How do I graph the plane passing through (5,-5,5) and perpendicular to 3x-2z+1=0 and 4x+3y+7=0?

To graph the plane, first graph the two given planes. Then, find the normal vector and use it to find three more points on the plane by adding and subtracting the coordinates of the given point. Connect these four points to form a quadrilateral, which represents the plane.

What is the distance from the plane passing through (5,-5,5) and perpendicular to 3x-2z+1=0 and 4x+3y+7=0 to the origin?

The distance from the plane to the origin can be found by using the formula |Ax0 + By0 + Cz0 + D|/√(A^2 + B^2 + C^2), where (x0,y0,z0) is any point on the plane and A, B, C, and D are the coefficients of the equation of the plane. In this case, (x0,y0,z0) can be taken as (5,-5,5) and the coefficients can be found by using the point-normal form of the equation of a plane.

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