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

[molly16]

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?

 

[HallsofIvy]

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 $(x_0, y_0, z_0)$ then the plane is given by $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. Of course, you use $(x_0, y_0, z_0)= (5, -5, 5).<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The equation for the line of intersection is r=(-1/3,-17/9,0) + s(6,8,9).<br /> <br /> I'm not sure what to do now. Can someone please explain? </div> </div> </blockquote> The "equation of the line of intersection" is irrelevant. The point you are given is not near that line.$

 

[molly16]

Ohhh I see! Thanks for the help!