Vectors, Planes, and intersections

[ParoXsitiC]

Homework Statement

The planes 3x+2y+z=6 and x+2y+5z=1- intersect along the line (x+2)/2 = (y-6)/(-7/2)= z.

A third plane passes through the origin and is perpendicular to the intersection of the first two planes, at what point do the three planes intersect?

Homework Equations

Intersection in parametric eq:

x = -2 + 2t
y = 6 - 7/2t
z = t

The Attempt at a Solution

At first glance I knew that a normal vector will be perpendicular to the plane, so I just took the vector from the intersection <2, -7/2, 1> and used that as the normal vector for the third plane. I knew it passes through origin so i used point (0,0,0)

I get

2x - 7/2y + z = D

I input (0,0,0) for (x,y,z) and get D= 0

Thus the equation for the third plane is

2x - 7/2y + z = 0


At this point I plot everything in maple and the intersection that was given to me doesn't even look like it's the true intersection, none the less my third plane doesn't look like its perpendicular to my intersection either.

We have never done 3 planes intersecting but I assume it similar to 2 planes, where I set a variable equal to zero such as x=0 and then solve the system of equations. Once the other 2 are found, we can then just find the third and that will be the point in the intersection. I am not sure how to find the vector

 

[LCKurtz]

Homework Statement

The planes 3x+2y+z=6 and x+2y+5z=1- intersect along the line (x+2)/2 = (y-6)/(-7/2)= z.

A third plane passes through the origin and is perpendicular to the intersection of the first two planes, at what point do the three planes intersect?

Homework Equations

Intersection in parametric eq:

x = -2 + 2t
y = 6 - 7/2t
z = t

The Attempt at a Solution

At first glance I knew that a normal vector will be perpendicular to the plane, so I just took the vector from the intersection <2, -7/2, 1> and used that as the normal vector for the third plane. I knew it passes through origin so i used point (0,0,0)

I get

2x - 7/2y + z = D

I input (0,0,0) for (x,y,z) and get D= 0

Thus the equation for the third plane is

2x - 7/2y + z = 0


At this point I plot everything in maple and the intersection that was given to me doesn't even look like it's the true intersection, none the less my third plane doesn't look like its perpendicular to my intersection either.

That might be caused by not using the "scaling=constrained" option in the Maple plot.

We have never done 3 planes intersecting but I assume it similar to 2 planes, where I set a variable equal to zero such as x=0 and then solve the system of equations. Once the other 2 are found, we can then just find the third and that will be the point in the intersection. I am not sure how to find the vector

I would multiply the equation of the third plane by 2 on both sides to get rid of the fractions. To find the intersection point you could either:

1. Solve the three plane equations in three unknowns for $x,y,z$ using substitution, row reduction, or determinants.

2. Probably easier, substitute the parametric equation of the original line of intersection into the third plane, solve for $t$, and use that $t$ to get $x,y,z$. Then check it trying $(x,y,z)$ in each plane.

 

[ParoXsitiC]

x+2y+5z=1- should be x+2y+5z=10

Also I had make sure scaling was constrained, but now trying to do it in maple it looks right. so must of been a small mistake.

None the less taking your advice I got

t = 100/69

solving for xyz:

x = 62/69
y = 64/69
z = 100/69

And it checks for all 3 planes, so I guess it's right :)

 

[LCKurtz]

x = 62/69
y = 64/69
z = 100/69

And it checks for all 3 planes, so I guess it's right :)

That's what I would guess too.