Intersect Two Perpendicular Planes?

[math12345]

The line of intersection of two perpendicular planes is r=(-1+5t)i +(5-t)j +(7-4t)k. One plane is 5x-3y+7z=29. Find the rectangular equation of the other plane.

The Attempt at a Solution

I think I need to find three points in the desired plane. (one point being (-1,5,7)?)

 

[icystrike]

Do know of any two lines that is parallel to the plane?

Then can u find the rectangular equation by:

r.n=D ? ( But first have to find the normal vector by ? )

 

[math12345]

okay, so does this seem right?

Direction vectors:
The first is the direction vector of the line = (5, -1, -4) ,
the second is the normal vector of the first plane = (5, -3, 7)
And (- 1, 5, 7) is a point of the second plane (and of the first plane)

So the equation of the second perpendicular plane is in parametric form

plane: (-1, 5, 7) + r(5, -1, -4) + s(5, - 3, 7)

To find the normal form of the plane, calculate the cross product of the two direction vectors,
and plug in the point that you have already.

the cross product of (5, -1, -4)x(5, -3, 7) = ( - 19, - 55, -10)
and (- 19, -55, -10)•(-1, 5, 7) = - 326
so the normal equation and FINAL ANSWER of the perpendicular plane is

19x+55y+10z=326

 

[icystrike]

Yes.. I've gotten the same ans too..