Plane Equations Homework: Parallel, Coincident, Distance

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Homework Statement



a)Determine whether the following two planes 3x-y+z=2 and 2x+2y-4z=5 are parallel, orthogonal,coincident (same) or none.

b)Find the equation of the plane that contains the point (2,3,-1) and parallel to the plane 5x-3y+2z=10

c)Find the distance from the point (3,5,-8) to the plane 6x-3y+2z=28


Homework Equations





The Attempt at a Solution

 
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Hi fazal! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
 
for A)
take the dot product of the two normal vectors. if its zero then they're perpendicular ?
take the cross product of the two normal vectors. if its zero then they're parallel ?

For them to be the same they'd have to be linear multiples of each other? how?

B)Parallel to the plane 5x-3y+2z=10 means a vector normal to the plane is (5, -3, 2). So the equation has the form 5x - 3y + 2z = d. To get d, substitute the point (2,3,-1).

therefore is the equation 5x-3y+2z=-1 after sub the above??
plse check for me...

c)For c, use the formula:

so the answer is:- 41/7
 
plse assist to check
 
fazal said:
for A)
take the dot product of the two normal vectors. if its zero then they're perpendicular ?
take the cross product of the two normal vectors. if its zero then they're parallel ?
What are the normal vectors for these planes? Are they parallel? Are they perpendicular?

For them to be the same they'd have to be linear multiples of each other? how?
For two vectors to be parallel, not "the same", one has to be a multiple of the other: <3, 2, 1> and <6, 4, 2> are parallel because <6, 4, 2>= 2<3, 2, 1>.

B)Parallel to the plane 5x-3y+2z=10 means a vector normal to the plane is (5, -3, 2). So the equation has the form 5x - 3y + 2z = d. To get d, substitute the point (2,3,-1).

therefore is the equation 5x-3y+2z=-1 after sub the above??
plse check for me...
Why would you need someone else to check for you? Is 5(2)- 3(3)+ 2(-1)= 1?

c)For c, use the formula:

so the answer is:- 41/7
Is that supposed to be negative? A distance is never negative.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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