Equation of plane perpendicular to another

In summary, the plane that passes through point P(-4, 2, 1) and is perpendicular to the plane x+5y+2z=3 has a normal vector that is parallell to (1,5,2). To find the equation for the plane, you must solve for a,b,c, which are the normal vector of the given plane.
  • #1
swearbynow
6
0
Find an equation of the plane that passes through point P(-4, 2, 1) and is perpendicular to the plane x+5y+2z=3
i really feel stupid with this question, i know how to do just about every other equation like this, just not with a plane perpendicular to another with a point, if anyone can just get me started i think i will be able to solve it
 
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  • #2
Well, the normal vector to the given plane should lie within the plane you're after.
 
  • #3
Can you write the general equation for a plane that goes through P?
 
  • #4
The normal vector to the given plane is parallell to (1,5,2)

It will be helpful to remember the parametric representation of a line going through the given point, and in the direction of that normal vector..
 
  • #5
nvm.
 
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  • #6
Well, that line, in its entirety, must lie in all planes that contain the given point and that are perpendicular to the given plane.

Those planes that go through the given point have the representation:
[tex]a(x+4)+b(y-2)+c(z-1)=0[/tex]
where a,b,c are fixed for any one, particular plane.
What equation for a,b,c can you now form for those planes within which that line lies?
 
  • #7
ok so would x+5y+2z=8 be a correct answer
 
  • #8
Eeh??

When inserting the line representation into the equation for the plane, you get the equation:
[tex](a+5b+2c)t=0[/tex]
What equation must therefore a,b,c fulfill?
 
  • #9
alright I am feeling stupid right now but i don't know what exactly your asking.

i was having a,b,c be the normal vector of the plane. 1,5,2. and then inserting them into the equation you had, 1(x+4) + 5(y-2) + 2(z-1)=0 and i got the answer above.
 
  • #10
I don't know why you deleted the line representation:
x=t-4,y=5t+2,z=2t+1

In its entirety, this line must lie in the plane(s) we are seeking:
a((t-4)+4)+b((5t+2)-2)+c((2t+1)-1)=0
In particular, it must hold for ALL values of t, showing that the constraint on the admissible planes is that their normal vectors must be perpendicular to the normal vector of the given plane, that is a+5b+2c=0.

For example, you may choose b=-1, c=1, hence a=3, yielding the admissible plane:
[tex]3(x+4)-(y-2)+(z-1)=0[/tex]
that is:
[tex]3x-y+z=-13[/tex]

The plane example you posted was parallell to the given plane, not perpendicular.
 
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  • #11
ok i think i understand now, thanks a lot
 
  • #12
ok i was lying before but now i do that's good thanks so much
 

1. What is the equation of a plane perpendicular to another plane?

The equation of a plane perpendicular to another plane can be found by taking the cross product of the normal vectors of the two planes. The resulting vector will be the normal vector of the perpendicular plane. Using this normal vector, the equation of the perpendicular plane can be written in the form ax + by + cz = d, where a, b, and c are the components of the normal vector and d is a constant.

2. How do you determine if two planes are perpendicular?

Two planes are perpendicular if their normal vectors are orthogonal, meaning their dot product is equal to zero. This can be checked by finding the normal vectors of the two planes and taking their dot product. If the dot product is equal to zero, the planes are perpendicular.

3. Can you have more than one plane perpendicular to another plane?

Yes, it is possible to have multiple planes that are perpendicular to a single plane. This is because there are infinite directions in which a plane can be perpendicular to another plane. For example, a plane can be perpendicular to another plane at any point along its surface, resulting in multiple perpendicular planes.

4. How does the equation of a perpendicular plane relate to the original plane?

The equation of a perpendicular plane is dependent on the equation of the original plane. This is because the normal vector of the perpendicular plane is determined by the normal vector of the original plane. Therefore, the equation of the perpendicular plane is a direct result of the original plane's equation.

5. Can a line be perpendicular to a plane?

Yes, a line can be perpendicular to a plane. This occurs when the line is parallel to the normal vector of the plane. In this case, the line and the plane will never intersect and will always be at a 90-degree angle to each other.

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