dot product

**EquinoX** · January 23rd 2009, 07:13 AM

What is the equation of a plane which is through (-2,3,2) and parallel to 3x + y + z = 4

**Danny** · January 23rd 2009, 09:35 AM

Originally Posted by EquinoX

What is the equation of a plane which is through (-2,3,2) and parallel to 3x + y + z = 4

The new plane has the same normal

$\vec{n} = < 3,1,1>$

The plane containing your point is

$3(x+2) + 1(x-3) + 1(x-2) = 0$

or

$3x+y+z = -1$

**EquinoX** · January 23rd 2009, 10:08 AM

isn't that perpendicular? instead of parralel? what would I have to change if the question asks for parallel

**Danny** · January 23rd 2009, 11:05 AM

Originally Posted by EquinoX

isn't that perpendicular? instead of parralel? what would I have to change if the question asks for parallel

No. The two parallel planes will have parallel normals.

**EquinoX** · January 23rd 2009, 11:06 AM

so how would I work this problem out if it's perpendicular?

**Danny** · January 23rd 2009, 11:45 AM

Originally Posted by EquinoX

so how would I work this problem out if it's perpendicular?

Planes are perpendicular is their normals are perpendicular. Find a vector lying on the plane and use this for the normal of the second plane (BTW - there is an infinite number of perpendicular planes since there are an infinite number of vectors lying on a given plane.)

**EquinoX** · January 23rd 2009, 01:15 PM

I got confused by your answer as in the book there's a question with this answer:

Find the equation of the plane perpendicular to -1 + 3j + 2k and passing through the point (1,0,4):

-(x-1) + 3(y-0) + 2(z-4) = 0

which is the same way you approach the parallel problem... can you explain this to me?

**Danny** · January 23rd 2009, 01:24 PM

Originally Posted by EquinoX

I got confused by your answer as in the book there's a question with this answer:

Find the equation of the plane perpendicular to -1 + 3j + 2k and passing through the point (1,0,4):

-(x-1) + 3(y-0) + 2(z-4) = 0

which is the same way you approach the parallel problem... can you explain this to me?

Ah, but in this question, you want a plane perpendicualr to

$- i + 3j + 2k$ or $<-1,3,2>$ - this is a vector!

So you want the line following this vector to be parallel to your normal. Hence, the bold numbers

$ \bold{-1}(x-1) + \bold{3}(y-0) +\bold{2}(z-4) = 0. $

**EquinoX** · January 23rd 2009, 01:57 PM

the answer to the question is not the bold numbers instead it is:

-x + 3y + 2 = 7

**Danny** · January 23rd 2009, 09:27 PM

Originally Posted by danny arrigo

Ah, but in this question, you want a plane perpendicualr to

$- i + 3j + 2k$ or $<-1,3,2>$ - this is a vector!

So you want the line following this vector to be parallel to your normal. Hence, the bold numbers

$ ***\;\;\; \bold{-1}(x-1) + \bold{3}(y-0) +\bold{2}(z-4) = 0. $

If you expand this (**), you get

$ -x + 3 y + 2z = 7. $

**earboth** · January 24th 2009, 07:36 AM

Originally Posted by EquinoX

the answer to the question is not the bold numbers instead it is:

-x + 3y + 2 = 7

I don't want to pick at you but I think it's high time that you read this here: http://www.mathhelpforum.com/math-he...l-posters.html

**EquinoX** · January 24th 2009, 10:04 AM

okay let me get this straighten out: my original question asks to solve a parallel problem and the example from the book that I present ask for perpendicular, and you solve both the same way.. why is that?

**earboth** · January 24th 2009, 10:24 PM

Originally Posted by EquinoX

... my original question asks to solve a parallel problem

Two planes $p_1$ and $p_2$ are parallel if the normal vectors $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$ are collinear. That means:

$\overrightarrow{n_2} = k\cdot \overrightarrow{n_1}\ ,\ k\in \mathbb{R}$

see attachment.

If you are looking for a plane perpendicular to a given plane, then there are an infinite number of those planes. Example: Open a door. Normally the door is perpendicular to the floor, but there isn't only one position the door can have.

see attachment.

and the example from the book that I present ask for perpendicular, and you solve both the same way.. why is that?

If you have a vector ( $\vec n$ )and a point Q with its stationary vector $\vec q$ you can derive the equation of a plane which is perpendicular to the vector and contains the given point: Let R denote an arbitrary point in the plane with its stationary vector $\vec r$ . Then the vector $\vec r - \vec q$ must be perpendicular to the vector $\vec n$ . That means: For all points in the plane the equation

$\vec n \cdot (\vec r - \vec q)=0$

is true and therefore this equation describes the plane completely (Point-normal-form of the equation of a plane)

see attachment

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