Find the cosine of the angle between the normals to the planes

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whig4life

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Having trouble figuring out this problem. Been working on it a while and getting no place. Looked through the book, Youtube, and even Google. If someone could show me how to set this one up, that'd be fantastic.

example: Find the cosine of the angle between the normals to the planes

x+y+2z=3 and 2x-y+2z=5
 
whig4life said:
example: Find the cosine of the angle between the normals to the planes
x+y+2z=3 and 2x-y+2z=5
Click to expand...

If each of \(\displaystyle M~\&~N\) is vector, the the cosine of the angle between them is \(\displaystyle \dfrac{M\cdot N}{\|M\|\|N\|}\).
 
Subhotosh Khan said:
Can you write the equations of the normals to those planes?
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No, the question makes no sense to me. Does it mean normal vector? Normal line? The vernacular of "normals" makes little sense to me. Please elaborate. I just am seeking to know how to approach and do this type of problem.
 
"Normal", in geometry, means "perpendicular". You asked about the angle between lines perpendicular to the two planes and pka told you that the cosine of the angle, \(\displaystyle \theta\), between two vectors, M and N, is given by \(\displaystyle cos(\theta)= \frac{|M||N|}{M\cdot N}\). When you said you did not understand that, Subhotosh Kahn asked if you knew how to write the equations of the "normals", that is the perpendicular lines you are interested in.

So, since you are asking about lines perpendicular to the given planes, the question is "Do you know how to write the equations of those perpendicular lines or (and this is actually easier) vectors in the directions of those perpendicular lines?"

If you do not even know how to write those vectors or lines, you will need to review. What do you know about the equations of planes and their perpendiculars?
 
HallsofIvy said:
"Normal", in geometry, means "perpendicular". You asked about the angle between lines perpendicular to the two planes and pka told you that the cosine of the angle, \(\displaystyle \theta\), between two vectors, M and N, is given by \(\displaystyle cos(\theta)= \frac{|M||N|}{M\cdot N}\). When you said you did not understand that, Subhotosh Kahn asked if you knew how to write the equations of the "normals", that is the perpendicular lines you are interested in.

So, since you are asking about lines perpendicular to the given planes, the question is "Do you know how to write the equations of those perpendicular lines or (and this is actually easier) vectors in the directions of those perpendicular lines?"

If you do not even know how to write those vectors or lines, you will need to review. What do you know about the equations of planes and their perpendiculars?
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I understand the formula but usually these problems start with a vector, not an equation of a plane. I cannot provide many more answers than that.
 
I have no idea what you mean by "these problems". The problem is about a plane. What I was hoping you knew is that if the equation of a plane is Ax+ By+ Cz= D, then any vector perpendicular to the plane is Ai+ Bj+ Ck and a line, perpendicular to the plane, through point \(\displaystyle (x_0, y_0, z_0)\) is given by \(\displaystyle x= At+ x_0\), \(\displaystyle y= Bt+ y_0\), and \(\displaystyle z= Ct+ z_0\).
 
The theory of it is great, I am wondering how to do it though. What are the operations I have to go through? I am as confused as ever.
 
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Subhotosh Khan said:
First you have to "find" the "normal" vectors (read the response by HallsOfIvy very carefully).

Also study your own thread:

http://www.freemathhelp.com/forum/t...-the-Equation-of-a-Plane-Through-Three-Points

understand what you did and why you did it!!
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Again, on the threat I started with vectors, not equations. It's that first step that is confusing me and stopping me from moving forward. How do I find the normal vectors from the equation of the plane?
 
Did you read my second response (#10):

if the equation of a plane is Ax+ By+ Cz= D, then any vector perpendicular to the plane is Ai+ Bj+ Ck and a line, perpendicular to the plane, through point \(\displaystyle (x_0, y_0, z_0)\), is given by \(\displaystyle x= At+ x_0\), \(\displaystyle y= Bt+ y_0\),\(\displaystyle z= Ct+ z_0\).

For your example, planes x+y+2z=3 and 2x-y+2z=5, the perpendicular (normal) vectors are <1, 1, 2> and <2, -1, 2>, respectively.
 
HallsofIvy said:
Did you read my second response (#10):

if the equation of a plane is Ax+ By+ Cz= D, then any vector perpendicular to the plane is Ai+ Bj+ Ck and a line, perpendicular to the plane, through point \(\displaystyle (x_0, y_0, z_0)\), is given by \(\displaystyle x= At+ x_0\), \(\displaystyle y= Bt+ y_0\),\(\displaystyle z= Ct+ z_0\).

For your example, planes x+y+2z=3 and 2x-y+2z=5, the perpendicular (normal) vectors are <1, 1, 2> and <2, -1, 2>, respectively.
Click to expand...


That is a much easier way to learn it thank you. Your above quoted answer would have been perfect if there has just been a little blerb along the lines "and notice that..." because I did not make the connection. Remember, I am still building on my math background.

I appreciate your patience with me as I learn this. In a few years, I will be helping some newcomer just as you are doing for me. Thank you.
 
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