(FP3)Vector: plane (OMG, I am completely blank in this..)Watch

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Thread starter 11 years ago
#1  for part (b), can anyone explain what is r, n and p and how to do the calculation?..

My teacher missed the vector plane part and so I am completely blank in this topic...

Thank you for your help.. you save my life 0
11 years ago
#2
n is the normal or prependicular to the plane.
p is a constant.

Have you come across

(r-a).n = 0 before?
0
Thread starter 11 years ago
#3
(Original post by silent ninja)
n is the normal or prependicular to the plane.
p is a constant.

Have you come across

(r-a).n = p before?
yes, but I have only seen the formula but I don't know how to use it..
what is "r"?..
0
11 years ago
#4
r= (x,y,z), it's a variable just like the r in a vector line . It's the position vector of any general point.

The vector product (of two vectors) gives us the perpendicular vector. In your question, since a, b and c lie in the same plane is perpendicular to the required plane. This is the normal, n.

From C4 work you'll remember that if the dot product of two vectors is zero, then they must be perpendicular. You can use this fact to define a plane.
The last thing we need to consider is a point that lies on the plane to use as our position vector, this is a. It can be any point, in your question for convenience use a given.
This image for a line equation helps as to why we subtract a, it's the same idea (otherwise the plan could be anywhere): So r-a is in the direction of the plane (AP in above case).

Putting all this together, a.n will give you the require constant.
0
Thread starter 11 years ago
#5
(Original post by silent ninja)
r= (x,y,z), it's a variable just like the r in a vector line . It's the position vector of any general point.

The vector product (of two vectors) gives us the perpendicular vector. In your question, since a, b and c lie in the same plane is perpendicular to the required plane. This is the normal, n.

From C4 work you'll remember that if the dot product of two vectors is zero, then they must be perpendicular. You can use this fact to define a plane.
The last thing we need to consider is a point that lies on the plane to use as our position vector, this is a. It can be any point, in your question for convenience use a given.
This image for a line equation helps as to why we subtract a, it's the same idea (otherwise the plan could be anywhere): So r-a is in the direction of the plane (AP in above case).

Putting all this together, a.n will give you the require constant.
thanks!

But why do you set (r-a).n=0 (p=0?)?

How to know that p=0?

And I found anothe problem: Why in part (b), there is no r?.. Only with n=p?..

I know I am stupid.. Hope you can help me >o<
0
11 years ago
#6
(-15,-10,-10) is the same as (3,2,2). They have taken out a factor of -5. It doesnt matter which direction the perpendicular goes in as long as it is perpendicular.
0
Thread starter 11 years ago
#7
(Original post by silent ninja)
(-15,-10,-10) is the same as (3,2,2). They have taken out a factor of -5. It doesnt matter which direction the perpendicular goes in as long as it is perpendicular.
Thanks, I just looked this out and I got another great problem about r above(I have just editted the post.. )

it's really confusing about the equation (r-a)n=p..
0
11 years ago
#8
I mistyped in my original post. It's (r-a).n=0

but this gives you
r.n = a.n
and a.n is a constant p.
0
11 years ago
#9
In the above question, they want it in cartesian form. So do everything normally up to the point

r.n = a.n

work out a.n, then replace r with r=(x,y,z).
Then expand the dot product of (x,y,z).n to give you a certesian equation.
0
Thread starter 11 years ago
#10
(Original post by silent ninja)
In the above question, they want it in cartesian form. So do everything normally up to the point

r.n = a.n

work out a.n, then replace r with r=(x,y,z).
Then expand the dot product of (x,y,z).n to give you a certesian equation.
Thanks! Very simple and clear!

You saved my little stupid life! Wish you every success in the future!! 0
11 years ago
#11
Cool, good luck with your exams 0
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