# P6 Chap 4CWatch

This discussion is closed.
#1
10 a) find an equation of a plane in the form r.n=p which contains the line L and the point with position vector a where

L ; r= t(2i+3j-k), a= i +4k
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13 years ago
#2
(Original post by ziya_the_king)
10 a) find an equation of a plane in the form r.n=p which contains the line L and the point with position vector a where

L ; r= t(2i+3j-k), a= i +4k
It contains the origin (put t=0) and is normal to the vectors

(2,3,-1) and (1,0,4)

i.e. normal to their cross-product = (12,-9,-3)

so r.(12,-9,-3) = 0

will do
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#3
(Original post by RichE)
It contains the origin (put t=0) and is normal to the vectors

(2,3,-1) and (1,0,4)

i.e. normal to their cross-product = (12,-9,-3)

so r.(12,-9,-3) = 0

will do
why does it contain the origin, why normal to vectors? and answer is r.(4i-3j-k)=0 ... i think i need more help to understand it!
0
13 years ago
#4
(Original post by ziya_the_king)
why does it contain the origin, why normal to vectors? and answer is r.(4i-3j-k)=0 ... i think i need more help to understand it!
divide my normal by 3 and you get the same equation

both answers are right - there isn't a unique equation for a plane - well there is but only up to scalar multiples
0
13 years ago
#5
(Original post by ziya_the_king)
why does it contain the origin, why normal to vectors? and answer is r.(4i-3j-k)=0 ... i think i need more help to understand it!
it contains the origin because the line L, r=t(2i+3j-k), passes through the origin. (when t = 0).
The vector component of L, (2i+3j-k), lies in the plane. The vector OA=(1,0,4) also lies in the plane since a is a point in the plane and the plane passes through the origin.
Since OA and (2i+3j-k) are vectors in the plane then their cross product gives n, the normal to the plane.
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#6
(Original post by Fermat)
it contains the origin because the line L, r=t(2i+3j-k), passes through the origin. (when t = 0).
Ok it does pass through the origin but what does dat show? How does it help us? ( U can see im struggling with vectors!!!)
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#7
There is also another part of da same qs which now the line is slightly in a different form:

l: r=4i+j-2k+t(-i+j+4k) , a= -i+j+2k
0
13 years ago
#8
(Original post by ziya_the_king)
There is also another part of da same qs which now the line is slightly in a different form:

l: r=4i+j-2k+t(-i+j+4k) , a= -i+j+2k
You need 2 vectors that lie on the plane, so that you can cross them to get 'n'
You also need a position vector of a point on the plane. With these, you can apply r.n = a.n

You've already got 1 vector- the direction vector of the line l: (-i+j+4k)
To get the other just use the rule that vector AB = b-a, where b=(4i+j-2k) and a=(-i+j+2k)

AB = 5i - 4k
Cross these two vectors to get 'n'

I get -4i + 16j - 5k.
So we can say n = (4i-16j+5k)

r.n = a.n
r.(4i-16j+5k) = (-i+j+2k).(4i-16j+5k)
r.(4i-16j+5k) = -10
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#9
(Original post by mockel)
To get the other just use the rule that vector AB = b-a, where b=(4i+j-2k) and a=(-i+j+2k)

why do we say b=(4i+j-2k) and then minus the position vector?...i know we have to get a vector in a plane but i dont know why u have chosen b as dat vector.
0
13 years ago
#10
In general, the equation of a line is r = a+tb, where 'a' is a position vector of a point on the line.
Since we know that this line lies on the plane, this point 'a' (4i+j-2k) must also lie on the plane.

So, we've now got 2 points on the plane: (4i+j-2k) and (-i+j+2k)

To get the normal vector to the plane, n, we use cross product between 2 vectors that lie on the plane (i.e. not points).
We have one vector already- the direction vector of the line-i+j+4k).
We now need another vector. Well, this can be obtained from the two points that we have, by doing AB = b-a.
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