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April 8th, 2014, 07:21 AM   #1
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Question vectors

Plane P contains point A(1,2,3) and is parallel to u=2i +3j and v=i +2j -k

Vector normal to plane is n=(3,a,b)

Find suitable values for a and b.

I get a=-2 b=-1, someone mentioned before they think the a value is wrong.

I found it simply by dot product of u and n being equal to zero, then using same principle to find b with dot product of v and n.

If anyone could confirm whether I am right or wrong, that would be great.
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April 8th, 2014, 10:11 AM   #2
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Quote:
Originally Posted by marks2014 View Post
I get a=-2 b=-1
I agree with these values.
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April 8th, 2014, 11:23 AM   #3
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the normal vector n = (3, a, b) with the parallel vectors u = (2, 3, 0) and v = (1, 2, -1) has to be orthogonal i.e. in a scalar product has to give zero or you have the system

6 + 3 a + 0 b = 0 and 3 + 2 a - b = 0

so a resolves from the first equation as -2 and b resolves from the second equation as 3 - 4 = b = -1. everything is fine till here...

The point A = (1, 2, 3) is not really on that plane P because (1, 2, 3) = g (2, 3, 0) + h (1, 2, -1) makes h = -3 from the third component and for g from the first two components you get two different values 1 = 2 g - 3 and 2 = 3 g - 6 so g has to be both 2 and 8/3 and it is conflict.
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April 8th, 2014, 11:26 AM   #4
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Originally Posted by dobri View Post
The point A = (1, 2, 3) is not really on that plane P because (1, 2, 3) = g (2, 3, 0) + h (1, 2, -1) makes h = -3 from the third component and for g from the first two components you get two different values 1 = 2 g - 3 and 2 = 3 g - 6 so g has to be both 2 and 8/3 and it is conflict.
Who said P contains (0, 0, 0)?
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April 8th, 2014, 11:36 AM   #5
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i say P contains A = (3, 5, -1)
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April 8th, 2014, 11:39 AM   #6
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But by saying that $P = \{g (2, 3, 0) + h (1, 2, -1)\mid g, h\in \Bbb R\}$ you are also saying that (0, 0, 0) lies in P.
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