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 April 8th, 2014, 06:21 AM #1 Member   Joined: Apr 2014 From: liverpool Posts: 37 Thanks: 4 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.
April 8th, 2014, 09:11 AM   #2
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Quote:
 Originally Posted by marks2014 I get a=-2 b=-1
I agree with these values.

 April 8th, 2014, 10:23 AM #3 Newbie   Joined: Apr 2014 From: Gevgelija Posts: 27 Thanks: 0 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.
April 8th, 2014, 10:26 AM   #4
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Quote:
 Originally Posted by dobri 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)?

 April 8th, 2014, 10:36 AM #5 Newbie   Joined: Apr 2014 From: Gevgelija Posts: 27 Thanks: 0 i say P contains A = (3, 5, -1)
 April 8th, 2014, 10:39 AM #6 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 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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