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 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108  
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  
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108  Quote:
 
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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