In order to span a vector space of dimension 3 you need three vectors. You only list 2.
2007-12-17 03:03:07
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answer #1
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answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6
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Vector space V is spanned by vectors
t = <2, -1, 2> and u = <1, -2, 2>.
a. Find an orthogonal basis in V
All the vectors in V or orthogonal to the normal vector n. Take the cross product.
n = t X u = <2, -1, 2> X <1, -2, 2> = <2, -2, -3>
Now find a vector in V that is orthogonal to both u and n. Take the cross product.
w = u X n = <10, 7, 2>
The orthogonal basis for V is t and w.
t = <2, -1, 2>
w = <10, 7, 2>
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b. Is vector v = <1, 4, 2> in V?
Solve for a and b.
at + bu = v
a<2, -1, 2> + b<1, -2, 2> = <1, 4, 2>
2a + b = 1
-a - 2b = 4
2a + 2b = 2
These equations lead to an inconsistency. There is no solution. So v does NOT lie in V.
Now that I think about it, you could also check to see if vector v lies in V. If it does, its cross product with the normal vector should equal zero. It does not. So again, v does NOT lie in V.
c. Not possible since v does not lie in V.
2007-12-18 10:08:37
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answer #2
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answered by Northstar 7
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All you're able to desire to do is learn the 7 or 8 axioms indoors the definition of a vector section. as an celebration, between the axioms says that scalar multiplication ought to desire to distribute over vector addition: r(v + w) = rv + rw, the area r is a scalar and v and w are vectors. indoors the case of R as a vector section over Q, r is a rational determination and v and w are any truly numbers. yet then r,v,w are all truly numbers and the distributive ingredients of multiplication of top numbers is a properly many circumstances happening fact. extra desirable, the various axioms of a vector section additionally shop on with trivially from the worry-loose arithmetic properties of top numbers. as an celebration, 0 (in R) is the additive id of the vector section, and addition in R is the two associative and commutative.
2016-12-11 07:35:27
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answer #3
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answered by Anonymous
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