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Advanced Math/Span of vectors


QUESTION: I am having difficulty understanding what to do for this problem. Any help would be greatly appreciated.

Let vector v = <1,4,2> and vector w = <3,1,-1>.
(a) Explain why the Span of vector v and vector w is a plane.
(b) Find conditions on x, y, z that ensure that <x,y,z> is in the span of vector v and vector w.
(c) Compute (vector v)cross product(vector w) and use it to find an equation for the plane
 pi: (x, y, z) = (0, 0, 0) + s<1,4,2> + <3,1,-1>
(d) Explain the link between (b) and (c).

ANSWER: Please show at least an attempt to figure this problem out. State what you know about planes and vectors and 3-D vs 2-D. Draw a picture. Even if you can't do it algebraically, what should you deduce about vectors v and w that would make them lie in a plane.

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QUESTION: I found that the span is two dimensional because the two vectors are linearly independent. This can be seen with the two vectors because neither is a multiple of the other.
As for it being a plane, if any two points are taken in the span, the line segment between them is entirely contained in the span. I think i got the idea for that part.

I am having difficulty understanding what conditions would ensure that <x,y,z> is in the span of vector v and vector w. By conditions do they mean the interval of values that a parameter in front of <x,y,z> must be equal to?
Also I have found the cross product between vector v and vector w and have also found the equation for the plane but I don't understand how there can be a link between the equation of the plane and the conditions that would ensure that <x,y,z> is in the span of vector v and vector w. Any help would be greatly appreciated.

ANSWER: Thanks for showing your progress on this. For a), the vectors are indeed linearly independent, which means they are not parallel (although parallel, non-coincident, lines in 3-D do form a plane). However, 2 lines that are not parallel may be skew in 3-D, which means they don't intersect. They are not skew if they intersect and it appears that for the information given in this problem, they intersect at the origin, (0,0,0).

For b), we need to make sure that the vector u = <x,y,z> lies in the plane. As per the definition of a plane, the line represented by u must be perpendicular to lines that are perpendicular to v and w, in other words the dot product of u with the normal to v and w must be zero

(x,y,z)・(vxw) = 0

so that

vxw = i(-6) + j((7) + k(-11)

and -6x + 7y - 11z = 0.

For c), you have already calculated the cross product as in b). So there's the connection.

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QUESTION: I am just curious where you get they intersect at the point (0,0,0) from. Is it from part c) ? So for a) what is being said is that since the span of v and w intersects with the plane it forms a plane? I am just trying to make sure about this statement. Thanks for any help.

Since the information doesn't specify points through which the vectors pass, which would unequivocally define a line, I assumed that the vectors originated at the origin. Three numbers along x,y,z, respectively, define a point which can be considered as the endpoint of a vector (just like in highschool).

Two non-parallel, intersecting vectors define a plane since therir cross-product is non-zero and generates a normal vector. The points of the plane are all points that lie on vectors perpendicular to this normal.  

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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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