Problem relating to vector spaces.?

Question

i have no clue how to solve this. help pls! thnkz.
determine whether the sets of vectors line in a plane, and if so, whether they lie on a line: v1=(1,2,1), v2=(-7,1,3), v3=(-14,-1,-4).

Answer 1

Calculate the cross product of v1 and v2:

v1xv2 = (5, -10, 15).

Since v1xv2 is, by definition, perpendicular to v1xv2, it is perpendicular to the unique plane in which those vecors lie. If v3 lies in that plane, then its scalar product with v1xv2 must be zero. Calculating that scalar product:

(v1xv2).v3 = (5, -10, 15).(-14,-1,-4) = -70 +8 - 60 = -122.

Since this is clearly not zero, it follows that v1, v2 and v3 do not lie in a plane.

Answer 2

Vector spaces are a mathematical abstraction. Sometimes, you can picture them, sometimes you can't. You've mentioned the vector space consisting of real polynomials of degree 2 (or less! this is important, as x^2 + 1 plus -x^2 is not a degree 2 polynomial!). Really though, that vector space is isomorphic to R^3 (the 3 dimensional Euclidean space), by setting {1, x, x^2} as basis vectors for the polynomial space, and mapping 1 -> (1, 0, 0), x -> (0, 1, 0), x^2 -> (0, 0, 1). So if it helps you picture it, you can look at it like R^3. The idea of the vector space is that it has these basis elements, and so you can always sort of break it down in that manner. But be careful, as not all vector spaces are finite-dimensional. For example, the space of all real continuous functions can be considered an infinite-dimensional vector space over the reals. Try picturing that one!

Answer 3

Take the cross product of any two vectors. Using the result, take the dot product with third vector. If the result is 0, the three vectors lie in a plane. This the method for calculating the volume of a parallelepiped. The volume is 0 if they lie in a plane.

The vectors lie in a line if and only if they are scalar multiples of each other.

Answer 4

The vectors all lie on a line only iff they are scalar multiples of each other.

v1 = av2 = bv3
where a and b are non-zero scalars.

This is clearly not the case for v1 and v2. So they don't all lie on a line.
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If the vectors are linearly independent then they do not all lie in a plane. If the determinant of the vectors is zero then they are not linearly independent and can all lie in a plane. If the determinant is zero then they all lie in a plane.

The determinant of the three vectors = -290 ≠ 0 so they do not all lie in a plane.