The general idea is that 3 vectors are in a plane iff they are not linearly independent. This can be checked in several ways:
No. For three vectors they must all lie in the same plane. Consider 2 vectors first. For them to resolve to zero, they must be in opposite direction and equal magnitude. So they will lie along the same line. For 3 vectors: take two of them. Any two vectors will lie in the same plane, and their resultant vector will also lie in that plane. Find the resultant of the first two vectors, and the third vector must be along the same line (equal magnitude, opposite direction), in order to result to zero. Since the third vector is along the same line as the resultant vector of the first two, then it must be in the same plane as the resultant of the first two. Therefore it lies in the same plane as the first two.
The smallest magnitude resulting from the addition of vectors with individual magnitudes of 4 and 3 is 1, obtained when the directions of the two component vectors are 180 degrees apart.
Of course it is! for example, [1, √3] + [-2, 0] + [1, - √3 ] = [0, 0]. Like this example, all other sets of such vectors will form an equilateral triangle on the graph.. Actually connecting the endpoints of the 3 vectors forms the equilateral triangle. The vectors are actually 120° apart.
2 inches, 3 inches, and 4 inches
The vectors can not be both equal, but they can have the same magnitude of 3, if they are at a 60 degree angle.
No. For three vectors they must all lie in the same plane. Consider 2 vectors first. For them to resolve to zero, they must be in opposite direction and equal magnitude. So they will lie along the same line. For 3 vectors: take two of them. Any two vectors will lie in the same plane, and their resultant vector will also lie in that plane. Find the resultant of the first two vectors, and the third vector must be along the same line (equal magnitude, opposite direction), in order to result to zero. Since the third vector is along the same line as the resultant vector of the first two, then it must be in the same plane as the resultant of the first two. Therefore it lies in the same plane as the first two.
The term collinear is used to describe vectors which are scalar multiples of one another (they are parallel; can have different magnitudes in the same or opposite direction). The term coplanar is used to describe vectors in at least 3-space. Coplanar vectors are three or more vectors that lie in the same plane (any 2-D flat surface).
Just one. A single 2-dimensional plane can be extended into 3-dimensions by the use of normal vectors to the plane, which by definition extend outwards of the plane at a right angle.
When we use the term planar, we're talking about two dimensions. We could also be designating a plane or flat, 2-dimensional construct. Planar figures are ones drawn in two dimensions and lying on a plane. Every single triangle is a planar figure, as is every circle, just to cite two examples.
The smallest magnitude resulting from the addition of vectors with individual magnitudes of 4 and 3 is 1, obtained when the directions of the two component vectors are 180 degrees apart.
Of course it is! for example, [1, √3] + [-2, 0] + [1, - √3 ] = [0, 0]. Like this example, all other sets of such vectors will form an equilateral triangle on the graph.. Actually connecting the endpoints of the 3 vectors forms the equilateral triangle. The vectors are actually 120° apart.
2 inches, 3 inches, and 4 inches
The vectors can not be both equal, but they can have the same magnitude of 3, if they are at a 60 degree angle.
1) Separate the vectors into components (if they are not already expressed as components). 2) Add each of the components separately. 3) If required, convert the vectors back to some other form. For twodimensional vectors, that would polar form.
You do the dot product of the vectors by multiplying their corresponding coordinates and adding them up altogether. For instance: <1,2,3> ∙ <-3,4,-1> = 1(-3) + 2(4) + 3(-1) = -3 + 8 - 3 = 2
(i) They are linearly dependent since the 2nd vector is twice the 1st vector. All 3 vectors lie in the x-z plane, so they don't span 3D space. (ii) They are linearly independent. Note that the cross-product of the first two is (-1,1,1). If the third vector is not perpendicular to the above cross-product, then the third vector does not lie in the plane defined by the first two vectors. (-1,1,1) "dot" (1,1,-1) = -1+1-1 = -1, not zero, so 3rd vector is not perpendicular to the cross product of the other two.
No it is not. It's possible to have to have a set of vectors that are linearly dependent but still Span R^3. Same holds true for reverse. Linear Independence does not guarantee Span R^3. IF both conditions are met then that set of vectors is called the Basis for R^3. So, for a set of vectors, S, to be a Basis it must be:(1) Linearly Independent(2) Span S = R^3.This means that both conditions are independent.