Showing that the scalar product between all pairs of them is equally zero.
Two vector are orthogonal (or perpendicular) if the scalar product between them is zero. Lets see an example in 2D:
If we have two vector, a = (a1, a2) and b = (b1,b2), if
a · b = sqrt(a1*b1 + a2*b2) = 0
then a and b are orthogonal.
To do the same with three vector (a, b, c) you have to show that all the combinations give zero, i.e.,
a · b = a · c = b · c= 0
(as the scalar product is commutative, you only need to probe this three cases).
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.
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).
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.
The multiplicative resultant is a three unit vector composed of a vector parallel to the 3 unit vector and a vector parallel to the product of the 3 unit and 4 unit vectors. R = (w4 + v4)(0 +v3) = (w40 - v4.v3) + (w4v3 + 0v4 + v4xv3) R = (0 - 0) + w4v3 + v4xv3 as v4.v3 =0 ( right angles or perpendicular)
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
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.
Yes - if you accept vectors pointing in opposite directions as "parallel". Example: 3 + 2 + (-5) = 0
-- A singe vector with a magnitude of zero produces a zero resultant.-- Two vectors with equal magnitudes and opposite directions produce a zero resultant.
yes, as long as they have 120 degrees separating them from each other, (360/3). all vectors must have total x and y component values of 0.