The magnitude of the vector from P = (x1, y1, z1) to Q = (x2, y2, z2)
is sqrt[(x2 - x1)2 + (y2- y1)2 + (z2 - z1)2]
(Pythagoras in 3-D).
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The magnitude of a vector can be found by taking the square root of each of the vector components squared. For example, if you had the vector 3i+4j, to find the magnitude, you take sqrt ( 3²+4² ) To get: sqrt ( 9+16 ) sqrt ( 25 ) = 5 Works the same in 3D or more, just put all the vector components in.
You need to take the magnitude of the cross-product of two position vectors. For example, if you had points A, B, C, and D, you could take the cross product of AB and BC, and then take the magnitude of the resultant vector.
No. If the vector is 2D then it's magnitude is (x^2+y^2)^0.5 where x and y are the components and ^0.5 means take the square root. In 3D this becomes (x^2+y^2+z^2)^0.5 etc. Thus the magnitude is always at least as big as one of the components. Here's an example of a 3D vector: (3,4,5) |(3,4,5)|=(3^2+4^2+5^2)^0.5=(9+16+25)^0.5=50^0.5=7.07... If y and z were 0: (3,0,0) |(3,0,0)|=(3^2+0^2+0^2)^0.5=(9+0+0)^0.5=9^0.5=3 ie the magnitude is the same size as x. You can also consider this geometrically. A vector is an arrow and the magnitude represents the length of the arrow. Vector addition is the 'adding' of these arrows so (3,4,5)=(3,0,0)+(0,4,0)+(0,0,5). Clearly the length of an arrow built of three smaller ones can't be less than any one of them.
here are the possible answers: A) A tridimensional vector B) A 4D vector C) A 5D vector D) An scalar number E) It is undefined
A vector can be resolved into infinitely many sets of components in both 2D and 3D space.