No, the magnitude of a vector (in Euclidean space) is the square root of the sum of the squares of its components. This value can never be greater than the value of one of its own components.
v = √((vx)2 + (vy)2 + (vz)2)
v2 = (vx)2 + (vy)2 + (vz)2
(vx)2 = -(vy)2 - (vz)2 + v2
vx = √(-(vy)2 - (vz)2 + v2)
Can vx > v?
Substituting:
√(-(vy)2 - (vz)2 + v2) > √((vx)2 + (vy)2 + (vz)2).
Simplified:
v2 > (vx)2 + 2(vy)2 + 2(vz)2.
Substituting again:
(vx)2 + (vy)2 + (vz)2 > (vx)2 + 2(vy)2 + 2(vz)2.
Simplifying again:
0 > (vy)2 + (vz)2.
This results in a fallacy, since 0 can't be greater than a positive number. This wouldn't work even if both vy and vz were 0.
Every number greater than 1 has 1 and the number itself as factors. If it has no other factors besides these two factors, it is a prime number. If it has more factors than 1 and itself, it is a composite number.
Larger by a factor of √2. If you draw a vector at 45°, then draw a vertical line from the end of that vector, you have a 45° right triangle. If you recall your trig. relationships, the hypotenuse of a 45° right triangle is √2 (or 1.414) times the length of either leg.
I think you meant to say that it's never greater than the least of the numbers.The easiest explanation is to simply remind you that the GCF is a factor of every numberin the group, and no factor of a number can be greater than the number itself.
< > = Greater than , less than and equal too
greater than > less than < pointy end towards the smaller value
no a vector cannot have a component greater than the magnitude of vector
can a vector have a component greater than the vector magnitude
No.
No.
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No.
No.
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No, a vector can not have any components greater than itself.
No.
No a vector may not have a component greater than its magnitude. When dealing with highschool phyics problems, the magnitude is usually the sum of two or more components and one component will offset the other, causing the magnitude to be less then its component