Answer:
Same direction to produce maximum magnitude and opposite direction to produce minimum magnitude
Explanation:
Let a be the angle between vectors A and B. Generally when we add A to B, we can split A into 2 sub vectors, 1 parallel to B and the other perpendicular to B.
Also let A and B be the magnitude of vector A and B, respectively.
We have the parallel component after addition be
Acos(a) + B
And the perpendicular component after addition be
Asin(a)
The magnitude of the resulting vector would be
[tex]\sqrt{(Acos(a) + B)^2 + (Asin(a))^2}[/tex]
[tex] = \sqrt{A^2cos^2a + B^2 + 2ABcos(a) + A^2sin^2a}[/tex]
[tex] = \sqrt{A^2(cos^2a + sin^2a) + B^2 + 2ABcos(a)}[/tex]
[tex] = \sqrt{A^2 + B^2 + 2ABcos(a)}[/tex]
As A and B are fixed, the equation above is maximum when cos(a) = 1, meaning a = 0 degree and vector A and B are in the same direction, and minimum with cos(a) = -1, meaning a = 180 degree and vector A and B are in opposite direction.
The resulting vector will be maximum when [tex]\theta[/tex] will be equal to zero, and both vectors A and B will be in the same direction.
What is Vector Quantity?
These are the quantities that have both magnitude and direction. The magnitude of the resulting vector can be calculated by the Pythagorean theorem.
Divide the vector A into its sub-components:
Parallel component to B,
[tex]A{\rm \ cos\theta }+ B[/tex]
Perpendicular Component,
[tex]A\rm \ sin\theta[/tex]
So, the magnitude of the resulting vector,
[tex]R = \sqrt {(A {\rm \ cos \theta} +B)^2 + A{\rm sin\theta ^2}\\[/tex]
After simplifying it we get,
[tex]R = \sqrt {A^2 +B^2 + 2AB{\rm cos \theta }\\[/tex]
Therefore, the resulting vector will be maximum when [tex]\theta[/tex] will be equal to zero, and both vectors A and B will be in the same direction.
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