Re: Numerical for you ....from vectors

faizankhan666@xxxxxxxxx wrote:
Question:

Two vectors of magnitude "a" and "b" make an angle "theeta" with each
other when placed tail to tail.Prove,by taking components along two
perpendicular axes that:


r=(a square +b square +2abcos(theeta)) (whole
under root)


gives the magnitude of the sum r of the two vectors.


This question is from "fundamentals of physics" "6th edition".


We shall call the two vectors a and b. There magnitudes, or L2-norm, will be denoted by |a| and |b|, respectively. Without loss of generality, we can choose the x-axis parallel to the vector a and the origin of the orthonormal Cartesian system at the tail of the vector a. Let the vector i and j be the unit vector of the x-axis and the y-axis, respectively. Now we can write a and b in component forms as

a = a_x i + a_y j = a_x i since a_y = 0 with our choice of axes,

b = b_x i + b_y j

Say that the vector c is the sum of a and b. Therefore

c = a + b = (a_x + b_x) i + b_y j

To simplify the computation we can look at the square of the magnitude of c so we do not have to carry on any square root until the final answer. So the square of the magnitude of the vector c is equal to the dot product of the vectors a and b, which can be written in component form as

|c|^2 = a . b = (a_x + b_x)^2 + (b_y)^2

From here, you can continue by expanding the first term and after some rearranging, you should notice that you have the sum (b_x)^2 + (b_y)^2 which is equal to |b|^2 and that you can express the component b_x as sin(theta - Pi/2) if the angle theta is greater than Pi/2. Using simple trigonometric identities about sin and cos, you should be able to complete the proof.

HTH,
Jean-Marc
.

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