Question

if an angle between two vectors of equal magnitude P is theta ,the magnitude of the difference of the vectors is

Answer 1

Vectors P and Q have equal magnitude P and angle between them is Ф.

Magnitude of vector P - Q = SqRoot[ P² + P² - 2 P P Cos Ф ]
        = √2 P SqRoot[ 1 - Cos Ф ] = √2 P SqRoot[ 2 Sin² Ф/2 ] = 2 P SinФ/2

Answer 2

Answer:

The magnitude of difference of two vectors is

$2Psin \frac{θ}{2}$

Explanation:

Given that,

The magnitude of both the vectors=P

The Angle between two vectors= θ

We are required to find the magnitude of difference between two vectors.

According to parallelogram law,The magnitude of difference between two vectors A and B making an angle θ is given as

$|A - B | =\sqrt{A {}^{2} + B {}^{2} - 2AB cosθ}$

Therefore,According to our given problem,the magnitude is

$\sqrt{P {}^{2} +P {}^{2} - 2P {}^{2} cosθ } \\ = \sqrt{2P {}^{2} (1 - cosθ)} \\ = \sqrt{2P {}^{2} .2sin {}^{2} \frac{θ}{2} } \\ = 2Psin \frac{θ}{2}$

Therefore,the magnitude of difference of two vectors is

$2Psin \frac{θ}{2}$

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