Physics, asked by varshashravankumar, 8 months ago

What is the magnitude of Ā =4i + 2j ?

Answers

Answered by MadMaxtheGreat

Answer:

What is the angle between vector A=4i+5j and vector B=3i+2k?

A⋅B=(4i+5j)⋅(3i+2k)=4×3=12 since all other terms are zero. Now divide by 42+52−−−−−−√×32+22−−−−−−√=533−−−√, because we really want the dot product of unit vectors in the directions of A and B, since that is the cosine of the angle between them.

This gives:

cosθ=12533√≈0.519777698

Take arccosine and get the desired angle:

θ≈arccos0.519777698≈58.68°.

cos a = (4*3+5*2)/sqrt((4^2+5^2)(3^2+2^2))=22/sqrt(41*13)=17.65 degrees =0.30805 radian

You divide dot product of the vectors by the product of their lengths.

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