Question

if the vectorA×vector B=vector B×vectorA then the angle between A and B

Answer 1

 assuming you’re looking for a formula to calculate the angle between (A+B) and (A-B). The angle between these two vectors will depend on the magnitudes of A and B, and also the angle between A and B, which we will call θ.θ. Now let ϕϕbe the angle between (A+B) and (A-B). Using the concept of dot product, we get:

(A+B).(A−B)=|A+B|∗|A−B|∗cosϕ(A+B).(A−B)=|A+B|∗|A−B|∗cos⁡ϕ

=>|A|2+A.B−A.B−|B|2=(|A|2+|B|2+2ABcosθ−−−−−−−−−−−−−−−−−−√)(|A|2+|B|2+2ABcosθ−−−−−−−−−−−−−−−−−−√)∗cosϕ=>|A|2+A.B−A.B−|B|2=(|A|2+|B|2+2ABcos⁡θ)(|A|2+|B|2+2ABcos⁡θ)∗cos⁡ϕ

=>|A|2−|B|2=(|A|2+|B|2+2ABcosθ)(|A|2+|B|2−2ABcosθ)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√∗cosϕ=>|A|2−|B|2=(|A|2+|B|2+2ABcos⁡θ)(|A|2+|B|2−2ABcos⁡θ)∗cos⁡ϕ

=>|A|2−|B|2=(|A|2+|B|2)2−(2ABcosθ)2−−−−−−−−−−−−−−−−−−−−−−−√∗cosϕ=>|A|2−|B|2=(|A|2+|B|2)2−(2ABcos⁡θ)2∗cos⁡ϕ

=>|A|2−|B|2=|A|4+|B|4+2|A|2|B|2–4|A|2|B|2cos2θ−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√∗cosϕ=>|A|2−|B|2=|A|4+|B|4+2|A|2|B|2–4|A|2|B|2cos2⁡θ∗cos⁡ϕ

=>|A|2−|B|2=|A|4+|B|4+2|A|2|B|2(1–2cos2θ)−−−−−−−−−−−−−−−−−−−−−−−−−−−√∗cosϕ=>|A|2−|B|2=|A|4+|B|4+2|A|2|B|2(1–2cos2⁡θ)∗cos⁡ϕ

=>|A|2−|B|2=|A|4+|B|4+2|A|2|B|2(−cos2θ)−−−−−−−−−−−−−−−−−−−−−−−−−−√∗cosϕ=>|A|2−|B|2=|A|4+|B|4+2|A|2|B|2(−cos⁡2θ)∗cos⁡ϕ

=>|A|2−|B|2=|A|4+|B|4−2|A|2|B|2cos2θ−−−−−−−−−−−−−−−−−−−−−−−√∗cosϕ=>|A|2−|B|2=|A|4+|B|4−2|A|2|B|2cos⁡2θ∗cos⁡ϕ

=>cosϕ=|A|2−|B|2|A|4+|B|4−2|A|2|B|2cos2θ−−−−−−−−−−−−−−−−−−−−−−−√=>cos⁡ϕ=|A|2−|B|2|A|4+|B|4−2|A|2|B|2cos⁡2θ

This is a huge, nasty expression

Answer 2

Answer:

Explanation:

By cross product we get a vector as a resultant on the other hand we get a scalar quantity from dot product and there is no possible angle between any vector and scalar quantity.

Angle can be found only if both are vectors.

For better understanding:

Keep a pen on the table and assume any number say 3, now try to find out the angle between pen and 3, you can't.

Now keep two pens on the table (not parallel) and try to find the angle between them, this time you can.

Hope this helps!!