Question

vectorC=vectorA×vectorB and vectorD=vectorB×vectorA.then angle between vevtorC and vectorD is what?​

Answer 1

Answer:

Explanation:

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