Question

Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A+B to be 100 times larger than the magnitude of A-B , what must be the angle between them?

Do you use the Law of cosines ؟

Answer 1

\((A+B).(A+B) = |A+B|^2 \\= A.A + B.B + 2 A.B = |A|^2 + |B|^2 + 2|A||B|cos \alpha\)

\((A-B).(A-B) = |A-B|^2 \\= A.A + B.B - 2 A.B= |A|^2 + |B|^2 - 2|A||B|cos \alpha\)

where \(\alpha \) is the angle between \(A\, { and }\, B\)

from these:
\[\dfrac{|A+B|^2 }{|A-B|^2 } = \dfrac{|A|^2 + |B|^2 + 2|A||B|cos \alpha}{|A|^2 + |B|^2 - 2|A||B|cos \alpha}\]

add in these:
\(A.A = B.B = |A|^2 = |B|^2\)
\(|A+B| = 100 |A-B|, \quad \dfrac{|A+B|^2}{|A-B|^2} = 10000 \)

and you have the backbone of proper vector solution

but you can also draw it and use the law of cosines....

Answer 2

that gives\[10000 = \dfrac{|A|^2(2 + 2cos \alpha)}{|A|^2(2 - 2cos \alpha)}\]
so\[10000(1- \cos \alpha) = 1 + cos \alpha\]
\[\cos \alpha = \dfrac{9999}{10001}, \qquad \alpha \approx 1.15^o\cdots\]