Angles between vectors

[RM5152]

Could someone please tell me whether my answers to this question look correct? Thanks

 

[Deleted member 4993]

Could someone please tell me whether my answers to this question look correct? Thanks View attachment 33172View attachment 33173

Correct.

However, why are you discarding \(\displaystyle a = -\frac{1}{\sqrt{3}}\) ?

 

[Steven G]

Here is how I would solve \(\displaystyle \dfrac{1}{4} = \dfrac{a^2}{1+a^2} \) Clearly the denominator on the rhs is 1 more than the numerator. I want the same to be true for 1/4. Since the denominator in 1/4 is 3 more than the numerator I divide both the numerator and denominator by 3. This gives me (1/3)/(4/3). Note that the denominator is 1 more than the numerator. Now I can equate the numerators or the denominators. This give me that a^2 = 1/3