- Two vectors lie with their tails in common in the same place. When the angle between them is increased by $20^\circ$, their scalar product has the same module but changes sign from positive to negative. The initial angle between the vectors was: A $0^\circ$. B $60^\circ$. C $70^\circ$. D $80^\circ$. E $90^\circ$.
My solution:
The scalar product is $$\mathbf{v}\bullet \mathbf{w}=vw\cos \theta$$ I will delete immediately the A) the B) because $\theta+20^\circ< 90^\circ$; also the C) and the E) are false because $$\theta+20^\circ= 90^\circ \implies \mathbf{v}\bullet \mathbf{w}=0$$ Hence the D) is true. In fact $$\alpha=\theta+20^\circ> 90^\circ$$ and $\cos \alpha<0$.
For another test on cross product I have a bit of difficulty.
- Two vectors lie with their tails in common in the same place. When the angle between them is increased by $20^\circ$, the module of their vector product doubles. The initial angle between the vectors was:
A $0^\circ$. B $18^\circ$. C $25^\circ$. D $45^\circ$. E $90^\circ$.
Possible solution (?)
$$|\mathbf{v}\times \mathbf{w}|=vw\sin \beta \implies 2vw\sin \beta=vw\sin(\beta+20^\circ)$$
Did I interpret the assignment correctly?
Your interpretations are correct indeed. The answer to the first question is $80^\circ$ because $\cos(100^\circ)=-\cos(80^\circ)$.
On the other hand, none of the options for the other question is exactly correct. The closest one is $18^\circ$, since$$\frac{\sin(38^\circ)}{\sin(18^\circ)}\approx1.992.$$