# Test into the book Halliday-Resnick on scalar product and cross product

1. 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.

1. 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$$\frac{\sin(38^\circ)}{\sin(18^\circ)}\approx1.992.$$
• Very kind Prof. with a lot of sincerity, I am very tired into my school, conferences online with Google Suite (reason Covid) I have my mind shakered. Why $\cos(100^\circ)=-\cos(80^\circ)$? Why in the 2nd part do you use the ratio of the two $\sin$? Is it possible to know if my consideration give the same your answer, please? +1 Oct 15, 2020 at 21:25
• For any angle $\theta$,$$\cos(90^\circ+\theta)=\cos(90^\circ)\cos(\theta)-\sin(90^\circ)\sin(\theta)=-\sin(\theta)$$and$$\cos(90^\circ-\theta)=\cos(90^\circ)\cos(\theta)+sin(90^\circ)\sin(\theta)=\sin(\theta);$$in particular, $\cos(90^\circ+\theta)=-\cos(90^\circ-\theta)$. And you were after an angle $\theta$ such that $\sin(20^\circ+\theta)=2\sin(\theta)$. However, this is the same thing as asserting that $\frac{\sin(20^\circ+\theta)}{\sin(\theta)}$. Oct 15, 2020 at 21:44