Solving for tan(y+2z) given \vec{F} and \vec{r}_1, \vec{r}_2

[sandy.bridge]

Homework Statement

Let
$$\vec{F}=(1/(x+2), 1, y)$$.
The field line that passes through (2, 3, 1) also passes through the point (3, y, z).
What is tan(y+2z)?
Let $$\vec{r}_1=(2, 3, 1), \vec{r}_2=(3, y, z)$$
Since the field line passes through both points, we have
$$\vec{F}(2, 3, 1)//(\vec{r}_2-\vec{r}_1)$$
hence when cross multiplied, it should equal zero!
However, I am not getting the right answer using this approach. Can someone pinpoint where I perhaps have made an error?
Thanks!

 

[lippyka]

The error is that you are not using the vector equation correctly. The vector equation $\vec{F}(\vec{r}_2-\vec{r}_1)=0$ means that the vector $\vec{F}$ evaluated at $\vec{r}_2$ must be equal to the negative of the vector $\vec{F}$ evaluated at $\vec{r}_1$. In other words, we have $$\vec{F}(3,y,z)=(1/(3+2), 1, y)=(-1/(2+2), -1, -3)=-\vec{F}(2,3,1)$$ from which you can solve for $y$ and $z$. Once you have those values, you can use them to calculate $\tan(y+2z)$.