Subject math GRE exam 0568 Q.47

Question

The question and its answer is given in the following picture:

I do not exactly from where the second equality in the following line came, could anyone explain it for me?

Answer 1

This solution doesn't look exactly right. First, let us define work, the work done by a force $\bf{F}$ along a path $C$ (a curve in the space), is equal to line integral

$$W = \int_C \mathbf{F} \cdot d\mathbf l.$$ Where, $d\mathbf{l}$ is a vector line element.

By definition, we have, as u can see here: wiki

$$\int_C \mathbf{F} \cdot d\mathbf l = \int_a^b \mathbf{F}(r(t)) \cdot \mathbf{r}'(t) dt,$$ where $\mathbf r$ is a parametrization of the curve $C$, such that $\mathbf r(a)$ and $\mathbf r(b)$ match the endpoints of C.

It's the right form of the second equality.

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By the way, the force is not equal to the vector $(-1,0,1)$, actually it is $\frac{1}{\sqrt{2}}(-1,0,1)$, since $F=|\mathbf F| =1.$