Triangle parameterisation I get how to answer the qs below, the problem is actually finding path $2$ $ \left( 2, 0, 0 \right) $ to $ \left( 0, 1, 0 \right) $ I get $(2-t)i+tj$ yet the answer for path 2 is... $$ (2-t)i+(t/2)j $$ Don't understand why, any help would be appreciated. * * * Some-more context... > Let $G$ be the vector field given by $$ G = 2{y i} + x^{2}{j} + z k $$ > Evaluate the line integral $$ I = \oint_{c} G \cdot dr $$ > where $C$ is given by the three sides of the triangle with verticies $ \left( 0, 0, 0 \right) $, $ \left( 2, 0, 0 \right) $ and $ \left( 0, 0, 0 \right) $, and the integration is preformed in the following direction: from $ \left( 0, 0, 0 \right) $ to $ \left( 2, 0, 0 \right) $ then to $ \left( 0, 1, 0 \right) $ and finally back to $ \left( 0, 0, 0 \right) $. You may evaluate the integral $I$... !enter image description here

A parametric equation for a line, whose 2 points $\overrightarrow a$ and $\overrightarrow b$ are given, is - $$\overrightarrow r = \overrightarrow a + (\overrightarrow b - \overrightarrow a)k $$

Putting in the 2 points given, we get (writing in coordinate form) - \begin{align} \overrightarrow r &= (2,0,0) + ((0,1,0)-(2,0,0))k \\\ &= (2,0,0) + (-2,1,0)k \end{align} Rescaling $ k = \frac{t}{2}$ to match with your values, we get - $$\overrightarrow r = (2-t)\hat i + \frac{t}{2}\hat j$$