## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k$ and $(1,-1,-1)$
Let us consider $P_1$ and $P_2$ as vectors and, a vector $u=\overrightarrow{P_1P_2}=P_2-P_1=2i-2j-2k$ and $|u|=\sqrt{(2)^2+(-2)^2+(-2)^2}= 2\sqrt {3}$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Now, $\hat{\textbf{u}}=\dfrac{2i-2j-2k}{2 \sqrt 3}=\dfrac{1}{\sqrt 3} i-\dfrac{1}{\sqrt 3}j-\dfrac{1}{\sqrt 3}k$ The mid-point can be found as: $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})$ Thus, $(\dfrac{0+2}{2},\dfrac{0+(-2)}{2},\dfrac{0+(-2)}{2})=(1,-1,-1)$