$\text { Calculate the direction of } \overrightarrow { P _ { 1 } P _ { 2 } } \text { and the midpoint of line segment } P _ { 1 } P _ { 2 } \text {. }$
-$\mathrm { P } _ { 1 } ( - 6 , - 7,4 )$ and $\mathrm { P } _ { 2 } ( - 5 , - 8,3 )$
A) $\frac { 6 } { \sqrt { 3 } } \mathbf { i } + \frac { 7 } { \sqrt { 3 } } \mathrm { j } + \frac { 4 } { \sqrt { 3 } } \mathbf { k } ; \left( - 3 , - \frac { 7 } { 2 } , 2 \right)$
B) $\frac { 1 } { \sqrt { 3 } } \mathrm { i } - \frac { 1 } { \sqrt { 3 } } \mathrm { j } - \frac { 1 } { \sqrt { 3 } } \mathbf { k } ; \left( - \frac { 11 } { 2 } , - \frac { 15 } { 2 } , \frac { 7 } { 2 } \right)$
C) $\frac { 1 } { 3 } \mathbf { i } - \frac { 1 } { 3 } \mathbf { j } - \frac { 1 } { 3 } \mathbf { k } ; \left( - \frac { 11 } { 2 } , - \frac { 15 } { 2 } , \frac { 7 } { 2 } \right)$
D) $\frac { 1 } { \sqrt { 3 } } \mathbf { i } - \frac { 1 } { \sqrt { 3 } } \mathbf { j } - \frac { 1 } { \sqrt { 3 } } \mathbf { k } ; \left( - \frac { 5 } { 2 } , - 4 , \frac { 3 } { 2 } \right)$

Question

Quizplus · Accepted Answer

The direction vector $\overrightarrow{P_1P_2}$ is found by subtracting the coordinates of $P_1$ from those of $P_2$:
$$\overrightarrow{P_1P_2} = \begin{pmatrix}-5 \ -8 \ 3\end{pmatrix} - \begin{pmatrix}-6 \ -7 \ 4\end{pmatrix} = \begin{pmatrix}1 \ -1 \ -1\end{pmatrix}$$
To find the midpoint, we average the coordinates of $P_1$ and $P_2$:
$$	ext{Midpoint} = \left(\frac{-6-5}{2}, \frac{-7-8}{2}, \frac{4+3}{2}ight) = \left(-\frac{11}{2}, -\frac{15}{2}, \frac{7}{2}ight)$$
The direction vector is a unit vector in the direction of $\begin{pmatrix}1 \ -1 \ -1\end{pmatrix}$, which simplifies to $\frac{1}{\sqrt{3}}\begin{pmatrix}1 \ -1 \ -1\end{pmatrix}$, which matches choice (B).

- P1(−6,−7,4)\mathrm { P } _ { 1 } ( - 6 , - 7,4 )P1​(−6,−7,4)

- $\mathrm { P } _ { 1 } ( - 6 , - 7,4 )$