Question

Find the vector and Cartesian equation of the line that passes through the points (3 , -2 , -5) and (3 , -2 , 6).

Answer 1

$\begin{array}{c}Let\\ A\equiv \left(3,-2,-5\right),B\equiv \left(3,-2,6\right)andO=\left(0,0,0\right)\\ \overrightarrow{OA}=\overrightarrow{a}=3\hat{i}-2\hat{j}-5\hat{k}\\ \overrightarrow{OB}=\overrightarrow{b}=3\hat{i}-2\hat{j}+6\hat{k}\\ Eq{u}^{n}ofthelinethroughA\left(\overrightarrow{a}\right)andB\left(\overrightarrow{b}\right)is\\ \overrightarrow{r}=\overrightarrow{a}+\lambda \left(\overrightarrow{b}-\overrightarrow{a}\right)\\ =3\hat{i}-2\hat{j}-5\hat{k}+\lambda \left(0\hat{i}-0\hat{j}+11\hat{k}\right)\\ \therefore \overrightarrow{r}=3\hat{i}-2\hat{j}-5\hat{k}+\lambda \left(11\hat{k}\right)\\ Eq{u}^{n}incartensianform\\ Eq{u}^{n}ofABis\frac{x-3}{3-3}=\frac{y+2}{-2+2}=\frac{z+5}{-5-6}\\ \Rightarrow \frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{-11}\end{array}$