Question

The equation ot the line passing through the point $(1,-2,3)$ and parallel to the line$x-y+2z=5$ and $3x+y+z=6$ is

• A. $\frac{x-1}{-3}=\frac{y+2}{5}=\frac{z-3}{4}$
• B. $\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{2}$
• C. $\frac{x-1}{3}=\frac{y+2}{1}=\frac{z-3}{6}$
• D. $\frac{x-1}{3}=\frac{y+2}{-1}=\frac{z-3}{2}$

Answer 1

The correct option is B $\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{2}$

$\begin{array}{c}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}=\hat{i}-2\hat{j}+3\hat{k}+\lambda \left(\hat{i}-j+2\hat{k}\right)\\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left(1+\lambda \right)\hat{i}+\left(2-\lambda \right)\hat{j}+\left(3+2\lambda \right)\hat{k}\\ \Rightarrow x=1+\lambda \\ \Rightarrow y=2-\lambda \\ \Rightarrow z=3+2\lambda \\ Then,cartesionequationoftheline\\ \frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{2}\end{array}$