Question

Find the direction cosines of the line $\frac{x+2}{2}=\frac{2y-5}{3};z=-1$. Also, find the vector equation of the line.

Answer 1

Equation of planes given,

$P_1:\frac{x+2}{2}=\frac{2y-5}{3}\Rightarrow 3x+6=4y-10\Rightarrow 3x-4y+16=0P_2:z=-1$

The required line is line of intersection of $P_1$ & $P_2$

let $(l,m,n)$ be directions of required line,it is perpendicular to the normals of $P_1$ and $P_2$

$\therefore 3l-4m+0n=0\therefore 0l+0m+n=0$

From cramers rule,

$\frac{l}{-4}=\frac{-m}{3}=\frac{n}{0}$

$\therefore$ Directions are $(-4,-3,0)$

Direction cosines of line are $(\frac{4}{5},\frac{3}{5},0)$

One point on the line $\overrightarrow{a}$ is $0\hat{i}+4\hat{j}-1\hat{k}$

Equation of line,

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\overrightarrow{r}=4\hat{j}-\hat{k}+\lambda (\frac{4}{5}\hat{i}+\frac{3}{5}\hat{j})$