Question

Find the direction cosines of the line $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$. Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line. [CBSE 2014]

Answer 1

The equation of the given line is $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$.
The given equation can be re-written as $\frac{x+2}{2}=\frac{y-{\displaystyle \frac{7}{2}}}{3}=\frac{z-5}{-6}$.

This line passes through the point $\left(-2,\frac{7}{2},5\right)$ and has direction ratios proportional to 2, 3, −6.

So, its direction cosines are

$\frac{2}{\sqrt{2^2+3^2+{\left(-6\right)}^2}},\frac{3}{\sqrt{2^2+3^2+{\left(-6\right)}^2}},\frac{-6}{\sqrt{2^2+3^2+{\left(-6\right)}^2}}$

Or $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$

The required line passes through the point having position vector $\overrightarrow{a}=-\hat{i}+2\hat{j}+3\hat{k}$ and is parallel to the vector $\overrightarrow{b}=2\hat{i}+3\hat{j}-6\hat{k}$.

So, its vector equation is

$\overrightarrow{r}=\left(-\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}-6\hat{k}\right)$