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Find the Cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$
Find the Cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$
CBSE | Class 12 | Exercise 11.2 | Maths | NCERT | Three Dimensional Geometry
Find the Cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

Solution:

It is given that
The points $(-2,4,-5)$
It is known that
Now, the Cartesian equation of a line through a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and having direction ratios $\mathrm{a}, \mathrm{b}, \mathrm{c}$ is
$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
Here, the point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ is $(-2,4,-5)$ and the direction ratio is given by: $\mathrm{a}=3, \mathrm{~b}=5, \mathrm{c}=6$
$\therefore$
Now the Cartesian equation of the required line is:
$\frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6} \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$

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