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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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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 $\frac{\large x+3}{\large 3} = \frac{\large y-4}{\large 5} = \frac{\large z+8}{\large 6}$.

$\begin{array}{1 1} \large\frac{x+2}{3}=\large\frac{y-4}{5}=\large\frac{z+5}{6} \\\large\frac{x+2}{-2}=\large\frac{y-4}{4}=\large\frac{z+5}{-5} \\ \large\frac{x-3}{-2}=\large\frac{y-5}{4}=\large\frac{z-6}{-5} \\\large\frac{x+3}{-2}=\large\frac{y-4}{4}=\large\frac{z+8}{-5} \end{array} $

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  • Cartesian equation of the line passing through a given point and parallel to the given vector is $ \large\frac{x-x_1}{a}=\large\frac{y-y_1}{b}=\large\frac{z-z_1}{c}$
Let $\overrightarrow a=(-2\hat i+4 \hat j-5\hat k)$
Hence $(-2,4,-5) \;is\; (x_1,y_1,z_1)$
It is given that the line is parallel to the line
$\large\frac{x+3}{3}=\large\frac{y-4}{5}=\large\frac{z+8}{6}$
Therefore the direction ratios $(a,b,c)$ are $(3,5,6)$
Hence the cartesian equation is $\large\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
Now substituting for $(x,y,z)\;and\; (a,b,c)$ get
$ \large\frac{x+2}{3}=\large\frac{y-4}{5}=\large\frac{z+5}{6} $
answered Jun 4, 2013 by meena.p
 

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