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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the vector and the cartesian equations of the lines that passes through the origin and $(5, -2, 3).$

$\overrightarrow r = t(5\hat i - 2\hat j + 3\hat k)$
$ \overrightarrow r = t (5\hat i + 2\hat j + 3\hat k)$
$\overrightarrow r = t (-5\hat i + 2\hat j + 3\hat k)$
$\overrightarrow r =t (5\hat i + 2\hat j - 3\hat k)$
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1 Answer

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Toolbox:
  • Vector equation of a line passing through a given point and parallel to a given vector is $ \overrightarrow r = \overrightarrow a+t \overrightarrow b$
  • Cartesian equation is $ \large\frac{x-x_1}{a}=\large\frac{y-y_1}{b}=\large\frac{z-z_1}{c}$
Given that the line passes through the orgin .
Hence $(x_1,y_1,z_1)=(0,0,0)$
and the direction ration of the parallel vector is $(5,-2,3)=(a,b,c)$
Hence the vector equation is
$ \overrightarrow r = (0\hat i +0 \hat j+0 \hat k)+t(5\hat i - 2\hat j + 3\hat k)$
$ \overrightarrow r = t (5\hat i - 2\hat j + 3\hat k)$
Cartesian equation is
$ \large\frac{x-0}{5}=\large\frac{y-0}{-2}=\large\frac{z-0}{3}$
i.e.,$\large\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}$
answered Jun 4, 2013 by meena.p
 

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