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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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The cartesian equation of a line is $\frac{\large x-5}{\large 3} = \frac{\large y+4}{\large 7} = \frac{\large z-6}{\large 2}$. Write its vector form.

$\begin{array}{1 1} \overrightarrow r = 5\hat i - 4\hat j + 6\hat k + t(3\hat i + 7\hat j + 2\hat k) \\ \overrightarrow r = 3\hat i +7\hat j + 2\hat k + t (5\hat i -4\hat j + 6\hat k) \\ \overrightarrow r = 5\hat i - 4\hat j + 6\hat k - t (5\hat i -4\hat j + 6\hat k) \\\overrightarrow r = 3\hat i+7\hat j + 2\hat k + t (3\hat i + 7\hat j + 2\hat k) \end{array} $

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  • Vector form of a line passing through a point and parallel to a given vector is $ \overrightarrow r = \overrightarrow a+t \overrightarrow v$ where $t$ is any real number.
Given line is $\large\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
Hence the point $(x,y,z)\;is \;(5,-4,6)$ and the direction ration of the parallel vector is $(3,7,2)$
Hence $\overrightarrow a = 5\hat i - 4\hat j + 6\hat k$ and the parallel vector is $ \overrightarrow v = 3\hat i + 7\hat j + 2\hat k$
Hence the required equatio is
$ \overrightarrow r = (5\hat i - 4\hat j + 6\hat k )+ t (3\hat i + 7\hat j + 2\hat k) $
where $t$ is any real number.
answered Jun 4, 2013 by meena.p
 

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