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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Write the vector equation of the line given by $\large \frac{x-5}{3} = \frac{y+4}{7}=\frac{z-6}{2} $

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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 equation 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 Nov 12, 2013 by sreemathi.v
 

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