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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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The cartesian equation of a line is $ \large\frac{x+3}{2} = \large\frac{y-5}{4}=\large\frac{z+6}{2} $ Find vector equation of line.

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  • The vector equation is $\overrightarrow r=\overrightarrow a+\lambda\overrightarrow b$
Given :
Cartesian equation of a line is $\large\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$
The vector equation is $\overrightarrow r=\overrightarrow a+\lambda\overrightarrow b$
Here $\overrightarrow a=-3\hat i+5\hat j-6\hat k$
$\overrightarrow b=2\hat i+4\hat j+2\hat k$
$\therefore \overrightarrow r=-3\hat i+5\hat j-6\hat k+\lambda(2\hat i+4\hat j+2\hat k)$
answered Nov 8, 2013 by sreemathi.v
 

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