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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3\hat{i} + 2\hat{j} -2\hat{k}$. .

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

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  • Equation of the line passing through a point and parallel to a vector is given by $\overrightarrow r =\overrightarrow a+t \overrightarrow v$
Let $ \overrightarrow {OA}= \hat i+2 \hat j+3 \hat k$ and $ \overrightarrow {v}= 3\hat i+2 \hat j-2 \hat k$
We know $\overrightarrow r =\overrightarrow a+t \overrightarrow v$
Substituting for $\overrightarrow a$ and $ \overrightarrow v$
Hence the vector equation of the line is
$\overrightarrow r=(\hat i+2 \hat j+ 3\hat k)+t(3 \hat i+2 \hat j-2 \hat k)$
Where $t$ is any real number
answered Jun 4, 2013 by meena.p
edited Apr 12 by meena.p
 

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