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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry

Find the vector equation of the line which is parallel to the vector $3\hat i+2\hat j+6\hat k$ and which passes through the point $(1,-2,3).$

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  • Vector equation of a line passing through a point and parallel to a given vector is $\overrightarrow {r}=\overrightarrow {a}+ \lambda \overrightarrow {b}$ where $\lambda \in R$
  • If two lines are parallel then their direction cosines are equal
Vector equation of a line passing through a point whose position vector is $\overrightarrow a$ and parallel to a given vector is
$\overrightarrow {r}=\overrightarrow {a}+ \lambda \overrightarrow {b}$
The given point is
$\overrightarrow a=\hat i-2 \hat j+3 \hat k$
The given parallel vector is
$\overrightarrow b=3\hat i+2 \hat j+6 \hat k$
Therefore equation of the line is
$\overrightarrow r=(\hat i-2 \hat j+3 \hat k)+\lambda (3 \hat i+2 \hat j+ 6 \hat k)$
answered Jun 10, 2013 by meena.p
 

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