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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $\hat{r} . (\hat{i} +2\hat{j} -5\hat{k}) +9 = 0$

$\begin{array}{1 1} (A) \overrightarrow l = ( \hat i + 2\hat j -5\hat k ) + \lambda ( \hat i + 2\hat j +3\hat k) \\ (B) \overrightarrow l = ( \hat i + 2\hat j + 3\hat k ) + \lambda ( \hat i + 2\hat j +3\hat k) \\ (C) \overrightarrow l = ( \hat i + 2\hat j - 5\hat k ) + \lambda ( \hat i + 2\hat j - 5\hat k) \\ (D) \overrightarrow l = ( \hat i + 2\hat j + 3\hat k ) + \lambda ( \hat i + 2\hat j - 5\hat k)\end{array} $

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1 Answer

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Toolbox:
  • Equation of a line passing through a point and $\perp$ to the given plane is given by
  • $\overrightarrow l=\overrightarrow r +\lambda\overrightarrow N,\lambda\in R$
Step 1:
The position vector $\overrightarrow r$ of the given point (1,2,3) is $\overrightarrow r=\hat i+2\hat j+3\hat k$
The direction ratios of the normal vector is $\overrightarrow N=\hat i+2\hat j-5\hat k$
Equation of a line passing through a point and $\perp$ to the given plane is given by $\overrightarrow l=\overrightarrow r+\lambda\overrightarrow N,\lambda\in R$
Step 2:
Substituting for $\overrightarrow r$ and $\overrightarrow N$
$\Rightarrow \overrightarrow l=(\hat i+2\hat j+3\hat k)+\lambda(\hat i+2\hat j-5\hat k)$
answered Jun 4, 2013 by sreemathi.v
 

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