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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 parallel to the planes $\hat{r}. (\hat{i} - \hat{j} + 2\hat{k}) = 5$ and $\hat{r} (3\hat{i} + \hat{j} + \hat{k} ) = 6$

$\begin{array}{1 1}(A) \overrightarrow r = \hat i + 2\hat j + 3\hat k + \lambda (3\hat i - 5\hat j - 4\hat k ) \\(B) \overrightarrow r = \hat i + 2\hat j + 3\hat k + \lambda (-3\hat i - 5\hat j - 4\hat k ) \\ (C) \overrightarrow r = \hat i + 2\hat j + 3\hat k + \lambda (-3\hat i + 5\hat j - 4\hat k ) \\ (D) \overrightarrow r = \hat i + 2\hat j + 3\hat k + \lambda (3\hat i + 5\hat j + 4\hat k )\end{array} $

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1 Answer

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Toolbox:
  • 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 planes are $\perp$ then we can written as $a_1a_2+b_1b_2+c_1c_2=0$
  • Where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are the direction cosines.
Step 1:
Let the direction of the line be $\overrightarrow b=b_1\hat i+b_2\hat j+b_3\hat k$
Equation of the line passing through $(1,2,1)$ having the direction $\overrightarrow b$ is
$\overrightarrow r=\hat i+2\hat j+3\hat k+\lambda \overrightarrow b$
$\quad\;=\hat i+2\hat j+3\hat k+\lambda( b_1\hat i+b_2\hat j+b_3\hat k ) $------(1)
Step 2:
Now the line $l_1$ and the plane $\overrightarrow r.(\hat i-\hat j+2\hat k)=5$ are parallel.
Therefore normal of the plane is perpendicular to this line $l_1$
$a_1a_2+b_1b_2+c_1c_2=0$
$b_1\times 1+b_2\times(-1)+b_3\times 2=0$
$b_1-b_2+2b_3=0$-----(2)
Step 3:
Also line $l_1$ and plane $\overrightarrow r.(3\hat i+\hat j+\hat k)=6$ are parallel.
Therefore normal of the plane is perpendicular to this line $l_1$
Therefore $(b_1\hat i+b_2\hat j+b_3\hat k).(3\hat i+\hat j+\hat k)=0$
(i.e)$3b_1+b_2+b_3=0$-----(3)
Step 4:
Now solving equ(2) and equ(3) we get
$\large\frac{b_1}{-1-2}=\frac{b_2}{6-1}=\frac{b_3}{1+3}$
(i.e)$\large\frac{b_1}{-3}=\frac{b_2}{5}=\frac{b_3}{4}$
(i.e)$\large\frac{b_1}{3}=\frac{b_2}{-5}=\frac{b_3}{-4}$
Hence $(3,-5,-4)$ are the direction ratio of the line $b$.
Therefore equation of the line (1) can be written as
$\overrightarrow r== \hat i + 2\hat j + 3\hat k + \lambda (3\hat i - 5\hat j - 4\hat k )$
answered Jun 5, 2013 by sreemathi.v
 

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