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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equations of the line passing through the point $(3,0,1)$ and parallel to the planes $x+2y=0$ and $3y-z=0.$

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  • Vector equation of a line passing through a point and parallel to a vector is $\overrightarrow r=\overrightarrow a+ \lambda \overrightarrow b$ where $\lambda \in R$
Step 1:
It is given that the line passes through the point $(3,0,1)$ and parallel to the planes $x+2y=0$ and $ 3y-z=0$
Let $\overrightarrow a=3 \hat i+\hat k$
Let $\overrightarrow n$ be the normal vector to the required plane.
Then $\overrightarrow n$ is perpendicular to the normal to the plane.$ x+2y=0$ and $ 3y-z=0$
(ie)It is $\perp$ to the vector $\overrightarrow n_1 =\hat i+2 \hat j$ and $\overrightarrow n_2= 3\hat j-\hat k$
Therefore $\overrightarrow n=\overrightarrow n_1 \times \overrightarrow n_2=\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 0 \\ 0 & 3 & -1 \end{vmatrix}$
On expanding we get,
$\overrightarrow n=\hat i(-2)-\hat j(-1)+\hat k(3)$
$\quad=-2 \hat i+\hat j+ 3\hat k$
Hence its direction ratios are $ (-2,1,3)$
Step 2:
Therefore equation of the line is
$\overrightarrow r=(3 \hat i+\hat k)+\lambda(-2\hat i+\hat j+3 \hat k)$
But we know $\overrightarrow r= x \hat i+y \hat j+z \hat k$
Substituting for $\overrightarrow r$ we get,
$(x-3) \hat i+y \hat j+(z-1) \hat k=-\lambda(-2 \hat i+\hat j+ 3 \hat k)$
This is the required equation of the line
answered Jun 11, 2013 by meena.p
 

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