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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{ i}- \hat{j} + 4\hat{k}$ and is in the direction$\hat{i} + 2\hat{j} - \hat{k}$.

$\begin{array}{1 1}(2 \hat i-\hat j+4 \hat k)+t(\hat i+2 \hat j -\hat k) \\ (2 \hat i-\hat j+4 \hat k)-t(\hat i+2 \hat j -\hat k) \\ (4 \hat i-\hat j+4 \hat k)+t(\hat i+2 \hat j -\hat k) \\ (2 \hat i-\hat j+4 \hat k)+t(\hat i+5 \hat j -6\hat k) \end{array} $
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  • Vector equation of a line passing through a point and parallel to a given vector is $ \overrightarrow r = \overrightarrow a+t \overrightarrow v$
  • Cartesian equation of the line is $ \large\frac{x-x_1}{a}=\large\frac{y-y_1}{b}=\large\frac{z-z_1}{c}$ where $ (a,b,c)$ are the direction ratio of the line.
Let $\overrightarrow a=2\hat i-\hat j+4\hat k$ and $\overrightarrow v=\hat i+2\hat j-\hat k$
We know the vector equation is $ \overrightarrow r = \overrightarrow a+t \overrightarrow v$
On substituting for $\overrightarrow a\;and\; \overrightarrow b$ we get,
$ \overrightarrow r = (2\hat i - \hat j + 4\hat k)+ t (\hat i + 2\hat j - \hat k)$ where t is any real number.
Cartesian equation of the line is $ \large\frac{x-x_1}{a}=\large\frac{y-y_1}{b}=\large\frac{z-z_1}{c}$
Here $(x_1,y_1,z_1)\;is\;(2,-1,4)$ and $(a,b,c) \;is\; (1,2,-1)$
Hence the Cartesian equation is $ \large\frac{x-2}{1}=\large\frac{y+1}{2}=\large\frac{z-4}{-1}$
answered Jun 4, 2013 by meena.p
 

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