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Home  >>  TN XII Math  >>  Vector Algebra
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Find the vector and cartesian equation of the line joining the points $(1 , -2 , 1 )$ and $(0 , -2 , 3 ).$

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  • Equation of a straight line passing through two given points $ \overrightarrow r = (1-t)\overrightarrow a+t\overrightarrow b$ ( vector equation ) where $\overrightarrow a \: and \: \overrightarrow b$ are the pv.s of the points on the line and t is a scalar.$ \large\frac{x-x_1}{x_2-x_1} = \large\frac{y-y1}{y2-y_1} = \large\frac{z-z_1}{z_2-z_1}$, (cartesian form ) where $ (x,y,z), (x_2, y_2, z_2)$ are the points on the line.
Vector equation : The line passing through the points $A(1, -2, 1) \: and \: B(0, -2, 3). \: \therefore \overrightarrow a = \overrightarrow i-2\overrightarrow j+\overrightarrow k\: and \: \overrightarrow b = -2\overrightarrow j+3\overrightarrow k$
The vector equation is $ \overrightarrow r = (1-t)\overrightarrow a+t\overrightarrow b$
$ \overrightarrow r = (1-t)(\overrightarrow i-2\overrightarrow j+\overrightarrow k)+t(-2\overrightarrow j+3\overrightarrow k), t$ is a scalar.
Cartesian equation : The line passing through $ A (1, -2, 1)$ and is parallel to $AB$ where $B(0, -2, 3) \therefore $ d.r.s of the line are $ (0 -1), (-2+2), (3-1)=-1, 0, 2$
The cartesian equation is $ \large\frac{x-1}{-1} = \large\frac{y+2}{0} = \large\frac{z-1}{2}$
answered Jun 12, 2013 by thanvigandhi_1
 

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