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Home  >>  TN XII Math  >>  Vector Algebra
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Show that $(2 , -1 ,3 ),(1 ,-1, 0 )$ and $(3, -1, 6 )$ are collinear.

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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.
Step 1
The equation of the line through $ A (2, -1, -3) \: and \: B(1, -1, 0)$ is
$ \large\frac{x-2}{1-2}=\large\frac{y+1}{-1+1}=\large\frac{z-3}{0-3}$
$ \large\frac{x-2}{-1} = \large\frac{y+1}{0} = \large\frac{z-3}{-3}$ (i)
 
 
 
Step 2
Subset the coordinates of $C(3, -1, 6)$ in (i)
$ \large\frac{3-2}{-1}=\large\frac{-1+1}{0}=\large\frac{6-3}{-3}$
$ -1\: = \: \large\frac{-1+1}{0}\: =\: -1$
therefore the points are collinear.
Note : $ \large\frac{y+1}{0} \Rightarrow$ that $ y=-1$ for all points on the line (i). This is true for the $ y$ coordinate of $C$ as well.

 

answered Jun 12, 2013 by thanvigandhi_1
edited Jun 24, 2013 by thanvigandhi_1
 

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