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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

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  • For two lines to be perpendicular , the product of their direction rations should be 0
  • $(ie)a_1a_2+b_1b_2+c_1c_2=0$
Let the given point be $A(1,-1,2) B(3,4,-2)$ and $ C(0,3,2)$ and $D(3,5,6)$
The direction ratios of the line joinig $A$ and $B$ is $(3,1),(4-(-1)),(-2-2)$
On simplifying we get,
$(2,5,-4)$
The direction ratios of the line joinig $C$ and $D$ is $(3-0),(5-3),(6-2)$
On simplifying we get,
$(3,2,4)$
For the lines $AB$ and $CD$ to be perpendicular
$a_1a_2+b_1b_2+c_1c_2=0$
On substituting for $a_1,b_1,c_1$ and $a_2,b_2,c_2$ we get,
$=(2 \times 3)+(5 \times 2)+(-4 \times 4)$
$=6+10-16$
$=0$
Hence $AB \perp CD$
answered Jun 4, 2013 by meena.p
 

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