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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

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  • If two lines are $\perp$ then the sum of the product of their direction cosines is $0$
  • (i.e) $a_1a_2+b_1b_2+c_1c_2=0$
Step 1:
Let $OA$ be the line joining the origin (0,0,0) and the point $A(2,1,1)$
Let $BC$ be the line joining the points $B(3,5,-1)$ and $(4,3,-1)$
The direction ratios of $OA$ are (2,1,1) and $BC$ are [(4-3),(3-5),(-1+1)]
Step 2:
It is given $OA$ is perpendicular to $BC$
Hence $a_1a_2+b_1b_2+c_1c_2=0$
Substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ we get
$2\times 1+1\times -2+1\times 0$
$\Rightarrow 2-2+0$
$\Rightarrow 0$
Hence $OA$ is $\perp$ $BC.$
answered Jun 3, 2013 by sreemathi.v

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