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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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O is the origin and A is (a,b,c).Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.

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  • Vector equation of a plane where $\overrightarrow n$ is the normal to the plane and passing through $\overrightarrow a$ is $(\overrightarrow r-\overrightarrow a).\overrightarrow n=0$
  • If $(a,b,c)$ are direction ration of given vector, then its direction cosines are $\bigg(\large\frac{a}{\sqrt {a^2+b^2+c^2}},\frac{b}{\sqrt {a^2+b^2+c^2}},\frac{c}{\sqrt {a^2+b^2+c^2}}\bigg)$
The given points are $A(a,b,c)$ and $O(0,0,0)$
$ \overrightarrow{OA}=(a\hat i+b \hat j+c\hat k)$
Where $a,b,c$ are the direction ratios
The direction cosines of $OA$ are
$\bigg(\large\frac{a}{\sqrt {a^2+b^2+c^2}},\frac{b}{\sqrt {a^2+b^2+c^2}},\frac{c}{\sqrt {a^2+b^2+c^2}}\bigg)$
Now it is given that the plane id perpendicular to $OA$
Hence $\overrightarrow n=\overrightarrow {OA}=a \hat i+b \hat j+c \hat k$
Therefore equation of the plane is $(\overrightarrow r-\overrightarrow a).\overrightarrow n=0$
Substituting the respective vectors we get,
Equation of the plane is $ [ \overrightarrow r- (a\hat i + b\hat j + c\hat k)].[ a\hat i + b\hat j + c\hat k]=0$
answered Jun 11, 2013 by meena.p

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