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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The direction ratio of the normal to the plane through $(1,0,0),(0,1,0)$, which makes $\large\frac{\pi}{4} $ with the plane $x+y=3$ is ?

$\begin{array}{1 1} (1, \sqrt 2 ,1 ) \\ (1,1, \sqrt 2 ) \\ (1,1,2) \\ (\sqrt 2 ,1, 1) \end{array} $

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  • Angle between two planes is given by $cos\theta=\large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|}$
Let the eqn. of the plane through $(1,0,0) \:\:and\:\:(0,1,0)$ be $ax+by+cz+d=0$
where $\overrightarrow n=(a,b,c)$ is unit normal vector to the plane.
$\Rightarrow\:\sqrt {a^2+b^2+c^2}=1$
subtracting these two we get $a-b=0$.......(i)
Also since this plane makes angle $\large\frac{\pi}{4}$ with $x+y=3$
$\Rightarrow\:\large\frac{1}{\sqrt 2}=\frac{(a,b,c).(1,1,0)}{1.\sqrt 2}$
Solving (i) and (ii) we get $a=b=\large\frac{1}{2}$
or $c=\pm\large\frac{1}{\sqrt 2}$
$\therefore $ $d.r.$ of normal is $(a,b,c)=(\large\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt 2})$
$=(1,1,\sqrt 2)$ (By multiplying by 2)
answered Dec 28, 2013 by rvidyagovindarajan_1

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