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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane through the intersection of the planes $ \overrightarrow{r}.\quad(\hat i+3 \hat j)-6=0\; and\;\overrightarrow{r}.(3\hat i-\hat j-4 \hat k)=0,$ whose perpendicular distance from origin is unity.

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  • Perpendicular distance of a plane from the orgin is $P= \bigg| \large\frac{d}{\sqrt{a^2+b^2+c^2}} \bigg|$
  • Equation of the plane passinfg through the line of intersection two plane is $(a_1x+b_1y+c_1z+d_1)+\lambda(a_2x+b_2y+c_2z+d_2)=0$
Let the equation of the plane passing through the line of intersection of the planes $x+3y-6=0$ and $3x-y-42=0$ be
$ x+3y-6+ \lambda (3x-y-4z)=0$
$ \Rightarrow x(1+3\lambda)+y(3-\lambda)+z(0-4\lambda)+(-6)=0$
It is given that the perpendicular distance of the above plane from origin is unity
(ie) $ \bigg| \large\frac{d}{\sqrt{a^2+b^2+c^2}} \bigg|=1$
Where $a=1+3 \lambda,\;b=3 -\lambda,\;c=-4 \lambda,\;d=-6$
$ \Rightarrow \bigg| \large\frac{-6}{\sqrt{(1+3\lambda)^2+(3-\lambda)^2+(-4\lambda)^2}} \bigg| = 1$
On simplifying we get,
$ \Rightarrow \bigg| \large\frac{-6}{\sqrt{1+6\lambda+9+\lambda^2-6\lambda+16 \lambda^2}} \bigg| $
$ \Rightarrow \bigg| \large\frac{-6}{\sqrt{26\lambda^2+10}} \bigg| = 1$
$(ie)\;\sqrt {26 \lambda^2+10}=6$
Squaring on both sides we get,
$26 \lambda^2+10=36$
$\quad\quad 26 \lambda=26$
$\qquad\quad \lambda^2=1$
$\qquad\quad \lambda=\pm 1$
Substituting the values of $\lambda$ in equ (1) we get when $\lambda=+1$
$4x+2y-4z-6=0$
or $2x+y-4z-3=0$
When $\lambda=-1$
$-2x+4y+4z-6=0$
or $x-2y-2z+3=0$
answered Jun 12, 2013 by meena.p
 

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