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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane through the points $(2,1,-1)$ and $(-1,3,4)$ and perpendicular to the plane $x-2y+4z=10.$

$\begin{array}{1 1} 18 x+17 y+4z-49=0 \\ 18 x-17 y+4z-49=0 \\ 18 x-17 y-4z-49=0 \\18 x+17 y+4z+49=0 \end{array} $

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  • Cortesian equation of a plane passing through $(x_1,y_1,z_1)$ having direction ratios proportional to $a,b,c$ for its normal is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
Step 1:
It is given that the plane passes through the points $(2,1,-1)$ and $(-1,3,4)$ and is perpendicular to the plane $x-2y+4z=10$
Let the equation of the plane through $(2,1,-1)$ be
$a(x-2)+b(y-1)+c(z+1)=0$-----(1)
If it passes through $(-1,3,4)$ then, substitute for x,y and z we get,
$a(-1-2)+b(3-1)+c(4+1)=0$
$=>-3a+2b+5c=0$----(2)
Step 2:
If this plane is perpendicular to the plane
$x-2y+4z=10$ then
$a-2b+4c=0$-----(3)
Step 3:
Now let us solve the equation (2) and (3)
$\large\frac{a}{\begin{vmatrix} 2 & 5 \\ -2 & 4 \end {vmatrix}}$$\quad=\large\frac{b}{\begin{vmatrix} 5 & -3 \\ 4 & 1 \end {vmatrix}}$$\quad=\large\frac{c}{\begin{vmatrix} -3 & 2 \\ 1 & -2 \end {vmatrix}}$
=>$ \large\frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}$
=>$ \large\frac{a}{18}=\frac{b}{17}=\frac{c}{4}$$=\lambda$(say)
(ie) $ a=18 \lambda, \;b=17 \lambda,\;c=4 \lambda$
Substituitng for $a,b,c$ in equ (1) we get,
$ 18 \lambda(x-2)+17 \lambda (y-1)+4 \lambda(z+1)=0$
$=>18 x+17 y+4z-49=0$
or $=>18 x+17 y+4z=49$
This is the required equation of the plane.
answered Jun 11, 2013 by meena.p
 

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