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In the following cases,find the distance of each of the given points from the corresponding given plane $(b)\;$Point (3,-2,1);Plane $(2x-y+2z+3=0)$

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  • Distance of a point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is
  • $D=\large\frac{\mid(ax_1+by_1+cz_1)+d\mid}{\sqrt{a^2+b^2+c^2}}$
Step 1:
Given point is $(3,-2,1)$
Equation of the given plane is $2x-y+2z+3=0$
Direction cosines are $(2,-1,2)$
Step 2:
Therefore distance of the point to the given plane is
$D=\large\frac{\mid 3\times 2+(-2\times -1)+1\times 2\mid+3}{\sqrt{2^2+(-1)^2+2^2}}$
$\Rightarrow \large\frac{\mid 6+2+2\mid+3}{\sqrt{4+1+4}}$
$\Rightarrow \large\frac{10+3}{\sqrt{9}}$
$\Rightarrow \large\frac{13}{3}$
Hence $D=\large\frac{13}{3}$
answered Jun 4, 2013 by sreemathi.v
edited Jun 4, 2013 by sreemathi.v

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