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In the following cases,find the distance of each of the given points from the corresponding given plane $(c)\;$Point (2,3,-5);Plane $(x+2y-2z-9=0)$

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1 Answer

  • 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 $(2,3,-5)$
The equation of the given plane is $x+2y-2z-9=0$
The direction cosines are $(1,2,-2)$
Step 2:
Therefore distance of the point to the given plane is
$D=\large\frac{\mid 2\times 1+(3\times 2)+(-5\times -2\mid-9}{\sqrt{1^2+2^2+(-2)^2}}$
On simplifying we get
$\Rightarrow \large\frac{\mid 2+6+10\mid-9}{\sqrt{1+4+4}}$
$\Rightarrow \large\frac{9}{\sqrt 9}$
$\Rightarrow \large\frac{9}{3}$
$\Rightarrow 3$
Hence $D=3$
answered Jun 4, 2013 by sreemathi.v

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