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In the following cases,find the distance of each of the given points from the corresponding given plane (d)Point(-6,0,0),Plane $2x-3y+6z-2=0$

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1 Answer

  • The distance between a point and a plane is given by
  • $D=\large\frac{\mid ax_1+by_1+cz_1+d\mid}{\sqrt{a^2+b^2+c^2}}$
Step 1:
Given point is $(-6,0,0)$
The equation of the given plane is $2x-3y+6z-2=0$
The direction cosines are $(2,-3,6)$
Step 2:
The distance between the point and the given plane is $D=\large\frac{\mid ax_1+by_1+cz_1+d\mid}{\sqrt{a^2+b^2+c^2}}$
$\Rightarrow \large\frac{\mid -6\times 2+0\times -3+0\times 6-2\mid}{\sqrt{2^2+(-3)^2+(6)^2}}$
$\Rightarrow \large\frac{-12-2}{\sqrt{49}}=\large\frac{14}{7}$$=2$
Hence $D=2$
answered Jun 3, 2013 by sreemathi.v

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