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The length of the normal from origin to the plane $\mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=9$ is equal to
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3 units
Given plane, $x+2 y-2 z=9$. $\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}$
length of the normal $=\frac{9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}$
$=\frac{9}{\sqrt{9}}=\frac{9}{3}=3$ units
length of the normal $=\frac{9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}$
$=\frac{9}{\sqrt{9}}=\frac{9}{3}=3$ units
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