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The shortest distance of the point $(2 , 10 ,1 )$ from the plane $\overrightarrow{r}.(\overrightarrow{3i}-\overrightarrow{j}+\overrightarrow{4k})=2\sqrt{26}$ is

\[\begin{array}{1 1 }(1)2\sqrt{26}&(2)\sqrt{26}\\(3)2&(4)\frac{1}{\sqrt{26}}\end{array}\]

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1 Answer

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The length of the perpendicular from the point $(x_1,y_1,z_1)$ to the plane $ax+y+cz+d=0$ is
$\large\frac{ax_1+y_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}$
The equation of the plane is
$\overrightarrow{r}.(\overrightarrow{3i}-\overrightarrow{j}+\overrightarrow{4k})=2\sqrt{26}$
Put $\overrightarrow{r} = x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$
then the cartesian equation of the plane is
$(x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}).(3\overrightarrow{i}-\overrightarrow{j}+4\overrightarrow{k})=2 \sqrt {26}$
$3x-y+4z=2 \sqrt {26}$
Then the length of the perpendicular from the point (2,10,1) to the plane $3x-y+4z-2 \sqrt {26}=0$
$\qquad= \pm \large\frac{3 \times 2 -10 +4 \times 1 -2 \sqrt {26}}{9+1+16}$
$\qquad= \pm \large\frac{6-10+4-2 \sqrt {26}}{\sqrt {9+1+16}}$
$\qquad= \large\frac{2 \sqrt {26}}{\sqrt {26}}$
$\qquad=2$
Hence 3 is the correct answer.
answered May 13, 2014 by meena.p
 

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