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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the length of the perpendicular drawn from the origin to the plane 2X-3Y+6Z+21=0.

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
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  • The distance of a point $P(x_1,y_1,z_1)$ from a plane $Ax+BY+CZ+D=0$ is given by $\large \frac{Ax_1+By_1+Cz_1+D}{\sqrt{A^2+B^2+C^2}}$
Given a place $2X+3Y+6Z+21=0$.
The distance of a point $P(x_1,y_1,z_1)$ from a plane $Ax+BY+CZ+D=0$ is given by $\large \frac{Ax_1+By_1+Cz_1+D}{\sqrt{A^2+B^2+C^2}}$
We need to calculate the distance from the origin $P(0,0,0)$ to the place.
$\Rightarrow$ Distance $=\Large \frac{2(0)-3(0)+6(0)+21}{\sqrt{2^2+3^2+6^2}}$
$\Rightarrow$ Distance $=\Large \frac{21}{\sqrt{49}}=\frac{21}{7}\normalsize =3\;$ units.
answered Mar 21, 2013 by sreemathi.v
edited Mar 21, 2013 by balaji.thirumalai
 

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