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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry

The distance of the plane $\overrightarrow{r}.\bigg(\frac{2}{7}\hat i+\frac{3}{7}\hat j-\frac{6}{7}\hat k\bigg)$=1 from the origin is

\[(A)\;1 \qquad (B)\;7 \qquad (C)\;\frac{1}{7} \qquad(D)\;None\;of\;these.\]

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1 Answer

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  • Distance of a plane from the orgin is distance = $ \bigg| \large\frac{d}{\sqrt{a^2+b^2+c^2}} \bigg|$
The given vector equation of the plane is $\overrightarrow r.\bigg(\large\frac{2}{7}$$\hat i+\large\frac{3}{7} \hat j-\large\frac{6}{7} $$\hat k \bigg)=1$
The cartesian equation this plane is $2x+3y-6z-7=0$
Now we know the distance of a plane is
$ \bigg| \large\frac{d}{\sqrt{a^2+b^2+c^2}} \bigg|$
Here $a=2,\;b=3,\;c=-6, \;and\; d=-7$
Substituting the respective values we get,
distance $= \bigg| \large\frac{-7}{\sqrt{2^2+3^2+(-6)^2}} \bigg|$
$\qquad\quad= \bigg| \large\frac{-7}{\sqrt{4+9+36}} \bigg|$
$\qquad\quad= \large\frac{-7}{7} $$=1$
Hence $A$ is the correct option
answered Jun 12, 2013 by meena.p
 

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