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In each of the following cases,determine the direction cosines of the normal to the plane and also the distance from origin(b) $x+y+z=1$

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$\begin{array}{1 1} d.c.=(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3})\; distance =1units \\d.c.=(1,1,1)\; distance =\frac{1}{\sqrt3}\;units \\d.c.=(\frac{1}{\sqrt3},\frac{1}{\sqrt3},-\frac{1}{\sqrt3}) distance\; =\frac{1}{\sqrt3}\;units \\ d.c.=(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3})\; distance \;=\frac{1}{\sqrt3}units \end{array} $

1 Answer

  • The equation of normal to a plane is $lx+my+nz=d$,where $l,m,n$ are the direction cosines and d is the distance of the perpendicular from the origin.
Step 1:
The given equation is $x+y+z=1$---(1)
The direction ratios of the normal are $(1,1,1)$.
Therefore $\sqrt{1^2+1^2+1^2}=\sqrt{3}$
Step 2:
Divide LHS and RHS of equ(1) by $\sqrt 3$,we get
$\large\frac{1}{\sqrt 3}$$x+\large\frac{1}{\sqrt 3}$$y+\large\frac{1}{\sqrt 3}$$z=\large\frac{1}{\sqrt 3}$
This equation of the form $lx+my+nz=d$,where $l,m$ and $n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.
Therefore the direction cosines are $\big(\large\frac{1}{\sqrt 3},\large\frac{1}{\sqrt 3},\large\frac{1}{\sqrt 3}\big)$ and distance from the origin is $\large\frac{1}{\sqrt 3}$units.
answered May 30, 2013 by sreemathi.v

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