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# In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin. $(a)\; z = 2$

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$\begin{array}{1 1} d.c.=(0,0,1) \; distance=2units. \\ d.c.=(1,1,1) \; distance=2units. \\ d.c.=(1,1,0) \; distance=2units. \\ d.c.=(1,0,1)\; distance=2units. \end{array}$

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• The equation of normal to the plane is $lx+my+nz=d$,where $l,m,n$ are the direction cosines of normal to the plane and $d$ is the perpendicular distance drawn from the origin.
Step 1:
Given equation of the plane is $z=2$
This can be written as $0x+0y+z=2$------(1)
Hence the direction ratios of the normal are $(0,0,1)$
Therefore $\sqrt{0^2+0^2+1^2}=1$
Step 2:
Dividing LHS and RHS of eq(1) by 1,we get
(i.e)$0x+0y+z=2$
This is of the form $lx+my+nz=d$,where $l,m,n$ are the direction cosines of the normal to the plane and $d$ is the distance of the perpendicular drawn from its origin.
Therefore the direction cosines are $0,0$ and $1$ and the distance from the origin is 2 units.
edited May 30, 2013

+1 vote