$\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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This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

$\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.

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