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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

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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Equation of plane Z=2, i.e.  $0x+0y+z=2$

The direction ratio of normal is 0,0,1

$\therefore \, \, \, \sqrt{0^2+0^2+1^2}=1$

Divide equation $0x+0y+z=2$ by 1 from both side

We get,      $0x+0y+z=2$

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

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