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find the equation of a plane which is at a distance of 3 root 3 units from the origin and the normal to which is equally inclined to the coordinate axes.

find the equation of a plane which is at a distance of 3 root 3 units from the origin and the normal to which is equally inclined to the coordinate axes.

Grade:12

3 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
8 years ago
Hi,

You know the direction cosine of normal to the plane and u know the point lying on that plane ( say, (3,3,3) in first quadrant) .So, you would get four such planes depending on the quadrant.

thanks................................................................................................
Ankit Jaiswal
165 Points
6 years ago
You know the direction cosine of normal to the plane and u know the point lying on that plane ( say, (3,3) in first quadrant) .So, you would get four such planes depending on the quadrant.
kush srivastava
13 Points
6 years ago
given tht direction ratios are (1,1,1) and perpendicular distance as given 3√3 so,, d/√(1²+1²+1²) =3√3=> d=9 so the required equation is x+y+z=9

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