Find the equation of the plane which is at a distance of 3?3 units from the origin and the normal to which is equally inclined to the coordinate axes. The answer given in the book is: x+y+z = 9

Asked by Manoj | 22nd May, 2013, 07:19: PM

Expert Answer:

The direction cosines of the normal are in the ratio 1:1:1 as the normal is equally inclined to the coordinate axes. Hence the direction cosines of normal will be 1/root3 , 1/root3 , 1/root3

equ. of plane is given as : lx+my+nz =p where l,m,n r direction cosines & p is the distance frm origin.

so , equ. is 1/roo3 x + 1/root3 y + 1/root3 z=3root 3

i.e. x+y+z=9

Answered by  | 23rd May, 2013, 02:42: AM

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