Write the vector equation of the plane 3x-y-7z=5. Find the equations of the planes through the intersection of the planes x+3y+6=0 and 3x-y-4z=0,whose perpendicular distance from the origin is equal to 1.

Asked by  | 6th Feb, 2012, 04:44: PM

Expert Answer:

Vector equation is :
r.(3i - j - 7k) = 5
Let the plane be
(x+ 3y + 6) + a(3x - y - 4z) = 0    ----------------- 1
This plane is at a distance of 1 from (0,0,0)
so 6/root((1+3a)2 + (3-a)2 + (4a)2)  =  1
On solving we get
 a = 1 , -1
Substitute in eq 1 to get the plane

Answered by  | 6th Feb, 2012, 09:04: PM

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