Question

the foot of the perpendicular drawn from the origin to a plane is (4,-2,-5). find the equation of the plane

Answer 1

Equation of plane is $\mathbf{4x-2y-5z=45}$

Step-by-step explanation:

1. Point origin (O)= (0,0,0)

2. Point on plane(A)=(a,b,c)=(4,-2,-5)

3. So equation of plane containing point A and direction cosine (l,m,n) is

   $\mathbf{l(x-a)+m(y-b)+n(z-c)=0}$      ...1)

   Where direction cosine

   l= (4-0)=4

   m= (-2-0)= -2

   n= (-5-0) = -5

4. On putting respective value in equation 1), we get

   $\mathbf{4(x-4)-2(y-(-2))-5(z-(-5))=0}$

   $\mathbf{4(x-4)-2(y+2)-5(z+5)=0}$

  On opening bracket

   $\mathbf{4x-2y-5z-16-4-25=0}$

   So equation of plane is

   $\mathbf{4x-2y-5z=45}$