6) a). Find the equation of the plane through the origin and perpendicular to x+y+z =...

a). Find the equation of the plane through the origin and perpendicular to x+y+z = 5 and 2x+y−2z = 7

b). Let A = (−1,3,0), B = (3,2,4) and C = (1,−1,5).
( I ) Find an equation for the plane that passes through these three points.

( II ) Find the area of the triangle determined by these three points.

Solutions

Expert Solution

6)a) Given:

x+y+z=5, 2x+y-2z=7

To find: Equation of plane perpendicular to these planes

Soln:

The normal vectors of the given planes are

The normal vector of the required plane is the vector product of these two normal vectors since the required plane is perpendicular to both these planes.

The normal vector of the required plane is given by

The equation of a plane having normal vector is given by where is a point in the plane.

It is given that the origin lies in the plane. So substituting the values of A, B, C we get

-3x+4y-z = 0.

milcah answered 1 year ago

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