Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular to the plane 4x+5y+6z=14. Write the equation in the form indicated.

The plane we want to find has general equation

with not equal to 0, and has normal vector

is perpendicular to both the normal vector of the other plane, which is , as well as the tangent vector to the line , which is .

This means the dot product of with either vector is 0, giving us

Suppose we fix . Then the system reduces to

and we get

Then one equation for the plane could be

or in standard form,

The solution is unique up to non-zero scalar multiplication, which is to say that any equation would be a valid answer. For example, suppose we instead let ; then we would have found and , but clearly dividing both sides of the equation

by 2 gives the same equation as before.

b

step-by-step explanation:

      2

  2 is to go somewhat to the top right of a because it is squared    

v= a h/3