Finding the Equation of a Parallel Plane: How Do I Use Vectors to Solve This?

[ineedhelpnow]

find the equation of the plane through the point (1,-1,-1) and parallel to the plane 5x-y-z=6.

$v= \left\langle -5,1,1 \right\rangle$ can someone explain this step? why do the signs need to be changed. how is this vector parallel to the given line?

$-5(x-1)+1(y+1)+1(z+1)=0$

$-5x+5+y+1+z+1=0$

$-5x+y+z=-7$

 

[Prove It]

find the equation of the plane through the point (1,-1,-1) and parallel to the plane 5x-y-z=6.

$v= \left\langle -5,1,1 \right\rangle$ can someone explain this step? why do the signs need to be changed. how is this vector parallel to the given line?

This vector is NOT parallel to the given line. What you were given was not even a line, it was a plane.

A plane has the same coefficients as its normal vector. So a vector normal to the plane is (5, -1, -1). But any scalar multiple of this is also a vector normal to the plane, including (-5, 1, 1).

But this isn't even the approach I would use. The plane you are looking for is essentially the same as the plane you are given, just translated by a certain amount.

That means the plane you want will be of the form 5x - y - z = d, where d is some constant.

You know the point (1, -1, -1) lies on the plane, so if you plug in the point (1, -1, -1) as x, y, z, you will be able to find d.

 

[ineedhelpnow]

But is the way I did it right?