Finding A Particular Equation

"Describe and find an equation for the surface generated by all points (x,y,z) that are four units from the plane 4x - 3y + z = 10."

Firstly, I found a point P on the plane: P = (0,0,10). Then I constructed a vector whose initial point is (0,0,10) and terminal point is Q = (x,y,z):

$\displaystyle \Large\vec{PQ} = \langle x, y,z-10 \rangle$

The vector normal to the plane is $\displaystyle \vec{n} = \langle 4,-3,1 \rangle$

$\displaystyle D = 4 = \frac{|\langle x,y,z-10 \rangle \cdot \langle 4,-3,1 \rangle|}{\sqrt{26}} \implies 4\sqrt{26} + 10 = 4x - 3y + z$

Does this appear to be the correct answer?

Re: Finding A Particular Equation

Quote:

Originally Posted by

**Bashyboy** "Describe and find an equation for the surface generated by all points (x,y,z) that are four units from the plane 4x - 3y + z = 10."

Firstly, I found a point P on the plane: P = (0,0,10). Then I constructed a vector whose initial point is (0,0,10) and terminal point is Q = (x,y,z):

$\displaystyle \Large\vec{PQ} = \langle x, y,z-10 \rangle$

The vector normal to the plane is $\displaystyle \vec{n} = \langle 4,-3,1 \rangle$

$\displaystyle D = 4 = \frac{|\langle x,y,z-10 \rangle \cdot \langle 4,-3,1 \rangle|}{\sqrt{26}} \implies 4\sqrt{26} + 10 = 4x - 3y + z$

Does this appear to be the correct answer?

No it does not.** Almost**, but not correct. You just cannot drop the absolute values.

If $\displaystyle A=|B|$ then $\displaystyle B=\pm A~.$

Re: Finding A Particular Equation

Note that there are **two** planes that are a given distance from a given distance.