The point (3,2,-1) lies equidistant from two parallel planes Psub1 and Psub2.
Psub1 has equation 2x-y+2z+4=0.
Find an equation for Psub2.
Thank you very much.
Do you know the formula for the distance between a point and a plane?.
The plane has equation $\displaystyle ax+by+cz+d=0$
The point: $\displaystyle (x_{0}, y_{0}, z_{0})$
$\displaystyle D=\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2 }+c^{2}}}$
You have:
$\displaystyle \frac{|2(3)+(-1)(2)+2(-1)+4|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}=2$
Now, use the formula again and solve for k as PH said:
$\displaystyle \frac{|2(3)+(-1)(2)+2(-1)+k|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}=2$
Jenny20, I am sorry, but I gave you the entire equation. Solve for k.
I don't what to seem 'short', but if you can't solve that, you shouldn't be in calc.
What it amounts to is solving $\displaystyle \frac{|6-2-2+k|}{3}=2$
4 is one of the answers. What's the other?.