1. ## distance question

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.

2. Originally Posted by Jenny20
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.
Hint, a parallel plane has equation,
$2x-y+2z+K=0$
All you need to do if find $K$.

3. Do you know the formula for the distance between a point and a plane?.

The plane has equation $ax+by+cz+d=0$

The point: $(x_{0}, y_{0}, z_{0})$

$D=\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2 }+c^{2}}}$

You have:

$\frac{|2(3)+(-1)(2)+2(-1)+4|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}=2$

Now, use the formula again and solve for k as PH said:

$\frac{|2(3)+(-1)(2)+2(-1)+k|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}=2$

4. Hi galactus,

solving what you gave : the result of k will be 4 again!

So how can we get the equation of Psub2?

5. No, one of the answers is 4. There's another.

6. Hi galactus,

Could you please show me how? Thank you very much.

7. 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 $\frac{|6-2-2+k|}{3}=2$

4 is one of the answers. What's the other?.