# Distance from a point to a plane

1. Mar 2, 2009

### nns91

1. The problem statement, all variables and given/known data

Find the distance from (3,0,10) to the plane 2x+3y+z=2

2. Relevant equations

D= absv(ax+by+cd)/ sqrt(a^2+b^2+c^2)

3. The attempt at a solution

I found vector n= 2i + 3j + k

Thus, the distance is 16/sqrt(14). Can you guys check the result for me ? I am not sure if I am right.

2. Mar 2, 2009

### Staff: Mentor

Your vector n is the normal to the plane. Can you show us how you got the distance you show?

3. Mar 2, 2009

### boaz

That's how I have solved this question :
[; ax+by+cz+d=0 \rightarrow 2x+3y+z-2=0 \\
(x_0,y_0,z_0) \rightarrow (3,0,10) \\
D=\frac{a\cdot x_0+b\cdot y_0+c\cdot z_0-d}{\sqrt{a^2+b^2+c^2}}
= \frac{2\cdot3+3\cdot0+1\cdot10-2}{\sqrt{2^2+3^2+1^2}}=\frac{14}{\sqrt{14}}=\sqrt{14} ;]

http://img222.imageshack.us/my.php?image=13210615.gif"

Last edited by a moderator: Apr 24, 2017
4. Mar 2, 2009

### Staff: Mentor

Of the two answers shown: 16/sqrt(14) and sqrt(14), I agree with the latter.