Distance between the point $(1,2,3)$ and the plane $-x+2y+3z+1=0$ I'm trying to find the distance between the point $(1,2,3)$ and the plane $-x+2y+3z+1=0$.
I've used a formula to find this directly and got an answer I believe to be correct ($13/\sqrt{14}$).
However, when attempting to solve this manually through finding the point on the plane closest to the other point (where the plane is orthogonal to the point), and then computing the distance between points I cannot get the same answer (I instead got $4.677072$).
Would someone be able to help me work through the steps of how to find the closest point, and then the distance to help me verify my solutions? Thank you.
My working for the second case:  I took a normal vector of $n=(-1,2,3)$ and from there assumed that I could calculate the nearest point on the plane as $(1-1 \lambda)+(2+2\lambda)+(3+3\lambda)=6+4\lambda$.
From there I took
\begin{align}6+4\lambda&=-1\\
\lambda &= 5/4\\
x&=1-5/4&\hspace{-1em}&=-0.25\\
y&=2+2(5/4)&\hspace{-1em}&=4.5\\
z&=3+3(5/4)&\hspace{-1em}&=6.75
\end{align}
From there I calculated the distance between these points, getting an incorrect solution.
 A: Your normal vector is correct. But I don't get what you're doing here:
$(1-1 \lambda)+(2+2\lambda)+(3+3\lambda)=6+4\lambda$. (???)
Note that the coordinates of the intersection point of the normal line (that passes through $(1,2,3)$ with the plane) has to satisfy both equation of line and plane. All you have to do is substitute the general coordinates into the equation of the plane and determine your $\lambda$.
Equation of plane is $-x + 2y + 3z + 1 =0$
So you have $-(1-\lambda)+2(2+2\lambda)+3(3+3\lambda)+1 = 0$
Solving, you get $\lambda = -\frac {13}{14}$.
So the point of intersection of that particular normal line with the plane is $(\frac{27}{14},\frac 17, \frac 3{14})$. This is the point on the plane closest to $(1,2,3)$.
Find the distance between that point and $(1,2,3)$. You should get the right answer.
A: The plane is $r\cdot n = -1 $, where $r$ denotes a point on the plane, and a normal vector $n$ is $$n=(-1,2,3).$$
Generally, the $b$ in $r\cdot \frac{n}{|n|} = b$ tells you that the origin is $-b$ units away from the plane. In our specific case, note that the plane equation can be rewritten
$$ r\cdot n=-1\iff (r+\frac{n}{|n|^2})\cdot n=0$$
The (signed) distance $d$ of a point $p=(1,2,3)$ to the plane $r\cdot n=0$ is the projection of $p$ to the line spanned by $n$,
$d = p\cdot \frac{n}{|n|}=\frac{-1+4+9}{\sqrt{14}}=\frac{12}{\sqrt{14}}.$
The distance of $p$ to the plane $r\cdot n=-1$ is therefore $d$ plus the signed normal distance going from $r\cdot n = 0$ to $r\cdot n = -1$, which is $+1/|n|$. Thus the final answer is $$d+\frac1{|n|}=\frac{13}{\sqrt{14}}\approx 3.47.$$
The website Math3d.org lets you graph the objects in question to verify which of your two answers are right. I've drawn the plane, a unit normal $N=(-1,2,3)/\sqrt{14}$, and the normal line passing through $(1,2,3)$:

If you rotate it just the right way, you can verify the result is indeed somewhere between $3$ and $4$:

Here's what I made so you can explore it yourself: (link)
A: From the equation of the plane, we know the equation of a vector that is perpendicular to the plane - it is given by the coefficients to $x, y, z$. So in this case, $n = (-1, 2, 3)$. From this, we can find the equation of the line passing through $(1, 2, 3)$ that is perpendicular to the plane. From there, calculate where this line intersects the plane, and calculate the distance between that point and (1, 2, 3).
