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.
Best Answer
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.