Distance between a point and a plan

Question

I've calculated the distance between the point $x=(1,1,1)$ and the plan

$$P=\text{Span}\left\{\left(\frac{1+\sqrt{5}}{2},0,1\right),\left(\frac{1-\sqrt{5}}{2},0,1\right)\right\}$$ I've found that $d\left(x,P\right)=1$. How can I check if my answer is correct or not ? To proceed, I've found the orthogonal projection on $P$.

Answer 1

I've found that $d\left(x,P\right)=1$.

So you already calculated it?

How can I check if my answer is correct or not ?

If you don't want to do calculations again, notice that $P$ is simply the $xz$-plane ($y=0$), so...

Answer 2

As a general method, to check we can

• find the normal vector $\vec n$ to $P$, by cross product
• find the intersection $Q$ of the line $(1,1,1)+t\cdot \vec n$ with $P$
• calculate the distance between $Q$ and $(1,1,1)$

in this case maybe you can find some good shortcut.

Answer 3

The plane that you mentioned is obviously $y=0$. Since $\bigl((1,0,0),(0,0,1)\bigr)$ is an orthonormal basis of that plane, the projection of $(1,1,1)$ is $(1,0,1)$ and therefore the distance is indeed $1$.