I'm a little confused with a task in my textbook:

Plane $A$ has equation $r*n=k$ (scalar product form) and Point $P$, outside $A$, has position vector $\vec p$.

a) Write down a vector equation of the line $l$ through $P$ which is perpendicular to $A$.

  • $l:\vec r = \vec p + \lambda \vec n$

b) Line $l$ intersects $A$ at $Q$. Show that $\vec {PQ}= (\frac{k-\vec p * \vec n}{\lvert \vec n \rvert ^2})\vec n$.

c) Hence show that the shortest distance from $P$ to $A$ is given by $\frac {\lvert \vec p * \vec n - k\rvert}{\lvert \vec n \rvert}$

I don't understand b) but I do understand a) and c), because part c) is very much the same as the Hesse normal form.

enter image description here

source: https://www.youtube.com/watch?v=v8Fy1C7yJq8

In the picture above:

  • plane $A$ is called $E$
  • point $Q$ is called $F$
  • Point $P$ is called $R$

The Hesse normal form makes use of the following properties:

  • $\cos(\alpha)=\frac d {\lvert \vec {PR} \rvert}$
  • $\cos(\alpha)=\frac {\vec n * \vec {PR}}{\lvert \vec n \rvert * \rvert \vec {PR} \lvert} = \frac {\vec {PR}}{\rvert \vec {PR} \lvert} * \vec n_0$

Hence: $$\frac d {\lvert \vec {PR} \rvert}=\frac {\vec {PR}}{\rvert \vec {PR} \lvert} * \vec n_0$$ $$d=\vec {PR}*\vec n_0$$ Or the Hesse normal form can be expressed in cartesian form which is the same as required to prove: $$d= \frac {\lvert n_1p_1+n_2p_2+n_3p_3-k \rvert}{\lvert \vec n\rvert} $$ $$d= \frac {\lvert \vec n * \vec p -k \rvert}{\lvert \vec n\rvert} $$

How can I show that the statement in b) is true? Can I use the same diagramm to prove it? If not, what other way is there? And how do you then get from b) to proving c)?


1 Answer 1


Let $\underline n=(n_1,n_2,n_3)$be the perpendicular vector to the plane $\pi\subset\mathbb R^3$, whose equation is $\pi\equiv\langle\underline n,\underline x- A\rangle= n_1(x-a_1)+n_2(y-a_2)+n_3(z-a_3)=0$.
Let $P$ be a point such that $P\notin \pi$, then the equation of the line for $P$ orthogonal to $\pi$ will be $$l\equiv\begin{cases}x=p_1+tn_1\\y=p_2+tn_2\\z=p_3+tn_3 \end{cases}.$$ The intersection of $l$ with $\pi$ is $$\{Q\}=l\cap\pi\equiv n_1(p_1+tn_1-a_1)+n_2(p_2+tn_2-a_2)+n_3(p_3+tn_3-a_3)=0\iff t(n_1^2+n_2^2+n_3^2)+p_1n_1-a_1n_1+p_2n_2-a_2n_2+p_3n_3-a_3n_3=0 \iff $$ $$t=\dfrac{a_1n_1+a_2n_2+a_3n_3-p_1n_1-p_2n_2-p_3n_3}{n_1^2+n_2^2+n_3^2}:=\alpha.$$ The formula is the same you wrote in your question. The difference is that the term $k$ in my case is given by $-(n_1a_1+n_2a_2+n_3a_3)$.
Substituting $t=\alpha$ in the equation of $l$ you find the coordinates of the intersection $Q$ and the vector $\vec{PQ}$ is simply $Q-P$.
Doing this for the general case seems to be tedious but in a normal problem it would be quite easy.


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