vector from a point P to a point Q in a plane in the direction of the normal vector

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.

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)?

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 $$Q-P$$.