distance of a point from a line in general This is exercise from M.Lawson's Book "Algebra & Geometry" about vectors.
The distance of a point from a line is defined to be the length of the perpendicular from the point to the line. Let the line in question have parametric equation $\pmb r=\pmb p+\lambda\pmb d$ and let the position vector of the point be $\pmb q$. Show that the distance of the point from the line is
$$\frac{||\pmb d \times(\pmb q-\pmb p)||}{||\pmb d||}.$$
Here $\lambda \in \Bbb{R}$.
Now I do not grasp at all how to deduce from basic vector arithmetic the distance of the point. I know that cross product of two vectors gives either area enclosed by the vectors ($a\times b = ||a||\;||b||\sin\theta$) or a vector which is perpendicular to the two vectors and length of this vector is the area given by the two vectors.
So, does $\frac{||\pmb d \times(\pmb q-\pmb p)||}{||\pmb d||}$ give length of $\pmb q-\pmb p$?
Below is picture of the vectors.

 A: In the picture below, $\require{cancel}\mathbf{q'}$ is the point of $\mathbf r$ which is closest to $\mathbf q$. Then the distance from $\mathbf q$ to $\mathbf r$ is $\lVert\mathbf q-\mathbf{q'}\rVert$. But it follows from the definition of sine that this distance is equal to $\sin(\theta)\lVert\mathbf q-\mathbf p\rVert$. On the other hand,$$\frac{\lVert\mathbf d\times(\mathbf q-\mathbf p)\rVert}{\lVert\mathbf d\rVert}=\frac{\sin(\theta){\cancel{\lVert\mathbf d\rVert}}\lVert\mathbf q-\mathbf p\rVert}{\cancel{\lVert\mathbf d\rVert}}=\sin(\theta)\lVert\mathbf q-\mathbf p\rVert.$$

A: Let the distance from the point to the line be denoted by $h$. Compute the area $A$ of the paralellogram with sides $\mathbf d$ and $\mathbf q-\mathbf p$ 
 
in two different ways: On the one hand, the norm of the cross product of two vectors is equal to the area of the paralellogram with those vectors as its sides, so we have $A=\lVert\mathbf d\times(\mathbf q-\mathbf p)\rVert$. On the other hand, the area is equal to the length of one of its sides times the altitude to one of the other vertices. If we take $\mathbf d$ as the side, the latter is exactly the distance that we’re looking for, i.e., $A=h\lVert d\rVert$. Equating and solving for $h$ gives your formula.
