
If you drop a perpendicular from a point to a line or plane, the point you reach on that line or plane is called the projection of the point onto the line or plane.
Suppose we have a point $P\text{'}$ , a line $L$ , and a plane $Q$ . Suppose $L$ is described by two points, ${P}_{1}$ and ${P}_{2}$ , on it, and $Q$ is described by a normal vector $\stackrel{\u27f6}{N}$ and a point ${P}_{3}$ on it.
In applying vector concepts to geometric situations, there is one basic fact that is fairly simple and extremely useful:
The projection of a vector $\stackrel{\u27f6}{A}$ on another vector $\stackrel{\u27f6}{B}$ is given by $\frac{(\stackrel{\u27f6}{A},\stackrel{\u27f6}{B})\stackrel{\u27f6}{B}}{(\stackrel{\u27f6}{B},\stackrel{\u27f6}{B})}$ .
Why is this so?
Because it is a vector in the direction of $\stackrel{\u27f6}{B}$ whose length is the length of $\stackrel{\u27f6}{A}$ multiplied by the cosine of the angle between $\stackrel{\u27f6}{A}$ and $\stackrel{\u27f6}{B}$ , which is exactly what this projection is.
How can we use this fact?
Suppose we want to find the projection of $P\text{'}$ onto $L$ . We can write $\stackrel{\u27f6}{OP\text{'}}$ as $\stackrel{\u27f6}{{P}_{1}P\text{'}}+\stackrel{\u27f6}{O{P}_{1}}$ , with $\stackrel{\u27f6}{{P}_{1}P\text{'}}$ a vector. If we project $\stackrel{\u27f6}{{P}_{1}P\text{'}}$ onto $\stackrel{\u27f6}{{P}_{1}{P}_{2}}$ , then ( $\stackrel{\u27f6}{O{P}_{1}}+$ this projection) will be on $L$ and is the point we want.
The answer is therefore
If instead we want the distance of $P\text{'}$ from $L$ we can subtract this point from $P\text{'}$ . The length of the resulting vector is our answer.
Suppose we want the distance of $P\text{'}$ to $Q$ . This will be the length of the projection of $\stackrel{\u27f6}{P\text{'}{P}_{3}}$ on $\stackrel{\u27f6}{N}$ .
The projection of $P\text{'}$ onto $Q$ can be obtained by adding the projection of $\stackrel{\u27f6}{P\text{'}{P}_{3}}$ on $\stackrel{\u27f6}{N}$ to $\stackrel{\u27f6}{OP\text{'}}$ .
The distance between two lines is just the length of the projection of a vector between a point on each on the cross product of vectors in the direction of each. This does not work if the two lines are parallel, since the cross product needed here will be the $\stackrel{\u27f6}{0}$ vector.
If the lines are parallel, you can form a vector from a point on one line to a point on the other, project it on a vector in the direction of the lines, and subtract this projection from it. The resulting vector will be normal to the lines and its length will be the desired distance.
Thus, if you know how to project one vector on another, and you keep your wits about you, you can answer all the geometric questions raised in Section 5.1.
Exercises:
5.6 Draw an appropriate picture and find a formula for calculating each
of the following:
Then calculate the answer with the following input:
$P\text{\'}=(1,2,3),{P}_{1}=(1,0,0),{P}_{2}=(1,1,1),{P}_{3}=(1,2,1),N=(2,1,4),{P}_{1}\text{\'}=(1,4,3),{P}_{2}\text{\'}=(2,3,4),{P}_{1}"=(6,0,3),{P}_{2}"=(6,2,6)$
.
5.7. The projection of a point $P\text{'}$ onto a line $L$ (containing ${P}_{1}$ and ${P}_{2}$ ).
5.8 The projection of a point $P\text{'}$ onto a plane $Q$ (having normal $\stackrel{\u27f6}{N}$ and containing ${P}_{3}$ ).
5.9 The distance between $P\text{'}$ and $Q$ .
5.10 The distance between $P\text{'}$ and $L$ .
5.11 The distance between two skew lines in 3 space, $L$ and $L\text{'}$ (containing ${P}_{1}\text{'}$ and ${P}_{2}\text{'}$ ).
5.12 The distance between two parallel lines in 3 space $L$ and $L"$ (containing ${P}_{1}"$ and ${P}_{2}"$ ).
