Distances and projection (HL)

The notation of projections will not appear in exams, but knowledge of scalar projections may still be assessed. Knowing why the formulas work makes them so much easier to remember.

The formulas should be used only after the intersections between the lines and/or planes have been characterized in the question, or through previous parts.

Contents

Projection

The vector projection of vector a\bm a onto vector b\bm b is

projba=acos⁡θ  b^=a⋅bb⋅bb\textbf{proj}_{\bm b}\bm a = \bm a \cos\theta \; \bm{\hat b} = \frac{\bm a \cdot \bm b}{\bm b \cdot \bm b} \bm b

For distances, we just need the magnitude, or the absolute value of the scalar projection, which is

D=∣acos⁡θ∣=∣a⋅b∣∣b∣D = \left\lvert\bm a \cos\theta\right\rvert = \frac{\left\lvert\bm a \cdot \bm b\right\rvert}{\lvert\bm b\rvert}

In most of the formulas below, we project a displacement vector of two points onto the direction of the distance.

Point and plane

Given some point (position vector) PP and vector equation of plane

r⋅n=a⋅n=ax+by+cz=d\bm r \cdot \bm n = \bm a \cdot \bm n = ax + by + cz = d

We want to project OP→−a\overrightarrow{OP} - \bm a onto n\bm n

D=∣(OP→−a)⋅n∣∣n∣=∣OP→⋅n−a⋅n∣∣n∣=∣OP→⋅n−d∣∣n∣D = \frac{\left\lvert(\overrightarrow{OP} - \bm a) \cdot \bm n\right\rvert}{\lvert\bm n\rvert} = \frac{\left\lvert\overrightarrow{OP} \cdot \bm n - \bm a \cdot \bm n\right\rvert}{\lvert\bm n\rvert} = \frac{\left\lvert\overrightarrow{OP}\cdot \bm n - d\right\rvert}{\lvert\bm n\rvert}

So we don’t actually need to find a point on the plane.

Note, distance between a line and a parallel plane is also this, where PP is a given point on the line.

Two parallel planes

Similar to above, but using both constants on the right side.

r⋅n=d1\bm r \cdot \bm n = d_1
r⋅n=d2\bm r \cdot \bm n = d_2
D=∣d2−d1∣∣n∣D = \frac{\left\lvert d_2 - d_1\right\rvert}{\lvert\bm n\rvert}

Two skew lines

Given two lines

r1=d1λ+a\bm {r_1} = \bm {d_1}\lambda + \bm a
r2=d2μ+b\bm {r_2} = \bm {d_2}\mu + \bm b

Even if the same parameter is given for both lines, you need to assume they are different if you want distance between two lines not two paths.

Distance is always perpendicular to the line. So we want to project a−b\bm a - \bm b onto the cross product d1×d2\bm{d_1} \times \bm{d_2}

D=∣(a−b)⋅(d1×d2)∣∣d1×d2∣D = \frac{\left\lvert(\bm a - \bm b) \cdot (\bm{d_1} \times \bm{d_2})\right\rvert}{\lvert \bm{d_1} \times \bm{d_2}\rvert}

Two parallel lines

Given two lines

r1=dλ+a\bm {r_1} = \bm {d}\lambda + \bm a
r2=dμ+b\bm {r_2} = \bm {d}\mu + \bm b

The cross product of scalar multiples of vectors, is 0⃗\vec 0. So we cannot use the skew lines formula. The two direction vectors we have is a−b\bm a - \bm b and d\bm d. Instead of finding cos⁡θ\cos \theta and adjacent side, we need to find the sin⁡θ\sin \theta and opposite side. So we use instead the cross product

D=∣(a−b)×d∣∣d∣D = \frac{\left\lvert(\bm a - \bm b) \times \bm d\right\rvert}{\lvert \bm d \rvert}

Two linear trajectories

Or nearest approach between two objects in linear, constant-velocity, motion.

This is different from distance of skew lines because the nearest approach may not necessarily be the shortest distance between the lines.

Given two trajectories

r1(t)=v1t+a\bm {r_1}(t) = \bm {v_1}t + \bm a
r2(t)=v2t+b\bm {r_2}(t) = \bm {v_2}t + \bm b

Where v1\bm {v_1} and v2\bm {v_2} are velocity vectors. The parameters are the same, and are often tt for time.

With ∣r2(t)−r1(t)∣\lvert\bm {r_2}(t) - \bm {r_1}(t)\rvert as the distance, solve

 d dt∣r2(t)−r1(t)∣=0\frac{\d}{\d t}\lvert\bm {r_2}(t) - \bm {r_1}(t)\rvert = 0

This is minimizing a square root function. However, mathematically speaking, it is easier to minimize the square of the distance

 d dt((r2x−r1x)2+(r2y−r1y)2+(r2z−r1z)2)=0\frac{\d}{\d t} \left((r_{2x} - r_{1x})^2 + (r_{2y} - r_{1y})^2 + (r_{2z} - r_{1z})^2\right) = 0

with each component a function of time. Then take the square root of the sum of squared differences.