P4 7.4–P4 7.6 - Position vectors and lines

P4 7.4–P4 7.6 - Position vectors and lines

Use position vectors to find displacements and distances, then describe a straight line with a vector equation. Decide whether lines coincide, are parallel, intersect or are skew by checking their directions and all three coordinates.

Displacement from position vectors

Two lines in space can have different directions and still miss each other. How can coordinates distinguish a genuine intersection from an apparent crossing in a drawing? First we need a way to locate points and describe the journeys between them.

Position vector

The position vector of a point AA is OA→\overrightarrow{OA}, measured from a chosen origin OO. If AA has coordinates (a1,a2,a3)(a_1,a_2,a_3), then its position vector is a=(a1a2a3)\mathbf a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}.

A point and a vector are related but different objects: A=(1,2,3)A=(1,2,3) locates a point, while OA→=(123)\overrightarrow{OA}=\begin{pmatrix}1\\2\\3\end{pmatrix} describes the displacement from the origin. A free arrow can be translated without changing it; a position vector specifically starts at the chosen origin.

Following OO to AA and then AA to BB reaches BB. Therefore a+AB→=b\mathbf a+\overrightarrow{AB}=\mathbf b.

Displacement between points

AB→=b−a.\overrightarrow{AB}=\mathbf b-\mathbf a.

Subtract the starting position from the finishing position. This result is not printed in the P1–P4 formula booklet: learn it.

The displacement does not depend on where we choose the origin, because an equal shift in both position vectors cancels in their difference.

Worked example 1: from A to B

Question. Points AA and BB have coordinates (1,2,−1)(1,2,-1) and (5,−1,3)(5,-1,3). Find AB→\overrightarrow{AB}.

Plan. Subtract A's position from B's, component by component.

AB→=(5−1−1−23−(−1))=(4−34).\overrightarrow{AB} =\begin{pmatrix}5-1\\-1-2\\3-(-1)\end{pmatrix} =\begin{pmatrix}4\\-3\\4\end{pmatrix}.

Answer. AB→=4i−3j+4k\overrightarrow{AB}=4\mathbf i-3\mathbf j+4\mathbf k.

Check. Adding this displacement to A's position gives (12−1)+(4−34)=(5−13)\begin{pmatrix}1\\2\\-1\end{pmatrix}+\begin{pmatrix}4\\-3\\4\end{pmatrix}=\begin{pmatrix}5\\-1\\3\end{pmatrix}, B's position.

Change only the journey's direction. The coordinates stay the same, but the subtraction reverses.

Worked example 2: from B to A

Question. For the same points, find BA→\overrightarrow{BA}.

Plan. The finishing point is now A, so subtract B's position from A's.

BA→=(1−52−(−1)−1−3)=(−43−4).\overrightarrow{BA} =\begin{pmatrix}1-5\\2-(-1)\\-1-3\end{pmatrix} =\begin{pmatrix}-4\\3\\-4\end{pmatrix}.

Answer. BA→=−4i+3j−4k\overrightarrow{BA}=-4\mathbf i+3\mathbf j-4\mathbf k.

Check. AB→+BA→=0\overrightarrow{AB}+\overrightarrow{BA}=\mathbf0, so the outward and return journeys cancel.

Reversing the order reverses the arrow, while leaving its length unchanged.

Watch out

AB→\overrightarrow{AB} is b−a\mathbf b-\mathbf a, not a+b\mathbf a+\mathbf b. The sum joins two journeys from the origin; it does not give the journey from A to B.

Identify the finishing point before calculating.

Distance between points

The distance between two points is the magnitude of their displacement vector.

Pythagoras combines the perpendicular coordinate changes. For A=(a1,a2,a3)A=(a_1,a_2,a_3) and B=(b1,b2,b3)B=(b_1,b_2,b_3), the result is symmetric: swapping A and B reverses each change, but squaring removes the signs.

Distance in three dimensions

AB=∣AB→∣=(b1−a1)2+(b2−a2)2+(b3−a3)2.AB=|\overrightarrow{AB}|= \sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}.

Distance is a nonnegative scalar, equal to zero precisely when the points coincide. This formula is not printed in the P1–P4 formula booklet: learn it.

Two dimensions are the case with no change in the third coordinate.

Worked example 3: a journey in one horizontal plane

Question. Find the distance between A=(1,−2,3)A=(1,-2,3) and B=(4,2,3)B=(4,2,3).

Plan. Find the three changes; the z-change is zero.

AB→=(340),AB=32+42+02=5.\overrightarrow{AB}=\begin{pmatrix}3\\4\\0\end{pmatrix}, \qquad AB=\sqrt{3^2+4^2+0^2}=5.

Answer. 55 units.

Check. Both points lie in the plane z=3z=3. The horizontal displacement is a 33-44-55 right-angled triangle.

Change only B's z-coordinate. The horizontal journey is unchanged, and there is now an additional perpendicular rise.

Worked example 4: adding a vertical change

Question. Find the distance between A=(1,−2,3)A=(1,-2,3) and C=(4,2,15)C=(4,2,15).

Plan. Combine the same horizontal displacement with the new z-change 1212.

AC→=(3412),AC=32+42+122=13.\overrightarrow{AC}=\begin{pmatrix}3\\4\\12\end{pmatrix}, \qquad AC=\sqrt{3^2+4^2+12^2}=13.

Answer. 1313 units.

Check. The horizontal length was 55, so a second use of Pythagoras gives 52+122=13\sqrt{5^2+12^2}=13.

This two-stage check explains the three-dimensional distance formula rather than merely substituting into it again.

Watch out

Distance is neither the sum of component changes nor the displacement vector itself. The vector (3412)\begin{pmatrix}3\\4\\12\end{pmatrix} has three signed entries; its length is the single number 1313.

Give a scalar length for this journey.

Vector line equations

A straight line is determined by a point on it and a nonzero direction vector.

Starting at position a\mathbf a, move any real multiple of b\mathbf b. Every resulting point lies on the line in direction b\mathbf b, and every point on that line can be reached this way.

Vector equation of a line

r=a+tb,t∈R,b≠0.\mathbf r=\mathbf a+t\mathbf b,\qquad t\in\mathbb R,\quad\mathbf b\ne\mathbf0.

Here r\mathbf r is the position vector of a general point. For distinct points with positions c\mathbf c and d\mathbf d, another useful form is

r=c+t(d−c).\mathbf r=\mathbf c+t(\mathbf d-\mathbf c).

These line forms are not printed in the P1–P4 formula booklet: learn them.

In the two-point form, t=0t=0 gives the first point and t=1t=1 the second. Restricting 0≤t≤10\le t\le1 gives the segment between them; unrestricted real tt gives the whole line. A parameter is not automatically time.

The same line has many vector equations: you can choose a different point on it or multiply its direction vector by any nonzero scalar. To test a point, one single parameter value must produce all its coordinates.

Worked example 5: a line through two points

Question. Find a vector equation of the line through A=(1,2,−1)A=(1,2,-1) and B=(5,0,5)B=(5,0,5). Determine whether P=(9,−2,11)P=(9,-2,11) lies on it.

Plan. Use AB→\overrightarrow{AB} as the direction, then test P in all three component equations.

r=(12−1)+t(4−26).\mathbf r=\begin{pmatrix}1\\2\\-1\end{pmatrix} +t\begin{pmatrix}4\\-2\\6\end{pmatrix}.

For P, 1+4t=91+4t=9 gives t=2t=2. Then

2−2(2)=−2,−1+6(2)=11.2-2(2)=-2,\qquad -1+6(2)=11.

Answer. The displayed equation describes the line, and P lies on it.

Check. Setting t=0t=0 gives A and t=1t=1 gives B. Also AP→=2AB→\overrightarrow{AP}=2\overrightarrow{AB}, so the membership result agrees with parallel displacements.

Continue this lesson.

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