Straight lines

Consider a straight line gg that passes through some point AA and with direction v⃗\vec v. The vector v⃗\vec v is called the direction vector of gg. The direction vector points into the same direction as the straight line. In other words, if we start at AA and move along v⃗\vec v, we are still on line gg.

Now, take another point BB in the 3d-space. How can we check if BB is on line gg?

As the figure above suggests, we have to check if the vector from AA to BB is collinear to the direction vector v⃗\vec v:

B∈g   if   AB→∥v⃗\boxed{B\in g\,\, \text{ if }\,\, \overrightarrow{AB} \parallel \vec v}

Let's consider what exactly this means for the coordinates of AA and BB. Since AB→∥v⃗\overrightarrow{AB} \parallel \vec v there is a value cc with

AB→=c⋅v⃗\overrightarrow{AB} = c \cdot \vec{v}

Note that we can also write the above expression as

B⃗=A⃗+c⋅v⃗\vec B = \vec A+ c \cdot \vec{v}

or in vector notation as

B⃗=A⃗+c⋅v⃗\vec B = \vec A + c \cdot \vec{v}

This last expression is also called the equation of a straight line, and says that BB lies on the straight line if I can reach it by starting at AA and following the direction v⃗\vec{v} (see sketch below, remember that A⃗\vec A and OA→\overrightarrow{OA} are just different notations for the position vector from OO to AA):

Exercise 1

Show that from the equation

AB→=c⋅v⃗\overrightarrow{AB} = c \cdot \vec{v}

follows

B⃗=A⃗+c⋅v⃗\vec B = \vec A+ c \cdot \vec{v}

(and vice versa).

Solution

The short answer is as follows:

Because of

AB→=B−A\overrightarrow{AB}=B-A

follows with

AB→=c⋅v⃗\overrightarrow{AB} = c \cdot \vec{v}

that

B−A=c⋅v⃗B-A = c \cdot \vec{v}

and adding on both sides AA we get

B=c⋅v⃗+A=A+c⋅v⃗B = c \cdot \vec{v}+A = A+c \cdot \vec{v}

The longer answer involves the coordinates:

We can write

AB→=c⋅v⃗\overrightarrow{AB} = c \cdot \vec{v}

as

(Bx−AxBy−AyBz−Az)=c(vxvyvz)=(c⋅vxc⋅vyc⋅vz)\left(\begin{array}{r} B_x-A_x\\ B_y-A_y\\ B_z-A_z \end{array}\right) = c \left(\begin{array}{r} v_x\\ v_y\\ v_z\end{array}\right) = \left(\begin{array}{r} c\cdot v_x\\ c\cdot v_y\\ c\cdot v_z\end{array}\right)

We thus obtain the 33 equations

Bx−Ax=c⋅vx B_x-A_x = c\cdot v_xBy−Ay=c⋅vyB_y-A_y = c\cdot v_yBz−Az=c⋅vzB_z-A_z = c\cdot v_z

If we take the coordinates of AA to the other side, we get the equations

Bx=Ax+c⋅vxBy=Ay+c⋅vyBz=Az+c⋅vz\begin{array}{lll} B_x &=& A_x+c\cdot v_x\\ B_y &=& A_y+c\cdot v_y\\ B_z &=& A_z+c\cdot v_z \end{array}

which is

B=A+c⋅v⃗B=A+c\cdot \vec{v}

or also

B⃗=A⃗+c⋅v⃗\vec B=\vec A+c\cdot \vec{v}
Example 1

The straight line gg passes through the point A(2∣1∣−3)A(2\vert 1\vert -3) and has the direction vector

v⃗=(3−14)\vec{v}=\left(\begin{array}{r} 3\\ -1\\ 4\end{array}\right)

Decide whether the point B(8∣−1∣5)B(8\vert -1\vert 5) lies on gg.

Solution

Method 1: (Collinearity) We need to test whether AB→∥v⃗\overrightarrow{AB} \parallel \vec v, hence whether there is a cc with AB→=c⋅v⃗\overrightarrow{AB} = c \cdot \vec{v}, thus if

(6−28)=(3c−c4c)\left(\begin{array}{r} 6\\ -2\\ 8\end{array}\right)=\left(\begin{array}{r} 3c\\ -c\\ 4c\end{array}\right)

so whether there is a cc such that all three equations are satisfied:

6=3c−2=−c8=4c\begin{array}{rll} 6&=&3c\\ -2&=&-c\\ 8&=&4c\end{array}

From the first equation it follows c=2c=2, and by checking the other two equations we see that for c=2c=2 all equations are satisfied. BB is therefore on the straight line.

Method 2: (equation of straight line) We need to test whether there is a value cc with

B⃗=A⃗+c⋅v⃗\vec B=\vec A+c\cdot\vec{v}

therefore with

(8−15)=(21−3)+c⋅(3−14)=(2+3c1−c−3+4c)\left(\begin{array}{r} 8\\ -1\\ 5\end{array}\right)=\left(\begin{array}{r} 2\\ 1\\ -3\end{array}\right)+c\cdot \left(\begin{array}{r} 3\\ -1\\ 4\end{array}\right) = \left(\begin{array}{r} 2+3c\\ 1-c\\ -3+4c\end{array}\right)

From which the three equations it follows:

8=2+3c−1=1−c5=−3+4c\begin{array}{rll} 8&=&2+3c\\ -1&=&1-c\\ 5&=&-3+4c\end{array}

From the first equation follows c=2c=2, and by checking the other two equations we see that for c=2c=2 they are also satisfied, so BB is on the straight line.

Exercise 2

  1. A straight line gg passes through the points U(1∣2∣−1)U(1\vert 2\vert -1) and V(5∣2∣10)V(5\vert 2\vert 10). Find a direction vector of gg. How many direction vectors are there for line gg? Characterise them.

  2. A line hh passes through the point Q(1∣3∣2.2)Q(1\vert 3\vert 2.2). Find its direction vector if hh is

    1. parallel to the xx-axis.
    2. parallel to the zz-axis.
    3. forms a right angle with the xzxz-plane.

  3. Consider the straight line hh that passes through the point Q(1∣2∣−1)Q(1\vert 2\vert -1) and has direction

    m⃗=(2−1.53)\vec{m}=\left(\begin{array}{r} 2\\ -1.5\\ 3 \end{array}\right)
    1. Is the point U(9∣−4∣11)U(9\vert -4\vert 11) on the line?
    2. Does the line pass through the origin?

  4. Consider the straight line gg that passes through the point A(−2∣3∣4)A(-2\vert 3\vert 4) and has direction

    v⃗=(21−2)\vec{v}=\left(\begin{array}{r} 2\\ 1\\ -2 \end{array}\right)

    Where does gg intersect with the xy-plane?

Solution
  1. v⃗=UV→=(4011)\vec v = \overrightarrow{UV} = \left(\begin{array}{r} 4\\ 0\\ 11 \end{array}\right)
  2. It is
    1. v⃗=(100)\vec v = \left(\begin{array}{r} 1\\ 0\\ 0 \end{array}\right)
    2. v⃗=(001)\vec v = \left(\begin{array}{r} 0\\ 0\\ 1 \end{array}\right)
    3. v⃗=(010)\vec v = \left(\begin{array}{r} 0\\ 1\\ 0 \end{array}\right)
  3. We have:
    1. Two methods:
      1. Method 1: (Collinearity) Find out if m⃗∥QU→\vec{m} \parallel \overrightarrow{QU}, that is, if there is a cc with QU→=c⋅m⃗\overrightarrow{QU}=c\cdot \vec{m} (see figure below) QU→=(9−1−4−211−(−1))=(8−612)=4m⃗\overrightarrow{QU}=\left(\begin{array}{r} 9-1\\ -4-2\\ 11-(-1) \end{array}\right) = \left(\begin{array}{r} 8\\ -6\\ 12 \end{array}\right) = 4\vec{m} so collinear. Thus, UU is on the line!
      2. Method 2: (equation of straight line) There must be a cc with: U⃗=Q⃗+c⋅m⃗\vec U = \vec Q+c\cdot \vec{m} Or written with the components: (9−411)=(1 2 −1)+c⋅(2 −1.53)=(1+2c 2−1.5c −1+3c)\left(\begin{array}{r} 9\\ -4\\ 11\end{array}\right) = \left(\begin{array}{r} 1\ 2\ -1 \end{array}\right)+c\cdot \left(\begin{array}{r} 2\ -1. 5\\ 3 \end{array}\right)=\left(\begin{array}{r} 1+2c\ 2-1.5c\ -1+3c \end{array}\right) We thus obtain the three equations 9=1+2c−4=2−1.5c11=−1+3c\begin{array}{rll} 9 &=& 1+2c\\ -4 &=& 2 -1.5c\\ 11 &=& -1+3c \end{array} We now need to test whether there exists a cc that satisfies all three equations. And indeed, this is the case for c=4c=4.
    2. Check whether O(0∣0∣0)O(0\vert 0\vert 0) lies on gg. We can again test whether m⃗∥QO→\vec{m} \parallel \overrightarrow{QO}, or whether there is a cc that satisfies the straight line equation O⃗=Q⃗+c⋅m⃗\vec O=\vec Q+c\cdot \vec{m}. This is not the case, thus the straight line does not pass through the origin.
  4. We denote the intersection between gg and the xyxy-plane by SS (see sketch below). Since SS is the intersection point, it lies both in the xyxy-plane and on gg. We do not know the coordinates SxS_x and SyS_y, but since SS lies on the xy-plane, Sz=0S_z=0 holds. So we have S(Sx∣Sy∣0)S(S_x\vert S_y\vert 0) Two methods to go on:
    1. Method 1 (collinearity): Since SS lies on gg the following must apply v⃗∥AS→\vec{v} \parallel \overrightarrow{AS} and thus there is a stretching factor cc with AS→=c⋅v⃗\overrightarrow{AS}=c\cdot \vec{v} With AS→=(Sx+2Sy−30−4)\overrightarrow{AS}=\left(\begin{array}{r} S_x+2\\ S_y-3\\ 0-4 \end{array}\right) follows (Sx+2Sy−30−4)=(2cc−2c)→Sx=2c−2→Sy=c+3→−4=−2c\left(\begin{array}{r} S_x+2\\ S_y-3\\ 0-4 \end{array}\right) = \left(\begin{array}{r} 2c\\ c\\ -2c \end{array}\right) \begin{array}{l} \rightarrow S_x=2c-2 \\ \rightarrow S_y=c+3 \\ \rightarrow -4=-2c\end{array}
    2. Method 2 (equation of straight line): Since SS lies on gg the following must hold for a cc (SxSy0)=(−2+2c3+c4−2c)\left(\begin{array}{r} S_x\\ S_y\\ 0 \end{array}\right) = \left(\begin{array}{r} -2+2c\\ 3+c\\ 4-2c \end{array}\right) and thus Sx=−2+2cSy=3+c0=4−2c\begin{array}{lll} S_x &=& -2+2c\\ S_y &=& 3+c\\ 0 &=& 4-2c \end{array} With both methods c=2c=2 follows (e.g. see the third equation in each case). So Sx=2⋅2−2=2S_x=2\cdot 2-2=2, and Sy=2+3=5S_y=2+3=5. The intersection point thus has the coordinates S(2∣5∣0)S(2\vert 5\vert 0).