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Chapter 10 · Vector Algebra

Example 27: Angle between AB→\overrightarrow{AB} and CD→\overrightarrow{CD} is π\pi⁠; they are collinear

ExamplesAngle between vectors, collinear vectors · Miscellaneous Examples
The question

If i^+j^+k^\hat i + \hat j + \hat k⁠, 2i^+5j^2\hat i + 5\hat j⁠, 3i^+2j^−3k^3\hat i + 2\hat j - 3\hat k and i^−6j^−k^\hat i - 6\hat j - \hat k are the position vectors of points A, B, C and D respectively, then find the angle between AB→\overrightarrow{AB} and CD→\overrightarrow{CD}⁠. Deduce that AB→\overrightarrow{AB} and CD→\overrightarrow{CD} are collinear.

Exam answer AB→=(2i^+5j^)−(i^+j^+k^)=i^+4j^−k^\begin{aligned} \overrightarrow{AB} &= (2\hat i + 5\hat j) - (\hat i + \hat j + \hat k) \\ &= \hat i + 4\hat j - \hat k \end{aligned} CD→=(i^−6j^−k^)−(3i^+2j^−3k^)=−2i^−8j^+2k^\begin{aligned} \overrightarrow{CD} &= (\hat i - 6\hat j - \hat k) \\ &\quad - (3\hat i + 2\hat j - 3\hat k) \\ &= -2\hat i - 8\hat j + 2\hat k \end{aligned} ∣AB→∣=1+16+1=32∣CD→∣=4+64+4=62\begin{aligned} |\overrightarrow{AB}| &= \sqrt{1 + 16 + 1} = 3\sqrt2 \\ |\overrightarrow{CD}| &= \sqrt{4 + 64 + 4} = 6\sqrt2 \end{aligned} AB→⋅CD→=(1)(−2)+(4)(−8)+(−1)(2)=−2−32−2=−36\begin{aligned} \overrightarrow{AB}\cdot\overrightarrow{CD} &= (1)(-2) + (4)(-8) \\ &\quad + (-1)(2) \\ &= -2 - 32 - 2 = -36 \end{aligned}

If θ\theta is the angle between AB→\overrightarrow{AB} and CD→\overrightarrow{CD}⁠, then

cos⁡θ=AB→⋅CD→∣AB→∣∣CD→∣=−36(32)(62)=−3636=−1\begin{aligned} \cos\theta &= \frac{\overrightarrow{AB}\cdot\overrightarrow{CD}}{|\overrightarrow{AB}||\overrightarrow{CD}|} \\ &= \frac{-36}{(3\sqrt2)(6\sqrt2)} = \frac{-36}{36} \\ &= -1 \end{aligned}

Since 0≤θ≤π0 \le \theta \le \pi⁠, θ=π\theta = \pi⁠. So the vectors are parallel and opposite in direction; indeed AB→=−12CD→\overrightarrow{AB} = -\frac12\overrightarrow{CD}⁠.

∴\therefore the angle between AB→\overrightarrow{AB} and CD→\overrightarrow{CD} is π\pi⁠, and AB→\overrightarrow{AB}⁠, CD→\overrightarrow{CD} are collinear.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The angle between two vectors comes from cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣\cos\theta = \frac{\vec u\cdot\vec v}{|\vec u||\vec v|} with 0≤θ≤π0 \le \theta \le \pi⁠. If cos⁡θ=±1\cos\theta = \pm1⁠, the angle is 0 or π\pi⁠: the vectors lie along parallel lines, which is what "collinear vectors" means.

Given, to find, and the plan
Given
A(1, 1, 1), B(2, 5, 0), C(3, 2, −3-3⁠), D(1, −6-6⁠, −1-1⁠), from the position vectors (B has no k^\hat k term, so its z-coordinate is 0).
To find
The angle θ\theta between AB→\overrightarrow{AB} and CD→\overrightarrow{CD}⁠; then show they are collinear.
Plan
(1) Find AB→\overrightarrow{AB} and CD→\overrightarrow{CD}⁠. (2) Their lengths. (3) Their scalar product. (4) cos⁡θ\cos\theta⁠, then θ\theta in [0,π][0, \pi]⁠. (5) Read off collinearity, and see what it does and does not say about the four points.

Basics you need

1Vector joining two points

AB→\overrightarrow{AB} = position vector of B −- position vector of A (Section 10.5.2). Basics: three numbers fix a point

2Angle between two vectors

cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣\cos\theta = \frac{\vec u\cdot\vec v}{|\vec u||\vec v|}⁠, 0≤θ≤π0 \le \theta \le \pi⁠, and u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3 (Section 10.6.1, Observation 6). Lesson: the dot product

3Collinear vectors

Vectors are collinear if they are parallel to the same line, whatever their lengths and directions (Section 10.3); equivalently v⃗=λu⃗\vec v = \lambda\vec u (Section 10.5.1, Remarks (i)). Lesson: six kinds of vectors

4Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

Solution

  1. The vector AB→\overrightarrow{AB}

    AB→=(2i^+5j^)−(i^+j^+k^)=i^+4j^−k^.\begin{aligned} \overrightarrow{AB} &= (2\hat i + 5\hat j) - (\hat i + \hat j + \hat k) \\ &= \hat i + 4\hat j - \hat k. \end{aligned}
    Position vector of B minus position vector of A, component by component: 2−1=12 - 1 = 1⁠, 5−1=45 - 1 = 4⁠, 0−1=−10 - 1 = -1⁠.
    AB→=OB→−OA→\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} (Section 10.5.2). Basics: three numbers fix a point

  2. The vector CD→\overrightarrow{CD}

    CD→=(i^−6j^−k^)−(3i^+2j^−3k^)=−2i^−8j^+2k^.\begin{aligned} \overrightarrow{CD} &= (\hat i - 6\hat j - \hat k) - (3\hat i + 2\hat j - 3\hat k) \\ &= -2\hat i - 8\hat j + 2\hat k. \end{aligned}
    D minus C: 1−3=−21 - 3 = -2⁠, −6−2=−8-6 - 2 = -8⁠, −1−(−3)=2-1 - (-3) = 2⁠. The order matters: DC→\overrightarrow{DC} would be the opposite vector and would give the opposite answer for the angle.
    CD→=OD→−OC→\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} (Section 10.5.2). Lesson: components

  3. Their lengths

    ∣AB→∣=(1)2+(4)2+(−1)2=18=32,∣CD→∣=(−2)2+(−8)2+(2)2=72=62.\begin{aligned} |\overrightarrow{AB}| &= \sqrt{(1)^2 + (4)^2 + (-1)^2} \\ &= \sqrt{18} = 3\sqrt2, \\ |\overrightarrow{CD}| &= \sqrt{(-2)^2 + (-8)^2 + (2)^2} \\ &= \sqrt{72} = 6\sqrt2. \end{aligned}
    18=9×218 = 9 \times 2 and 72=36×272 = 36 \times 2⁠, so the roots simplify. The product of the lengths is (32)(62)=18×2=36(3\sqrt2)(6\sqrt2) = 18 \times 2 = 36⁠.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  4. The scalar product

    AB→⋅CD→=(1)(−2)+(4)(−8)+(−1)(2)=−2−32−2=−36.\begin{aligned} \overrightarrow{AB}\cdot\overrightarrow{CD} &= (1)(-2) + (4)(-8) \\ &\quad + (-1)(2) \\ &= -2 - 32 - 2 = -36. \end{aligned}
    Multiply matching components and add. The result is negative, so the angle is already known to be obtuse before we divide.
    u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3 (Section 10.6.1). Lesson: the dot product

  5. The angle

    cos⁡θ=AB→⋅CD→∣AB→∣∣CD→∣=−36(32)(62)=−3636=−1.\begin{aligned} \cos\theta &= \frac{\overrightarrow{AB}\cdot\overrightarrow{CD}}{|\overrightarrow{AB}||\overrightarrow{CD}|} \\ &= \frac{-36}{(3\sqrt2)(6\sqrt2)} = \frac{-36}{36} = -1. \end{aligned}

    The only angle in [0,π][0, \pi] with cosine −1-1 is π\pi⁠, so θ=π\theta = \pi⁠.

    The angle between two vectors is always taken in [0,π][0, \pi]⁠, and on that interval cosine takes each value in [−1,1][-1, 1] exactly once; cos⁡π=−1\cos\pi = -1⁠. An angle of π\pi means the vectors point in exactly opposite directions.
    cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, 0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Definition 2 and Observation 6). Lesson: Examples 26–30

  6. Deduce that they are collinear

    Since θ=π\theta = \pi⁠, AB→\overrightarrow{AB} and CD→\overrightarrow{CD} are parallel and opposite in direction, so they are collinear. In components:

    CD→=−2i^−8j^+2k^=−2(i^+4j^−k^)=−2 AB→,\begin{aligned} \overrightarrow{CD} &= -2\hat i - 8\hat j + 2\hat k \\ &= -2(\hat i + 4\hat j - \hat k) = -2\,\overrightarrow{AB}, \end{aligned}

    that is, AB→=−12CD→\overrightarrow{AB} = -\frac12\overrightarrow{CD}⁠, a scalar multiple, which is the book's alternative argument.

    The four points themselves are not on one line: AC→=2i^+j^−4k^\overrightarrow{AC} = 2\hat i + \hat j - 4\hat k is not a multiple of AB→=i^+4j^−k^\overrightarrow{AB} = \hat i + 4\hat j - \hat k (the i^\hat i parts would need the factor 2, the j^\hat j parts the factor 14\frac14⁠). So the lines AB and CD are two different parallel lines.

    A(1, 1, 1) B(2, 5, 0) C(3, 2, −3) D(1, −6, −1) AB, length 3√2 CD, length 6√2 ≈ 3.93 apart CD = −2 AB, θ = π
    A, B, C, D drawn true to scale in the plane that holds the two parallel lines: CD→\overrightarrow{CD} is twice as long as AB→\overrightarrow{AB} and points the opposite way, and the lines are about 3.93 units apart.
    "Collinear vectors" means vectors parallel to the same line, whatever their lengths and directions; it is a statement about directions, not about where the vectors sit. The vectors can be slid parallel to themselves onto one line, but the points A, B, C, D stay on two separate lines.
    Collinear vectors (Section 10.3); v⃗=λu⃗\vec v = \lambda\vec u (Section 10.5.1, Remarks (i)). Lesson: six kinds of vectors
Result

The angle between AB→\overrightarrow{AB} and CD→\overrightarrow{CD} is θ=π\theta = \pi⁠. Hence AB→\overrightarrow{AB} and CD→\overrightarrow{CD} are collinear (parallel and opposite, CD→=−2 AB→\overrightarrow{CD} = -2\,\overrightarrow{AB}⁠), although the points A, B, C, D do not all lie on one line.

(Textbook answer: AB→=i^+4j^−k^\overrightarrow{AB} = \hat i + 4\hat j - \hat k⁠, ∣AB→∣=32|\overrightarrow{AB}| = 3\sqrt2⁠, CD→=−2i^−8j^+2k^\overrightarrow{CD} = -2\hat i - 8\hat j + 2\hat k⁠, ∣CD→∣=62|\overrightarrow{CD}| = 6\sqrt2⁠, cos⁡θ=−3636=−1\cos\theta = \frac{-36}{36} = -1⁠; "Since 0≤θ≤π0 \le \theta \le \pi⁠, it follows that θ=π\theta = \pi", so the vectors are collinear; alternatively AB→=−12CD→\overrightarrow{AB} = -\frac12\overrightarrow{CD}⁠.)

Check it with numbers

∣AB→∣=18≈4.242641|\overrightarrow{AB}| = \sqrt{18} \approx 4.242641 and ∣CD→∣=72≈8.485281|\overrightarrow{CD}| = \sqrt{72} \approx 8.485281⁠, whose product is exactly 36, so cos⁡θ=−3636=−1\cos\theta = -\frac{36}{36} = -1⁠.

With the cross product (Section 10.6.3): AB→×CD→=0⃗\overrightarrow{AB}\times\overrightarrow{CD} = \vec 0⁠, because CD→=−2 AB→\overrightarrow{CD} = -2\,\overrightarrow{AB}⁠; but AB→×AC→=−15i^+2j^−7k^≠0⃗\overrightarrow{AB}\times\overrightarrow{AC} = -15\hat i + 2\hat j - 7\hat k \ne \vec 0⁠, so C is off the line AB. The distance between the two lines is ∣AB→×AC→∣∣AB→∣=27818=1393≈3.929942\frac{|\overrightarrow{AB}\times\overrightarrow{AC}|}{|\overrightarrow{AB}|} = \frac{\sqrt{278}}{\sqrt{18}} = \frac{\sqrt{139}}{3} \approx 3.929942⁠.

Common mistakes
  1. Confusing collinear vectors with collinear points. AB→\overrightarrow{AB} and CD→\overrightarrow{CD} are collinear vectors, but A, B, C, D are not on one line.
  2. Subtracting the wrong way. Using DC→=2i^+8j^−2k^\overrightarrow{DC} = 2\hat i + 8\hat j - 2\hat k gives cos⁡θ=+1\cos\theta = +1 and θ=0\theta = 0⁠; the question asks about CD→\overrightarrow{CD}⁠.
  3. Misreading cos⁡θ=−1\cos\theta = -1⁠. It means θ=π\theta = \pi (opposite directions), not π2\frac{\pi}{2}⁠; perpendicular vectors have cos⁡θ=0\cos\theta = 0⁠.
What to take away
  1. cos⁡θ=±1\cos\theta = \pm1 exactly when the vectors are collinear: θ=0\theta = 0 same way, θ=π\theta = \pi opposite ways.
  2. A quicker test is v⃗=λu⃗\vec v = \lambda\vec u⁠; here λ=−2\lambda = -2⁠.
  3. Collinear vectors may lie on different parallel lines; collinear points lie on one line.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two pointsSteps 1, 2Section 10.5.2 (Lesson: components)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 3Section 10.5.1
u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3Step 4Section 10.6.1 (Lesson: the dot product)
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, 0≤θ≤π0 \le \theta \le \piStep 5Section 10.6.1, Definition 2 and Observation 6
Collinear vectors; v⃗=λu⃗\vec v = \lambda\vec uStep 6Section 10.3; Section 10.5.1, Remarks (i)
a⃗×b⃗=0⃗\vec a\times\vec b = \vec 0 for parallel vectors; height of a parallelogram =∣a⃗×b⃗∣∣a⃗∣= \frac{|\vec a\times\vec b|}{|\vec a|}CheckSection 10.6.3, Observations 2 and 9
The book's working and answerResultExample 27, Miscellaneous Examples (Lesson: Examples 26–30)
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