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Chapter 10 · Vector Algebra

Example 17: ∣a⃗−b⃗∣|\vec a-\vec b| when ∣a⃗∣=2|\vec a|=2⁠, ∣b⃗∣=3|\vec b|=3⁠, a⃗⋅b⃗=4\vec a\cdot\vec b=4

ExamplesMagnitude through the scalar product · Section 10.6.1
The question

Find ∣a⃗−b⃗∣|\vec a - \vec b|⁠, if two vectors a⃗\vec a and b⃗\vec b are such that ∣a⃗∣=2|\vec a| = 2⁠, ∣b⃗∣=3|\vec b| = 3 and a⃗⋅b⃗=4\vec a\cdot\vec b = 4⁠.

Exam answer

Using ∣r⃗∣2=r⃗⋅r⃗|\vec r|^2 = \vec r\cdot\vec r and b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b⁠,

∣a⃗−b⃗∣2=(a⃗−b⃗)⋅(a⃗−b⃗)=a⃗⋅a⃗−a⃗⋅b⃗−b⃗⋅a⃗+b⃗⋅b⃗=∣a⃗∣2−2(a⃗⋅b⃗)+∣b⃗∣2=(2)2−2(4)+(3)2=4−8+9=5\begin{aligned} |\vec a - \vec b|^2 &= (\vec a - \vec b)\cdot(\vec a - \vec b) \\ &= \vec a\cdot\vec a - \vec a\cdot\vec b \\ &\quad - \vec b\cdot\vec a + \vec b\cdot\vec b \\ &= |\vec a|^2 - 2(\vec a\cdot\vec b) + |\vec b|^2 \\ &= (2)^2 - 2(4) + (3)^2 \\ &= 4 - 8 + 9 = 5 \end{aligned}

As a magnitude is non-negative,

∴ ∣a⃗−b⃗∣=5\therefore\ |\vec a - \vec b| = \sqrt5⁠.

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

The one idea this problem is built on

We do not know the components of a⃗\vec a and b⃗\vec b⁠, only their lengths and their scalar product. The bridge is ∣r⃗∣2=r⃗⋅r⃗|\vec r|^2 = \vec r\cdot\vec r⁠: square the length we want, write it as a scalar product, and expand it like (x−y)2(x - y)^2:

∣a⃗−b⃗∣2=∣a⃗∣2−2 a⃗⋅b⃗+∣b⃗∣2.|\vec a - \vec b|^2 = |\vec a|^2 - 2\,\vec a\cdot\vec b + |\vec b|^2.
Given, to find, and the plan
Given
∣a⃗∣=2|\vec a| = 2⁠, ∣b⃗∣=3|\vec b| = 3⁠, a⃗⋅b⃗=4\vec a\cdot\vec b = 4⁠.
To find
∣a⃗−b⃗∣|\vec a - \vec b|⁠.
Plan
(1) Write ∣a⃗−b⃗∣2=(a⃗−b⃗)⋅(a⃗−b⃗)|\vec a - \vec b|^2 = (\vec a - \vec b)\cdot(\vec a - \vec b)⁠. (2) Expand by distributivity and commutativity. (3) Substitute the given numbers. (4) Take the non-negative square root, and check with the cosine rule.

Basics you need

1Length from a scalar product

r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2⁠, since the angle of a vector with itself is 0 (Section 10.6.1, Observation 3). Lesson: the dot product

2Expanding scalar products

a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b + \vec c) = \vec a\cdot\vec b + \vec a\cdot\vec c (Property 1), a⃗⋅b⃗=b⃗⋅a⃗\vec a\cdot\vec b = \vec b\cdot\vec a (Observation 7), and numbers come out of either factor (Property 2), all in Section 10.6.1. Basics: dot products obey the same rule

3The angle formula

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} (Section 10.6.1, Observation 6), used here only for the check. Basics: cos θ is a shadow

Solution

  1. Square the length and write it as a scalar product

    ∣a⃗−b⃗∣2=(a⃗−b⃗)⋅(a⃗−b⃗).|\vec a - \vec b|^2 = (\vec a - \vec b)\cdot(\vec a - \vec b).
    A length cannot be found from the given data directly, but its square can: the scalar product of a vector with itself is the square of its length. This turns the question into algebra with the three given numbers.
    r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2 (Section 10.6.1, Observation 3). Lesson: the dot product

  2. Expand

    (a⃗−b⃗)⋅(a⃗−b⃗)=a⃗⋅a⃗−a⃗⋅b⃗−b⃗⋅a⃗+b⃗⋅b⃗=∣a⃗∣2−2(a⃗⋅b⃗)+∣b⃗∣2.\begin{aligned} &(\vec a - \vec b)\cdot(\vec a - \vec b) \\ &= \vec a\cdot\vec a - \vec a\cdot\vec b - \vec b\cdot\vec a + \vec b\cdot\vec b \\ &= |\vec a|^2 - 2(\vec a\cdot\vec b) + |\vec b|^2. \end{aligned}
    Distributivity multiplies out the brackets into four scalar products, just like (x−y)(x−y)(x - y)(x - y)⁠. Commutativity, b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b⁠, joins the two middle terms into −2(a⃗⋅b⃗)-2(\vec a\cdot\vec b)⁠, and a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2⁠, b⃗⋅b⃗=∣b⃗∣2\vec b\cdot\vec b = |\vec b|^2⁠.
    Property 1, Property 2 and Observation 7 (Section 10.6.1). Basics: dot products obey the same rule

  3. Substitute

    ∣a⃗−b⃗∣2=(2)2−2(4)+(3)2=4−8+9=5.\begin{aligned} |\vec a - \vec b|^2 &= (2)^2 - 2(4) + (3)^2 \\ &= 4 - 8 + 9 = 5. \end{aligned}
    Only the three given numbers appear: ∣a⃗∣=2|\vec a| = 2⁠, a⃗⋅b⃗=4\vec a\cdot\vec b = 4 and ∣b⃗∣=3|\vec b| = 3⁠. The scalar product is counted twice, 2×4=82 \times 4 = 8⁠.
    The given data. Lesson: Examples 13–21

  4. Take the square root, and check

    A length is never negative, so ∣a⃗−b⃗∣=5|\vec a - \vec b| = \sqrt5.

    Check with the cosine rule: cos⁡θ=42×3=23\cos\theta = \frac{4}{2 \times 3} = \frac23⁠, and the side opposite θ\theta in the triangle with sides 2 and 3 is

    22+32−2⋅2⋅3⋅23=13−8=5.\begin{aligned} &\sqrt{2^2 + 3^2 - 2 \cdot 2 \cdot 3 \cdot \tfrac23} \\ &= \sqrt{13 - 8} = \sqrt5. \end{aligned}
    θ O A B a, length 2 b, length 3 a − b, length √5 cos θ = 4/(2 × 3) = 2/3 4 − 8 + 9 = 5
    Drawn to scale, one scale on both axes: a⃗=OA→\vec a = \overrightarrow{OA} of length 2 and b⃗=OB→\vec b = \overrightarrow{OB} of length 3, with cos⁡θ=23\cos\theta = \frac23⁠. Then a⃗−b⃗=BA→\vec a - \vec b = \overrightarrow{BA}⁠, of length 5\sqrt5⁠. Here the shadow of b⃗\vec b on a⃗\vec a is 3×23=23 \times \frac23 = 2⁠, exactly ∣a⃗∣|\vec a|⁠, so BA is perpendicular to OA.
    a⃗−b⃗\vec a - \vec b runs from the tip of b⃗\vec b to the tip of a⃗\vec a (since b⃗+(a⃗−b⃗)=a⃗\vec b + (\vec a - \vec b) = \vec a⁠), so ∣a⃗−b⃗∣|\vec a - \vec b| is the third side of the triangle OAB, and the expansion in Step 2 is the cosine rule written with vectors. In this example (a⃗−b⃗)⋅a⃗=∣a⃗∣2−a⃗⋅b⃗=4−4=0(\vec a - \vec b)\cdot\vec a = |\vec a|^2 - \vec a\cdot\vec b = 4 - 4 = 0⁠, so the triangle has a right angle at A and AB2=32−22=5\text{AB}^2 = 3^2 - 2^2 = 5 as well.
    Triangle law (Section 10.4); cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} (Section 10.6.1, Observation 6). Lesson: adding arrows, head to tail
Result

∣a⃗−b⃗∣=5|\vec a - \vec b| = \sqrt5⁠.

(Textbook answer: ∣a⃗−b⃗∣2=∣a⃗∣2−2(a⃗⋅b⃗)+∣b⃗∣2|\vec a - \vec b|^2 = |\vec a|^2 - 2(\vec a\cdot\vec b) + |\vec b|^2 =(2)2−2(4)+(3)2= (2)^2 - 2(4) + (3)^2⁠, therefore ∣a⃗−b⃗∣=5|\vec a - \vec b| = \sqrt5⁠.)

Check it with numbers

A concrete pair with the given data: a⃗=2i^\vec a = 2\hat i and b⃗=2i^+5 j^\vec b = 2\hat i + \sqrt5\,\hat j⁠. Then ∣a⃗∣=2|\vec a| = 2⁠, ∣b⃗∣=4+5=3|\vec b| = \sqrt{4 + 5} = 3⁠, a⃗⋅b⃗=4\vec a\cdot\vec b = 4⁠, and a⃗−b⃗=−5 j^\vec a - \vec b = -\sqrt5\,\hat j⁠, of length 5≈2.236068\sqrt5 \approx 2.236068⁠.

The angle: cos⁡θ=23≈0.666667\cos\theta = \frac23 \approx 0.666667⁠, so θ≈48.19∘\theta \approx 48.19^\circ⁠; and 4+9−8=54 + 9 - 8 = 5⁠.

Common mistakes
  1. Subtracting the lengths. ∣a⃗−b⃗∣|\vec a - \vec b| is not ∣a⃗∣−∣b⃗∣=−1|\vec a| - |\vec b| = -1⁠; a length cannot even be negative.
  2. Forgetting the factor 2. 4−4+9=94 - 4 + 9 = 9 would give 3; the two middle terms −a⃗⋅b⃗−b⃗⋅a⃗-\vec a\cdot\vec b - \vec b\cdot\vec a make −2(a⃗⋅b⃗)-2(\vec a\cdot\vec b)⁠.
  3. Stopping at the square. ∣a⃗−b⃗∣2=5|\vec a - \vec b|^2 = 5⁠; the length is 5\sqrt5⁠, not 5.
What to take away
  1. To find a length from scalar data, square it: ∣r⃗∣2=r⃗⋅r⃗|\vec r|^2 = \vec r\cdot\vec r⁠.
  2. ∣a⃗±b⃗∣2=∣a⃗∣2±2 a⃗⋅b⃗+∣b⃗∣2|\vec a \pm \vec b|^2 = |\vec a|^2 \pm 2\,\vec a\cdot\vec b + |\vec b|^2⁠.
  3. This is the cosine rule in vector form.

Every fact used, and where it comes from

FactUsed inSource
r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2Steps 1, 2Section 10.6.1, Observation 3 (Lesson: the dot product)
Distributivity; scalars come out; commutativityStep 2Section 10.6.1, Properties 1 and 2, Observation 7 (Basics: dot products obey the same rule)
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}Step 4Section 10.6.1, Observation 6
Triangle law: b⃗+(a⃗−b⃗)=a⃗\vec b + (\vec a - \vec b) = \vec aStep 4Section 10.4
The book's working and answerResultExample 17, Section 10.6.2 (Lesson: Examples 13–21)
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