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Chapter 10 · Vector Algebra

Exercise 10.3 · Q13: a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a for unit vectors with a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0

Exercise 10.3Expanding scalar products · Section 10.6.1
The question

If a⃗,b⃗,c⃗\vec a, \vec b, \vec c are unit vectors such that a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠, find the value of a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a⁠.

Exam answer

Given ∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec a| = |\vec b| = |\vec c| = 1 and a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠.

Expanding, and using x⃗⋅y⃗=y⃗⋅x⃗\vec x\cdot\vec y = \vec y\cdot\vec x and x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2⁠,

(a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗)=0⇒∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0⇒3+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0\begin{aligned} &(\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) = 0 \\ &\Rightarrow |\vec a|^2 + |\vec b|^2 + |\vec c|^2 \\ &\quad + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0 \\ &\Rightarrow 3 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0 \end{aligned}

∴ a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32\therefore\ \vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\dfrac32⁠.

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

The one idea this problem is built on

The sum is the zero vector, so its scalar product with itself is 0. Expanding that square produces exactly the three lengths (each 1) and twice the sum we want:

∣a⃗+b⃗+c⃗∣2=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗).\begin{aligned} &|\vec a + \vec b + \vec c|^2 \\ &= |\vec a|^2 + |\vec b|^2 + |\vec c|^2 \\ &\quad + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a). \end{aligned}
Given, to find, and the plan
Given
∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec a| = |\vec b| = |\vec c| = 1 and a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠.
To find
a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a⁠.
Plan
(1) Dot the given sum with itself. (2) Expand the nine terms. (3) Put in the lengths and solve. (4) Confirm by a second method. (5) See the picture: an equilateral triangle.

Basics you need

1Unit vector

A unit vector has magnitude 1 (Section 10.3). Lesson: six kinds of vectors

2Expanding scalar products

Distributivity (Property 1), x⃗⋅y⃗=y⃗⋅x⃗\vec x\cdot\vec y = \vec y\cdot\vec x (Observation 7) and x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2 (Observation 3), Section 10.6.1. Basics: dot products obey the same rule

3Scalar product and angle

x⃗⋅y⃗=∣x⃗∣∣y⃗∣cos⁡θ\vec x\cdot\vec y = |\vec x||\vec y|\cos\theta (Section 10.6.1). Used in the picture of Step 5. Basics: cos θ is a shadow

Solution

  1. Dot the sum with itself

    (a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗)=0⃗⋅0⃗=0.(\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) = \vec 0\cdot\vec 0 = 0.
    We need only the sum of the three scalar products, not each one separately, and that sum appears when the sum vector is multiplied by itself. Since a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠, that product is ∣0⃗∣2=0|\vec 0|^2 = 0⁠.
    x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2 (Section 10.6.1, Observation 3). Lesson: the dot product

  2. Expand the nine terms

    (a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗)=a⃗⋅a⃗+a⃗⋅b⃗+a⃗⋅c⃗+b⃗⋅a⃗+b⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗+c⃗⋅b⃗+c⃗⋅c⃗=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗).\begin{aligned} &(\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) \\ &= \vec a\cdot\vec a + \vec a\cdot\vec b + \vec a\cdot\vec c \\ &\quad + \vec b\cdot\vec a + \vec b\cdot\vec b + \vec b\cdot\vec c \\ &\quad + \vec c\cdot\vec a + \vec c\cdot\vec b + \vec c\cdot\vec c \\ &= |\vec a|^2 + |\vec b|^2 + |\vec c|^2 \\ &\quad + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a). \end{aligned}
    Each of the three terms of the first bracket meets each of the three of the second: nine products. The three "squares" become lengths squared. The other six pair up because the order does not matter: b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b⁠, c⃗⋅b⃗=b⃗⋅c⃗\vec c\cdot\vec b = \vec b\cdot\vec c⁠, a⃗⋅c⃗=c⃗⋅a⃗\vec a\cdot\vec c = \vec c\cdot\vec a⁠. This is the vector form of (x+y+z)2=x2+y2+z2+2(xy+yz+zx)(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)⁠.
    Section 10.6.1, Property 1, Observations 3 and 7. Basics: dot products obey the same rule

  3. Put in the lengths and solve

    0=1+1+1+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗),0 = 1 + 1 + 1 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a), a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32.\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac32.

    So a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac32.

    Each unit vector has ∣a⃗∣2=∣b⃗∣2=∣c⃗∣2=1|\vec a|^2 = |\vec b|^2 = |\vec c|^2 = 1⁠. Moving 3 across gives twice the sum equal to −3-3⁠, and halving gives −32-\frac32⁠.
    Unit vector (Section 10.3). Lesson: six kinds of vectors

  4. A second method: dot with each vector

    Take the scalar product of a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0 with a⃗\vec a⁠, then with b⃗\vec b⁠, then with c⃗\vec c⁠:

    1+a⃗⋅b⃗+a⃗⋅c⃗=0,b⃗⋅a⃗+1+b⃗⋅c⃗=0,c⃗⋅a⃗+c⃗⋅b⃗+1=0.\begin{aligned} 1 + \vec a\cdot\vec b + \vec a\cdot\vec c &= 0, \\ \vec b\cdot\vec a + 1 + \vec b\cdot\vec c &= 0, \\ \vec c\cdot\vec a + \vec c\cdot\vec b + 1 &= 0. \end{aligned}

    Adding the three lines: 3+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=03 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0⁠, the same equation.

    Each line uses distributivity once and a⃗⋅a⃗=1\vec a\cdot\vec a = 1 (and so on). In the sum, every mixed product appears twice, once in each order.
    Section 10.6.1, Property 1. Lesson: Examples 13–21

  5. The picture: an equilateral triangle

    Placed head to tail, a⃗\vec a⁠, b⃗\vec b⁠, c⃗\vec c return to the start, so they form a triangle with all three sides 1: an equilateral triangle. Each pair of the vectors, placed tail to tail, makes an angle of 120∘120^\circ⁠. Indeed, from c⃗=−(a⃗+b⃗)\vec c = -(\vec a + \vec b)⁠,

    1=∣a⃗+b⃗∣2=1+1+2 a⃗⋅b⃗⇒ a⃗⋅b⃗=−12,\begin{aligned} 1 &= |\vec a + \vec b|^2 = 1 + 1 + 2\,\vec a\cdot\vec b \\ \Rightarrow\ \vec a\cdot\vec b &= -\frac12, \end{aligned}

    and in the same way b⃗⋅c⃗=c⃗⋅a⃗=−12\vec b\cdot\vec c = \vec c\cdot\vec a = -\frac12⁠. So the sum is 3×(−12)=−323 \times \left(-\frac12\right) = -\frac32⁠.

    a b c a + b + c = 0 a b 120° tail to tail a·b = cos 120° = −1/2
    Left: a⃗\vec a⁠, b⃗\vec b⁠, c⃗\vec c head to tail close up into an equilateral triangle of side 1. Right: a⃗\vec a and b⃗\vec b tail to tail make 120∘120^\circ⁠, so a⃗⋅b⃗=cos⁡120∘=−12\vec a\cdot\vec b = \cos 120^\circ = -\frac12⁠. One scale throughout.
    a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=cos⁡θ=−12\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta = \cos\theta = -\frac12 gives θ=120∘\theta = 120^\circ⁠. The angle between two vectors is measured tail to tail, which is the outside angle of the triangle at that corner: 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ⁠.
    x⃗⋅y⃗=∣x⃗∣∣y⃗∣cos⁡θ\vec x\cdot\vec y = |\vec x||\vec y|\cos\theta (Section 10.6.1); triangle law (Section 10.4). Basics: cos θ is a shadow
Result

a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac32⁠.

(Textbook answer: −32-\frac32⁠.)

Check it with numbers

A concrete triple: a⃗=12(i^−j^)\vec a = \frac{1}{\sqrt2}(\hat i - \hat j)⁠, b⃗=12(j^−k^)\vec b = \frac{1}{\sqrt2}(\hat j - \hat k)⁠, c⃗=12(k^−i^)\vec c = \frac{1}{\sqrt2}(\hat k - \hat i)⁠. Each has length 1+12=1\sqrt{\frac{1 + 1}{2}} = 1⁠, and their sum is 12(i^−j^+j^−k^+k^−i^)=0⃗\frac{1}{\sqrt2}(\hat i - \hat j + \hat j - \hat k + \hat k - \hat i) = \vec 0⁠.

a⃗⋅b⃗=12(0−1+0)=−12\vec a\cdot\vec b = \frac12(0 - 1 + 0) = -\frac12⁠, b⃗⋅c⃗=12(0+0−1)=−12\vec b\cdot\vec c = \frac12(0 + 0 - 1) = -\frac12⁠, c⃗⋅a⃗=12(−1+0+0)=−12\vec c\cdot\vec a = \frac12(-1 + 0 + 0) = -\frac12⁠. The sum is −1.5=−32-1.5 = -\frac32⁠.

In decimals, 12≈0.707107\frac{1}{\sqrt2} \approx 0.707107⁠, and the only nonzero product in a⃗⋅b⃗\vec a\cdot\vec b is (−0.707107)(0.707107)≈−0.500000(-0.707107)(0.707107) \approx -0.500000⁠.

Common mistakes
  1. Forgetting the 2. The six mixed terms pair up into 2(… )2(\dots)⁠; dropping the 2 gives −3-3⁠.
  2. Squaring term by term. ∣a⃗+b⃗+c⃗∣2|\vec a + \vec b + \vec c|^2 is not ∣a⃗∣2+∣b⃗∣2+∣c⃗∣2|\vec a|^2 + |\vec b|^2 + |\vec c|^2⁠; that would give 0=30 = 3⁠.
  3. Assuming the vectors are perpendicular. Mutually perpendicular unit vectors give a sum of length 3\sqrt3⁠, not 0⃗\vec 0⁠.
What to take away
  1. To use u⃗=0⃗\vec u = \vec 0⁠, take u⃗⋅u⃗=0\vec u\cdot\vec u = 0 (or dot u⃗\vec u with a suitable vector).
  2. ∣a⃗+b⃗+c⃗∣2|\vec a + \vec b + \vec c|^2 =∣a⃗∣2+∣b⃗∣2+∣c⃗∣2= |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)+\, 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a)⁠.
  3. Three unit vectors adding to 0⃗\vec 0 make an equilateral triangle, 120∘120^\circ apart in pairs.

Every fact used, and where it comes from

FactUsed inSource
A unit vector has magnitude 1Steps 3, 5Section 10.3 (Lesson: six kinds of vectors)
x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2Steps 1, 2, 4Section 10.6.1, Observation 3
Distributivity; x⃗⋅y⃗=y⃗⋅x⃗\vec x\cdot\vec y = \vec y\cdot\vec xSteps 2, 4Section 10.6.1, Property 1 and Observation 7 (Basics: dot products obey the same rule)
x⃗⋅y⃗=∣x⃗∣∣y⃗∣cos⁡θ\vec x\cdot\vec y = |\vec x||\vec y|\cos\thetaStep 5Section 10.6.1, Definition 2 (Basics: cos θ is a shadow)
Triangle law of additionStep 5Section 10.4 (Lesson: adding arrows, head to tail)
Scalar product in componentsCheckSection 10.6.1 (Lesson: the dot product)
The answerResultNCERT answer key, Exercise 10.3 Q13 (Lesson: the dot product exercises)
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