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Chapter 10 · Vector Algebra

Example 29: a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a when a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0⁠, ∣a⃗∣=3|\vec a|=3⁠, ∣b⃗∣=4|\vec b|=4⁠, ∣c⃗∣=2|\vec c|=2

ExamplesScalar products from a vector sum · Miscellaneous Examples
The question

Three vectors a⃗\vec a⁠, b⃗\vec b and c⃗\vec c satisfy the condition a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠. Evaluate the quantity μ=a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\mu = \vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a⁠, if ∣a⃗∣=3|\vec a| = 3⁠, ∣b⃗∣=4|\vec b| = 4 and ∣c⃗∣=2|\vec c| = 2⁠.

Exam answer

Since a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠,

(a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗)=0(\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) = 0

Expanding, and using 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 and r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2⁠,

∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0\begin{aligned} &|\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 \end{aligned} ⇒9+16+4+2μ=0⇒2μ=−29\begin{aligned} &\Rightarrow 9 + 16 + 4 + 2\mu = 0 \\ &\Rightarrow 2\mu = -29 \end{aligned}

∴ μ=−292\therefore\ \mu = -\dfrac{29}{2}⁠.

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

The one idea this problem is built on

The scalar product of the zero vector with anything is 0. So we can dot the condition a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0 with a⃗\vec a⁠, with b⃗\vec b⁠, with c⃗\vec c (or with a⃗+b⃗+c⃗\vec a + \vec b + \vec c itself) and get equations in which the known lengths and the unknown products a⃗⋅b⃗\vec a\cdot\vec b⁠, b⃗⋅c⃗\vec b\cdot\vec c⁠, c⃗⋅a⃗\vec c\cdot\vec a appear together.

Given, to find, and the plan
Given
a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠, ∣a⃗∣=3|\vec a| = 3⁠, ∣b⃗∣=4|\vec b| = 4⁠, ∣c⃗∣=2|\vec c| = 2⁠.
To find
μ=a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\mu = \vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a⁠.
Plan
(1)–(3) Dot the condition with a⃗\vec a⁠, b⃗\vec b and c⃗\vec c in turn, as the book does. (4) Add the three equations. (5) Get the same answer in one line by squaring ∣a⃗+b⃗+c⃗∣|\vec a + \vec b + \vec c|⁠, and see the triangle behind it.

Basics you need

1Scalar product with the zero vector

r⃗⋅0⃗=0\vec r\cdot\vec 0 = 0 for every r⃗\vec r (Section 10.6.1, Definition 2). Lesson: the dot product

2Rules of the scalar product

a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b + \vec c) = \vec a\cdot\vec b + \vec a\cdot\vec c⁠, a⃗⋅b⃗=b⃗⋅a⃗\vec a\cdot\vec b = \vec b\cdot\vec a and a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 (Section 10.6.1, Property 1, Observations 7 and 3). Basics: dot products obey the same rule

3Head to tail

a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0 means that placing the vectors head to tail brings you back to the start: they form a triangle (Section 10.4). Lesson: adding arrows, head to tail

Solution

  1. Dot the condition with a⃗\vec a

    a⃗⋅(a⃗+b⃗+c⃗)=a⃗⋅0⃗=0⇒a⃗⋅a⃗+a⃗⋅b⃗+a⃗⋅c⃗=0⇒a⃗⋅b⃗+a⃗⋅c⃗=−∣a⃗∣2⇒a⃗⋅b⃗+a⃗⋅c⃗=−9…(1)\begin{aligned} &\vec a\cdot(\vec a + \vec b + \vec c) = \vec a\cdot\vec 0 = 0 \\ &\Rightarrow \vec a\cdot\vec a + \vec a\cdot\vec b + \vec a\cdot\vec c = 0 \\ &\Rightarrow \vec a\cdot\vec b + \vec a\cdot\vec c = -|\vec a|^2 \\ &\Rightarrow \vec a\cdot\vec b + \vec a\cdot\vec c = -9 \quad\ldots(1) \end{aligned}
    Distribute a⃗\vec a over the sum and move a⃗⋅a⃗=∣a⃗∣2=9\vec a\cdot\vec a = |\vec a|^2 = 9 to the right-hand side. The book's first displayed line here reads "a⃗+b⃗+c⃗=0⃗=0\vec a + \vec b + \vec c = \vec 0 = 0"; its next line, a⃗⋅a⃗+a⃗⋅b⃗+a⃗⋅c⃗=0\vec a\cdot\vec a + \vec a\cdot\vec b + \vec a\cdot\vec c = 0⁠, shows the line meant is a⃗⋅(a⃗+b⃗+c⃗)=0\vec a\cdot(\vec a + \vec b + \vec c) = 0⁠, as written above.
    a⃗⋅0⃗=0\vec a\cdot\vec 0 = 0⁠; distributivity; a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 (Section 10.6.1). Lesson: the dot product

  2. Dot the condition with b⃗\vec b

    b⃗⋅(a⃗+b⃗+c⃗)=0⇒a⃗⋅b⃗+b⃗⋅b⃗+b⃗⋅c⃗=0⇒a⃗⋅b⃗+b⃗⋅c⃗=−∣b⃗∣2⇒a⃗⋅b⃗+b⃗⋅c⃗=−16…(2)\begin{aligned} &\vec b\cdot(\vec a + \vec b + \vec c) = 0 \\ &\Rightarrow \vec a\cdot\vec b + \vec b\cdot\vec b + \vec b\cdot\vec c = 0 \\ &\Rightarrow \vec a\cdot\vec b + \vec b\cdot\vec c = -|\vec b|^2 \\ &\Rightarrow \vec a\cdot\vec b + \vec b\cdot\vec c = -16 \quad\ldots(2) \end{aligned}
    The same move with b⃗\vec b⁠; commutativity lets us write b⃗⋅a⃗\vec b\cdot\vec a as a⃗⋅b⃗\vec a\cdot\vec b⁠, so the products match the ones in μ\mu⁠.
    b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b (Section 10.6.1, Observation 7). Basics: dot products obey the same rule

  3. Dot the condition with c⃗\vec c

    a⃗⋅c⃗+b⃗⋅c⃗=−∣c⃗∣2=−4.…(3)\vec a\cdot\vec c + \vec b\cdot\vec c = -|\vec c|^2 = -4. \quad\ldots(3)
    Similarly, c⃗⋅(a⃗+b⃗+c⃗)=0\vec c\cdot(\vec a + \vec b + \vec c) = 0 gives c⃗⋅a⃗+c⃗⋅b⃗+∣c⃗∣2=0\vec c\cdot\vec a + \vec c\cdot\vec b + |\vec c|^2 = 0⁠.
    Same rules as Steps 1 and 2. Lesson: the dot product

  4. Add the three equations

    2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=−9−16−4=−29,2μ=−29,μ=−292.\begin{aligned} &2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) \\ &= -9 - 16 - 4 = -29, \\ &2\mu = -29, \quad \mu = -\frac{29}{2}. \end{aligned}
    Each of the three products appears in exactly two of the equations: a⃗⋅b⃗\vec a\cdot\vec b in (1) and (2), b⃗⋅c⃗\vec b\cdot\vec c in (2) and (3), c⃗⋅a⃗\vec c\cdot\vec a in (1) and (3). So adding them gives twice μ\mu⁠.
    Adding equations side by side. Lesson: Examples 26–30

  5. The same answer in one line, and the triangle behind it

    0=∣a⃗+b⃗+c⃗∣2=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2μ=9+16+4+2μ,\begin{aligned} &0 = |\vec a + \vec b + \vec c|^2 \\ &= |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2\mu \\ &= 9 + 16 + 4 + 2\mu, \end{aligned}

    so again 2μ=−292\mu = -29⁠. Geometrically, a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0 says the three vectors, placed head to tail, close up into a triangle with sides 3, 4 and 2. Such a triangle exists because 2+3>42 + 3 > 4⁠.

    a, |a| = 3 b, |b| = 4 c, |c| = 2 a + b + c = 0 head to tail, it closes μ = −29/2
    With one scale: a⃗\vec a (length 3), then b⃗\vec b (length 4) from its tip, then c⃗\vec c (length 2) from the tip of b⃗\vec b⁠, which ends where a⃗\vec a began. The values of a⃗⋅b⃗\vec a\cdot\vec b⁠, b⃗⋅c⃗\vec b\cdot\vec c⁠, c⃗⋅a⃗\vec c\cdot\vec a are fixed by this triangle.
    Squaring the sum produces all three mixed products at once, each twice; this is the expansion (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) carried over to scalar products. It is the quickest route and a check on Steps 1 to 4.
    ∣r⃗∣2=r⃗⋅r⃗|\vec r|^2 = \vec r\cdot\vec r⁠; triangle law (Section 10.4). Basics: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2
Result

μ=a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−292\mu = \vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac{29}{2}⁠.

(Textbook answer: a⃗⋅b⃗+a⃗⋅c⃗=−9\vec a\cdot\vec b + \vec a\cdot\vec c = -9 ... (1), a⃗⋅b⃗+b⃗⋅c⃗=−16\vec a\cdot\vec b + \vec b\cdot\vec c = -16 ... (2), a⃗⋅c⃗+b⃗⋅c⃗=−4\vec a\cdot\vec c + \vec b\cdot\vec c = -4 ... (3); adding, 2μ=−292\mu = -29⁠, i.e., μ=−292\mu = \frac{-29}{2}⁠. The book's first displayed line is printed as "a⃗+b⃗+c⃗=0⃗=0\vec a + \vec b + \vec c = \vec 0 = 0"; the line that leads to its next step is a⃗⋅(a⃗+b⃗+c⃗)=0\vec a\cdot(\vec a + \vec b + \vec c) = 0⁠.)

Check it with numbers

μ=−292=−14.5\mu = -\frac{29}{2} = -14.5⁠.

Solving (1), (2), (3) one product at a time: (1)+(2)−(3)(1) + (2) - (3) gives 2 a⃗⋅b⃗=−212\,\vec a\cdot\vec b = -21⁠, so a⃗⋅b⃗=−212\vec a\cdot\vec b = -\frac{21}{2}⁠; likewise b⃗⋅c⃗=−112\vec b\cdot\vec c = -\frac{11}{2} and c⃗⋅a⃗=32\vec c\cdot\vec a = \frac32⁠. Their sum is −21−11+32=−292\frac{-21 - 11 + 3}{2} = -\frac{29}{2}⁠.

A sample (the triangle in the figure): a⃗=3i^\vec a = 3\hat i⁠, b⃗=−72i^+152j^\vec b = -\frac72\hat i + \frac{\sqrt{15}}{2}\hat j⁠, c⃗=12i^−152j^\vec c = \frac12\hat i - \frac{\sqrt{15}}{2}\hat j⁠. Then a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠, ∣b⃗∣2=49+154=16|\vec b|^2 = \frac{49 + 15}{4} = 16⁠, ∣c⃗∣2=1+154=4|\vec c|^2 = \frac{1 + 15}{4} = 4⁠, and a⃗⋅b⃗=−212\vec a\cdot\vec b = -\frac{21}{2}⁠, b⃗⋅c⃗=−74−154=−112\vec b\cdot\vec c = -\frac74 - \frac{15}{4} = -\frac{11}{2}⁠, c⃗⋅a⃗=32\vec c\cdot\vec a = \frac32⁠.

Common mistakes
  1. Forgetting the 2. ∣a⃗+b⃗+c⃗∣2=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2μ|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2\mu⁠; without the 2 you get μ=−29\mu = -29⁠.
  2. A sign slip when moving ∣a⃗∣2|\vec a|^2⁠. a⃗⋅a⃗+a⃗⋅b⃗+a⃗⋅c⃗=0\vec a\cdot\vec a + \vec a\cdot\vec b + \vec a\cdot\vec c = 0 gives a⃗⋅b⃗+a⃗⋅c⃗=−9\vec a\cdot\vec b + \vec a\cdot\vec c = -9⁠, not +9+9⁠.
  3. Thinking the vectors are perpendicular. a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0 says nothing about right angles; here every product is nonzero.
What to take away
  1. From a vector equation, take scalar products with suitable vectors to get number equations.
  2. If a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠, then a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a =−12(∣a⃗∣2+∣b⃗∣2+∣c⃗∣2)= -\frac12(|\vec a|^2 + |\vec b|^2 + |\vec c|^2)⁠.
  3. Vectors adding to 0⃗\vec 0 close up head to tail into a triangle (or lie along a line).

Every fact used, and where it comes from

FactUsed inSource
r⃗⋅0⃗=0\vec r\cdot\vec 0 = 0Steps 1–3Section 10.6.1, Definition 2 (Lesson: the dot product)
Distributivity; commutativity; a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2Steps 1–3, 5Section 10.6.1, Property 1, Observations 7 and 3
(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)Step 5Algebra (Basics: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2)
Triangle law of additionStep 5Section 10.4 (Lesson: adding arrows)
The book's working and answerResultExample 29, Miscellaneous Examples (Lesson: Examples 26–30)
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