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Chapter 10 · Vector Algebra

Exercise 10.3 · Q7: Expand (3a⃗−5b⃗)⋅(2a⃗+7b⃗)(3\vec a-5\vec b)\cdot(2\vec a+7\vec b)

Exercise 10.3Expanding scalar products · Section 10.6.1
The question

Evaluate the product (3a⃗−5b⃗)⋅(2a⃗+7b⃗)(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b)⁠.

Exam answer

Using the distributive law of the scalar product,

(3a⃗−5b⃗)⋅(2a⃗+7b⃗)=3a⃗⋅2a⃗+3a⃗⋅7b⃗−5b⃗⋅2a⃗−5b⃗⋅7b⃗=6 a⃗⋅a⃗+21 a⃗⋅b⃗−10 b⃗⋅a⃗−35 b⃗⋅b⃗\begin{aligned} &(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b) \\ &= 3\vec a\cdot 2\vec a + 3\vec a\cdot 7\vec b \\ &\quad - 5\vec b\cdot 2\vec a - 5\vec b\cdot 7\vec b \\ &= 6\,\vec a\cdot\vec a + 21\,\vec a\cdot\vec b \\ &\quad - 10\,\vec b\cdot\vec a - 35\,\vec b\cdot\vec b \end{aligned}

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

=6∣a⃗∣2+21 a⃗⋅b⃗−10 a⃗⋅b⃗−35∣b⃗∣2\begin{aligned} &= 6|\vec a|^2 + 21\,\vec a\cdot\vec b \\ &\quad - 10\,\vec a\cdot\vec b - 35|\vec b|^2 \end{aligned}

∴\therefore the product is 6∣a⃗∣2+11 a⃗⋅b⃗−35∣b⃗∣26|\vec a|^2 + 11\,\vec a\cdot\vec b - 35|\vec b|^2⁠.

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

The one idea this problem is built on

The scalar product obeys the same rules as multiplying numbers: it distributes over addition, scalars come out, and the order does not matter. So (3a⃗−5b⃗)⋅(2a⃗+7b⃗)(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b) expands exactly like (3a−5b)(2a+7b)=6a2+11ab−35b2(3a - 5b)(2a + 7b) = 6a^2 + 11ab - 35b^2⁠, with one change: a vector "squared" is a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2⁠.

Given, to find, and the plan
Given
Two vectors a⃗\vec a and b⃗\vec b (any vectors).
To find
(3a⃗−5b⃗)⋅(2a⃗+7b⃗)(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b) in terms of ∣a⃗∣|\vec a|⁠, ∣b⃗∣|\vec b| and a⃗⋅b⃗\vec a\cdot\vec b⁠.
Plan
(1) Expand into four scalar products. (2) Bring the scalars out. (3) Use b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b and r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2⁠. (4) Collect like terms, and check with a concrete pair.

Basics you need

1Distributivity

a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b + \vec c) = \vec a\cdot\vec b + \vec a\cdot\vec c (Section 10.6.1, Property 1). Basics: dot products obey the same rule

2Scalars come out

(λa⃗)⋅b⃗=λ(a⃗⋅b⃗)=a⃗⋅(λb⃗)(\lambda\vec a)\cdot\vec b = \lambda(\vec a\cdot\vec b) = \vec a\cdot(\lambda\vec b) (Section 10.6.1, Property 2). Lesson: the dot product

3Order and squares

b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b (Observation 7) and a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 (Observation 3), Section 10.6.1. Lesson: Examples 13–21

Solution

  1. Expand into four products

    (3a⃗−5b⃗)⋅(2a⃗+7b⃗)=(3a⃗)⋅(2a⃗)+(3a⃗)⋅(7b⃗)+(−5b⃗)⋅(2a⃗)+(−5b⃗)⋅(7b⃗).\begin{aligned} &(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b) \\ &= (3\vec a)\cdot(2\vec a) + (3\vec a)\cdot(7\vec b) \\ &\quad + (-5\vec b)\cdot(2\vec a) + (-5\vec b)\cdot(7\vec b). \end{aligned}
    By distributivity, each of the two terms of the first bracket meets each of the two terms of the second, giving 2×2=42 \times 2 = 4 scalar products, exactly as in multiplying out (3a−5b)(2a+7b)(3a - 5b)(2a + 7b)⁠. The minus sign belongs to −5b⃗-5\vec b⁠, so it travels with both products that contain it.
    Section 10.6.1, Property 1. Basics: dot products obey the same rule

  2. Bring the scalars out

    =6 a⃗⋅a⃗+21 a⃗⋅b⃗−10 b⃗⋅a⃗−35 b⃗⋅b⃗.\begin{aligned} &= 6\,\vec a\cdot\vec a + 21\,\vec a\cdot\vec b \\ &\quad - 10\,\vec b\cdot\vec a - 35\,\vec b\cdot\vec b. \end{aligned}
    Numbers in front of the vectors multiply together and come out of the scalar product: 3×2=63 \times 2 = 6⁠, 3×7=213 \times 7 = 21⁠, (−5)×2=−10(-5) \times 2 = -10 and (−5)×7=−35(-5) \times 7 = -35⁠.
    Section 10.6.1, Property 2. Lesson: the dot product

  3. Use the order rule and the squares

    =6∣a⃗∣2+21 a⃗⋅b⃗−10 a⃗⋅b⃗−35∣b⃗∣2.\begin{aligned} &= 6|\vec a|^2 + 21\,\vec a\cdot\vec b \\ &\quad - 10\,\vec a\cdot\vec b - 35|\vec b|^2. \end{aligned}
    The scalar product does not depend on the order, so b⃗⋅a⃗\vec b\cdot\vec a is the same number as a⃗⋅b⃗\vec a\cdot\vec b⁠, and the two middle terms are like terms. A vector dotted with itself makes angle 0 with itself, and cos⁡0=1\cos 0 = 1⁠, so a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 and b⃗⋅b⃗=∣b⃗∣2\vec b\cdot\vec b = |\vec b|^2⁠.
    Section 10.6.1, Observations 7 and 3. Lesson: Examples 13–21

  4. Collect like terms

    (3a⃗−5b⃗)⋅(2a⃗+7b⃗)=6∣a⃗∣2+11 a⃗⋅b⃗−35∣b⃗∣2.\begin{aligned} &(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b) \\ &= 6|\vec a|^2 + 11\,\vec a\cdot\vec b - 35|\vec b|^2. \end{aligned}

    So the product is 6∣a⃗∣2+11 a⃗⋅b⃗−35∣b⃗∣26|\vec a|^2 + 11\,\vec a\cdot\vec b - 35|\vec b|^2.

    21−10=1121 - 10 = 11⁠, so the middle terms combine to 11 a⃗⋅b⃗11\,\vec a\cdot\vec b⁠. The answer is a number (a scalar), as every scalar product is; it cannot be simplified further without knowing ∣a⃗∣|\vec a|⁠, ∣b⃗∣|\vec b| and a⃗⋅b⃗\vec a\cdot\vec b⁠.
    a⃗⋅b⃗\vec a\cdot\vec b is a real number (Section 10.6.1, Observation 1). Lesson: the dot product
Result

(3a⃗−5b⃗)⋅(2a⃗+7b⃗)(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b) =6∣a⃗∣2+11 a⃗⋅b⃗−35∣b⃗∣2= 6|\vec a|^2 + 11\,\vec a\cdot\vec b - 35|\vec b|^2⁠.

(Textbook answer: 6∣a⃗∣2+11 a⃗⋅b⃗−35∣b⃗∣26|\vec a|^2 + 11\,\vec a\cdot\vec b - 35|\vec b|^2⁠.)

Check it with numbers

Take a⃗=2i^−j^+k^\vec a = 2\hat i - \hat j + \hat k and b⃗=i^+3j^−2k^\vec b = \hat i + 3\hat j - 2\hat k⁠. Then ∣a⃗∣2=4+1+1=6|\vec a|^2 = 4 + 1 + 1 = 6⁠, ∣b⃗∣2=1+9+4=14|\vec b|^2 = 1 + 9 + 4 = 14 and a⃗⋅b⃗=2−3−2=−3\vec a\cdot\vec b = 2 - 3 - 2 = -3⁠, so the formula gives

6(6)+11(−3)−35(14)=36−33−490=−487.\begin{aligned} &6(6) + 11(-3) - 35(14) \\ &= 36 - 33 - 490 = -487. \end{aligned}

Directly: 3a⃗−5b⃗=i^−18j^+13k^3\vec a - 5\vec b = \hat i - 18\hat j + 13\hat k and 2a⃗+7b⃗=11i^+19j^−12k^2\vec a + 7\vec b = 11\hat i + 19\hat j - 12\hat k⁠, and their scalar product is

(1)(11)+(−18)(19)+(13)(−12)=11−342−156=−487.\begin{aligned} &(1)(11) + (-18)(19) + (13)(-12) \\ &= 11 - 342 - 156 = -487. \end{aligned}

The two agree.

Common mistakes
  1. Dropping the cross terms. 6∣a⃗∣2−35∣b⃗∣26|\vec a|^2 - 35|\vec b|^2 misses 21 a⃗⋅b⃗−10 a⃗⋅b⃗21\,\vec a\cdot\vec b - 10\,\vec a\cdot\vec b⁠; the cross terms do not cancel here (they did in (a⃗+b⃗)⋅(a⃗−b⃗)(\vec a + \vec b)\cdot(\vec a - \vec b) only because the coefficients matched).
  2. Losing the sign of −5b⃗-5\vec b⁠. Taking +10 b⃗⋅a⃗+10\,\vec b\cdot\vec a gives 31 a⃗⋅b⃗31\,\vec a\cdot\vec b instead of 11 a⃗⋅b⃗11\,\vec a\cdot\vec b⁠.
  3. Writing a⃗⋅a⃗=a⃗\vec a\cdot\vec a = \vec a⁠. a⃗⋅a⃗\vec a\cdot\vec a is the number ∣a⃗∣2|\vec a|^2⁠, not a vector.
What to take away
  1. Scalar products of sums expand like ordinary products: distribute, then bring the scalars out.
  2. b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b makes the cross terms like terms; a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 replaces the squares.
  3. The result is a number in terms of ∣a⃗∣|\vec a|⁠, ∣b⃗∣|\vec b| and a⃗⋅b⃗\vec a\cdot\vec b⁠.

Every fact used, and where it comes from

FactUsed inSource
Distributivity a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b + \vec c) = \vec a\cdot\vec b + \vec a\cdot\vec cStep 1Section 10.6.1, Property 1 (Basics: dot products obey the same rule)
(λa⃗)⋅b⃗=λ(a⃗⋅b⃗)=a⃗⋅(λb⃗)(\lambda\vec a)\cdot\vec b = \lambda(\vec a\cdot\vec b) = \vec a\cdot(\lambda\vec b)Step 2Section 10.6.1, Property 2
b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b⁠; a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2Step 3Section 10.6.1, Observations 7 and 3
a⃗⋅b⃗\vec a\cdot\vec b is a real numberStep 4Section 10.6.1, Observation 1
Scalar product in componentsCheckSection 10.6.1 (Lesson: the dot product)
The answerResultNCERT answer key, Exercise 10.3 Q7 (Lesson: the dot product)
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