Chapter 10 · Vector Algebra

Exercise 10.4 · Q7: Distributive law $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$

Exercise 10.4Vector product in components, distributive law · Section 10.6.3

The question

Let the vectors $\vec a, \vec b, \vec c$ be given as $a_1\hat i + a_2\hat j + a_3\hat k$ ⁠, $b_1\hat i + b_2\hat j + b_3\hat k$ ⁠, $c_1\hat i + c_2\hat j + c_3\hat k$ ⁠. Then show that $\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$ ⁠.

Exam answer

\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \\ &= (a_2b_3 - a_3b_2)\hat i \\ &\quad - (a_1b_3 - a_3b_1)\hat j \\ &\quad + (a_1b_2 - a_2b_1)\hat k \quad\ldots(\text{i}) \end{aligned}

Similarly,

\begin{aligned} \vec a\times\vec c &= (a_2c_3 - a_3c_2)\hat i \\ &\quad - (a_1c_3 - a_3c_1)\hat j \\ &\quad + (a_1c_2 - a_2c_1)\hat k \quad\ldots(\text{ii}) \end{aligned}

Now $\vec b + \vec c = (b_1 + c_1)\hat i$ $+ (b_2 + c_2)\hat j + (b_3 + c_3)\hat k$ ⁠. So

\begin{aligned} &\vec a\times(\vec b + \vec c) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 + c_1 & b_2 + c_2 & b_3 + c_3 \end{vmatrix} \\ &= [a_2(b_3 + c_3) - a_3(b_2 + c_2)]\hat i \\ &\quad - [a_1(b_3 + c_3) - a_3(b_1 + c_1)]\hat j \\ &\quad + [a_1(b_2 + c_2) - a_2(b_1 + c_1)]\hat k \end{aligned}

Opening the brackets and grouping the $b$ terms and the $c$ terms,

\begin{aligned} &\vec a\times(\vec b + \vec c) \\ &= (a_2b_3 - a_3b_2)\hat i \\ &\quad - (a_1b_3 - a_3b_1)\hat j \\ &\quad + (a_1b_2 - a_2b_1)\hat k \\ &\quad + (a_2c_3 - a_3c_2)\hat i \\ &\quad - (a_1c_3 - a_3c_1)\hat j \\ &\quad + (a_1c_2 - a_2c_1)\hat k \\ &= \vec a\times\vec b + \vec a\times\vec c \quad [\text{by (i), (ii)}] \end{aligned}

$\therefore\ \vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$ ⁠. Hence proved.

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

The one idea this problem is built on

Every component of $\vec a\times\vec b$ is a difference of two products, each with one factor from $\vec b$ ⁠. Replacing $\vec b$ by $\vec b + \vec c$ and multiplying out splits each component into a " $b$ part" plus a " $c$ part". Collecting the $b$ parts gives $\vec a\times\vec b$ ⁠; collecting the $c$ parts gives $\vec a\times\vec c$ ⁠.

Given, to show, and the plan

Given: $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ ⁠, $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$ ⁠, $\vec c = c_1\hat i + c_2\hat j + c_3\hat k$ ⁠.
To show: $\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$ ⁠.
Plan: (1) Write $\vec a\times\vec b$ and $\vec a\times\vec c$ in components. (2) Write $\vec b + \vec c$ and $\vec a\times(\vec b + \vec c)$ in components. (3) Multiply out and regroup. (4) Compare, and check with numbers.

Basics you need

1Vector product in components

$\vec a\times\vec b = (a_2b_3 - a_3b_2)\hat i$ $- (a_1b_3 - a_3b_1)\hat j$ $+ (a_1b_2 - a_2b_1)\hat k$ ⁠, the expansion of the determinant with $\hat i, \hat j, \hat k$ on top (Section 10.6.3). Basics: a determinant measures a signed area

2Adding vectors in components

$\vec b + \vec c = (b_1 + c_1)\hat i + (b_2 + c_2)\hat j$ $+ (b_3 + c_3)\hat k$ (Section 10.5.1). Lesson: components

3Ordinary algebra

$p(q + r) = pq + pr$ ⁠, and terms can be added in any order. Basics: expanding brackets

Solution

$\vec a\times\vec b$ and $\vec a\times\vec c$ in components
$\begin{aligned} \vec a\times\vec b &= (a_2b_3 - a_3b_2)\hat i - (a_1b_3 - a_3b_1)\hat j \\ &\quad + (a_1b_2 - a_2b_1)\hat k, \\ \vec a\times\vec c &= (a_2c_3 - a_3c_2)\hat i - (a_1c_3 - a_3c_1)\hat j \\ &\quad + (a_1c_2 - a_2c_1)\hat k. \end{aligned}$
These are the determinant formula with $\vec b$ ⁠, then $\vec c$ ⁠, in the bottom row. The second line is the first with every $b$ replaced by $c$ ⁠.

The determinant form of the vector product (Section 10.6.3). Lesson: the cross product
$\vec a\times(\vec b + \vec c)$ in components

The components of $\vec b + \vec c$ are $b_1 + c_1$ ⁠, $b_2 + c_2$ ⁠, $b_3 + c_3$ ⁠, so they form the bottom row:
$\begin{aligned} &\vec a\times(\vec b + \vec c) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 + c_1 & b_2 + c_2 & b_3 + c_3 \end{vmatrix} \\ &= [a_2(b_3 + c_3) - a_3(b_2 + c_2)]\hat i \\ &\quad - [a_1(b_3 + c_3) - a_3(b_1 + c_1)]\hat j \\ &\quad + [a_1(b_2 + c_2) - a_2(b_1 + c_1)]\hat k. \end{aligned}$
The same formula as in Step 1, with $b_1 + c_1$ in place of $b_1$ ⁠, and so on. Brackets keep each sum together.

Adding vectors in components (Section 10.5.1). Lesson: components
Multiply out and regroup

Take the $\hat i$ component:
$\begin{aligned} &a_2(b_3 + c_3) - a_3(b_2 + c_2) \\ &= a_2b_3 + a_2c_3 - a_3b_2 - a_3c_2 \\ &= (a_2b_3 - a_3b_2) + (a_2c_3 - a_3c_2). \end{aligned}$
In the same way, the $\hat j$ bracket is $(a_1b_3 - a_3b_1) + (a_1c_3 - a_3c_1)$ and the $\hat k$ bracket is $(a_1b_2 - a_2b_1) + (a_1c_2 - a_2c_1)$ ⁠.

Each bracket is opened by $p(q + r) = pq + pr$ ⁠, and then the four terms are put in a different order: the two with a $b$ first, the two with a $c$ second. The minus sign in front of the $\hat j$ bracket applies to both groups, so it stays in front of each.

Distributive law for numbers; rearranging a sum. Basics: expanding brackets
Compare with Step 1
$\begin{aligned} &\vec a\times(\vec b + \vec c) \\ &= [(a_2b_3 - a_3b_2)\hat i - (a_1b_3 - a_3b_1)\hat j \\ &\qquad + (a_1b_2 - a_2b_1)\hat k] \\ &\quad + [(a_2c_3 - a_3c_2)\hat i - (a_1c_3 - a_3c_1)\hat j \\ &\qquad + (a_1c_2 - a_2c_1)\hat k] \\ &= \vec a\times\vec b + \vec a\times\vec c. \end{aligned}$
The first square bracket is exactly $\vec a\times\vec b$ from Step 1 and the second is exactly $\vec a\times\vec c$ ⁠. Since the components agree one by one, the vectors are equal. This is Property 3 (i) of Section 10.6.3, proved here from the component formula. (The book itself obtains the determinant formula, just before Example 22, by expanding the brackets with this law; so within the book the exercise shows that the formula and the law agree, and with the determinant formula taken as the starting point it is a complete proof.)

Vectors are equal iff their components are equal (Section 10.5.1). Lesson: the cross product, exercises
Check with numbers

Take $\vec a = \hat i + 2\hat j + \hat k$ ⁠, $\vec b = 2\hat i - \hat j + \hat k$ ⁠, $\vec c = \hat j + 2\hat k$ ⁠. Then $\vec b + \vec c = 2\hat i + 3\hat k$ ⁠, and
$\begin{aligned} \vec a\times(\vec b + \vec c) &= \hat i(6 - 0) - \hat j(3 - 2) \\ &\quad + \hat k(0 - 4) \\ &= 6\hat i - \hat j - 4\hat k, \\ \vec a\times\vec b &= \hat i(2 + 1) - \hat j(1 - 2) \\ &\quad + \hat k(-1 - 4) \\ &= 3\hat i + \hat j - 5\hat k, \\ \vec a\times\vec c &= \hat i(4 - 1) - \hat j(2 - 0) \\ &\quad + \hat k(1 - 0) \\ &= 3\hat i - 2\hat j + \hat k. \end{aligned}$
And $(3\hat i + \hat j - 5\hat k) + (3\hat i - 2\hat j + \hat k) = 6\hat i - \hat j - 4\hat k$ ⁠.

A numerical example cannot prove a general law, but it is a quick test that the algebra above has no slip; the general proof is Steps 1 to 4.

The determinant form of the vector product (Section 10.6.3). Basics: a determinant measures a signed area

Result

$\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$ for all vectors $\vec a, \vec b, \vec c$ ⁠. Hence proved.

(Textbook answer: the answer key has no entry for this "show that" question.)

Check it with numbers

With the sample of Step 5, component by component: $\hat i$ ⁠: $6 = 3 + 3$ ⁠; $\hat j$ ⁠: $-1 = 1 + (-2)$ ⁠; $\hat k$ ⁠: $-4 = -5 + 1$ ⁠.

A second sample, $\vec a = \hat k$ ⁠, $\vec b = \hat i$ ⁠, $\vec c = \hat j$ ⁠: $\hat k\times(\hat i + \hat j) = \hat j - \hat i$ ⁠, and $\hat k\times\hat i + \hat k\times\hat j = \hat j + (-\hat i)$ ⁠, the same vector.

Common mistakes

Proving it with one numerical example. A sample only checks; the proof must use general components $a_1, \ldots, c_3$ ⁠.
Dropping the minus sign in front of the $\hat j$ bracket. It belongs to both $(a_1b_3 - a_3b_1)$ and $(a_1c_3 - a_3c_1)$ ⁠.
Using the law to prove itself. Writing " $\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$ by distributivity" assumes what is to be shown; the components do the work.

What to take away

Each component of $\vec a\times\vec b$ is linear in the components of $\vec b$ ⁠, so sums in $\vec b$ split.
To prove a vector identity in components, compare the $\hat i$ ⁠, $\hat j$ ⁠, $\hat k$ parts one at a time.
Check a general identity on a numerical sample before trusting the algebra.

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\times\vec b$ as a determinant in components	Steps 1, 2, 5	Section 10.6.3 (Basics: a determinant measures a signed area)
Sum of vectors in components; equal vectors have equal components	Steps 2, 4	Section 10.5.1 (Lesson: components)
$p(q + r) = pq + pr$ for numbers	Step 3	Algebra (Basics: expanding brackets)
Distributivity of the vector product (the statement proved)	Step 4	Section 10.6.3, Property 3 (i) (Lesson: the cross product)
$\hat k\times\hat i = \hat j$ ⁠, $\hat k\times\hat j = -\hat i$	Check	Section 10.6.3, Observations 4 and 7
The question	Result	Exercise 10.4 Q7 (no answer-key entry) (Lesson: the cross product, exercises)

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Exercise 10.4 · Q7: Distributive law a⃗×(b⃗+c⃗)=a⃗×b⃗+a⃗×c⃗\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec ca×(b+c)=a×b+a×c

Basics you need

Solution

a⃗×b⃗\vec a\times\vec ba×b and a⃗×c⃗\vec a\times\vec ca×c in components

a⃗×(b⃗+c⃗)\vec a\times(\vec b + \vec c)a×(b+c) in components

Multiply out and regroup

Compare with Step 1

Check with numbers

Every fact used, and where it comes from

Exercise 10.4 · Q7: Distributive law $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$

$\vec a\times\vec b$ and $\vec a\times\vec c$ in components

$\vec a\times(\vec b + \vec c)$ in components