Chapter 10 · Vector Algebra

Exercise 10.4 · Q4: $(\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)$

Exercise 10.4Properties of the vector product · Section 10.6.3

The question

Show that

(\vec a - \vec b)\times(\vec a + \vec b) = 2(\vec a\times\vec b)

Exam answer

Using the distributive law of the vector product,

\begin{aligned} &(\vec a - \vec b)\times(\vec a + \vec b) \\ &= \vec a\times\vec a + \vec a\times\vec b \\ &\quad - \vec b\times\vec a - \vec b\times\vec b \end{aligned}

Now $\vec a\times\vec a = \vec 0$ ⁠, $\vec b\times\vec b = \vec 0$ and $\vec b\times\vec a = -\vec a\times\vec b$ ⁠. So

\begin{aligned} &(\vec a - \vec b)\times(\vec a + \vec b) \\ &= \vec 0 + \vec a\times\vec b + \vec a\times\vec b - \vec 0 \\ &= 2(\vec a\times\vec b) \end{aligned}

$\therefore\ (\vec a - \vec b)\times(\vec a + \vec b) = 2(\vec a\times\vec b)$ ⁠. Hence proved.

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

The one idea this problem is built on

The vector product can be expanded over brackets like ordinary multiplication, with two differences: order matters, $\vec b\times\vec a = -\vec a\times\vec b$ ⁠, and a vector times itself is $\vec 0$ ⁠. So the "square" terms vanish and the two cross terms, instead of cancelling, add up.

Given, to show, and the plan

Given: Any two vectors $\vec a$ and $\vec b$ ⁠.
To show: $(\vec a - \vec b)\times(\vec a + \vec b) = 2(\vec a\times\vec b)$ ⁠.
Plan: (1) Expand the product with the distributive law, keeping the order of every factor. (2) Use $\vec a\times\vec a = \vec b\times\vec b = \vec 0$ ⁠. (3) Use $\vec b\times\vec a = -\vec a\times\vec b$ and collect. (4) Check with numbers and see what it means for areas.

Basics you need

1Distributive law

$\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c$ and $\lambda(\vec a\times\vec b) = (\lambda\vec a)\times\vec b = \vec a\times(\lambda\vec b)$ (Section 10.6.3, Property 3). Lesson: the cross product

2A vector crossed with itself

$\vec a\times\vec a = \vec 0$ ⁠: the angle is 0 and $\sin 0 = 0$ (Section 10.6.3, Observation 2). Basics: cos θ is a shadow, sin θ is a height

3Swapping the order changes the sign

$\vec b\times\vec a = -\vec a\times\vec b$ (Section 10.6.3, Observation 6). Lesson: Examples 22–25

4Expanding brackets in algebra

$(x - y)(x + y) = x^2 + xy - yx - y^2$ ⁠, term by term, before any simplifying. Basics: expanding brackets

Solution

Expand, keeping the order
$\begin{aligned} &(\vec a - \vec b)\times(\vec a + \vec b) \\ &= (\vec a - \vec b)\times\vec a + (\vec a - \vec b)\times\vec b \\ &= \vec a\times\vec a - \vec b\times\vec a + \vec a\times\vec b - \vec b\times\vec b. \end{aligned}$
The first line uses the distributive law on the right-hand bracket. For the left-hand bracket we need $(\vec u + \vec v)\times\vec w = \vec u\times\vec w + \vec v\times\vec w$ ⁠, which follows from the stated law and the sign rule: $(\vec u + \vec v)\times\vec w = -\vec w\times(\vec u + \vec v)$ $= -\vec w\times\vec u - \vec w\times\vec v$ $= \vec u\times\vec w + \vec v\times\vec w$ ⁠. Also $(-\vec b)\times\vec a = -(\vec b\times\vec a)$ by the scalar rule with $\lambda = -1$ ⁠. In every term the factor from the first bracket stays on the left.

Distributivity and the scalar rule (Section 10.6.3, Property 3). Lesson: the cross product
The "square" terms are zero
$\vec a\times\vec a = \vec 0, \qquad \vec b\times\vec b = \vec 0.$
A vector is parallel to itself, so the angle between them is 0 and $|\vec a\times\vec a| = |\vec a||\vec a|\sin 0 = 0$ ⁠. (If $\vec a = \vec 0$ ⁠, the product is $\vec 0$ by definition.) This is the big difference from numbers, where $x \cdot x = x^2$ is not 0.

$\vec a\times\vec a = \vec 0$ (Section 10.6.3, Observation 2). Lesson: the cross product
The cross terms add up

Since $\vec b\times\vec a = -\vec a\times\vec b$ ⁠, the term $-\vec b\times\vec a$ equals $+\vec a\times\vec b$ ⁠. So
$\begin{aligned} &(\vec a - \vec b)\times(\vec a + \vec b) \\ &= \vec 0 + \vec a\times\vec b + \vec a\times\vec b - \vec 0 \\ &= 2(\vec a\times\vec b). \end{aligned}$
With numbers, $-yx + xy$ would cancel to 0, because $yx = xy$ ⁠. The vector product is not commutative: reversing the order turns the vector round, so $-\vec b\times\vec a$ points the same way as $\vec a\times\vec b$ and the two terms double instead of cancelling.

$\vec b\times\vec a = -\vec a\times\vec b$ (Section 10.6.3, Observation 6). Lesson: Examples 22–25
Check with numbers, and what it means

Take $\vec a = \hat i + 2\hat j + 3\hat k$ ⁠, $\vec b = 2\hat i - \hat j + \hat k$ ⁠. Then $\vec a - \vec b = -\hat i + 3\hat j + 2\hat k$ and $\vec a + \vec b = 3\hat i + \hat j + 4\hat k$ ⁠, and
$\begin{aligned} &(\vec a - \vec b)\times(\vec a + \vec b) \\ &= \hat i(12 - 2) - \hat j(-4 - 6) \\ &\quad + \hat k(-1 - 9) \\ &= 10\hat i + 10\hat j - 10\hat k, \end{aligned}$ $\begin{aligned} \vec a\times\vec b &= \hat i(2 + 3) - \hat j(1 - 6) \\ &\quad + \hat k(-1 - 4) \\ &= 5\hat i + 5\hat j - 5\hat k. \end{aligned}$
So $(\vec a - \vec b)\times(\vec a + \vec b) = 2(\vec a\times\vec b)$ here, as proved.

The vectors $\vec a + \vec b$ and $\vec a - \vec b$ are the diagonals of the parallelogram on $\vec a$ and $\vec b$ ⁠. Taking lengths, the identity says the parallelogram built on the two diagonals has twice the area of the original parallelogram.

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

Result

$(\vec a - \vec b)\times(\vec a + \vec b) = 2(\vec a\times\vec b)$ for all vectors $\vec a$ and $\vec b$ ⁠. Hence proved.

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

Check it with numbers

With the sample of Step 4: $10\hat i + 10\hat j - 10\hat k = 2(5\hat i + 5\hat j - 5\hat k)$ ⁠, component by component.

With $\vec a = \hat i$ ⁠, $\vec b = \hat j$ ⁠: $(\hat i - \hat j)\times(\hat i + \hat j) = \hat i\times\hat j - \hat j\times\hat i = \hat k + \hat k = 2\hat k = 2(\hat i\times\hat j)$ ⁠.

Common mistakes

Treating the product as commutative. Writing $\vec b\times\vec a = \vec a\times\vec b$ makes the cross terms cancel and gives $\vec 0$ ⁠, which is wrong.
Writing $\vec a\times\vec a = |\vec a|^2$ ⁠. That is the scalar product $\vec a\cdot\vec a$ ⁠; the vector product of a vector with itself is $\vec 0$ ⁠.
Using the "difference of squares" pattern. $(\vec a - \vec b)\times(\vec a + \vec b)$ is not $\vec a\times\vec a - \vec b\times\vec b$ ⁠; that pattern needs commutativity.

What to take away

Expand vector products like brackets, but keep every factor in its order.
$\vec a\times\vec a = \vec 0$ and $\vec b\times\vec a = -\vec a\times\vec b$ do the simplifying.
$(\vec a - \vec b)\times(\vec a + \vec b) = 2(\vec a\times\vec b)$ ⁠: the parallelogram on the diagonals has twice the area.

Every fact used, and where it comes from

Fact	Used in	Source
Distributive law and the scalar rule for the vector product	Step 1	Section 10.6.3, Property 3 (Lesson: the cross product)
$\vec a\times\vec a = \vec 0$	Step 2	Section 10.6.3, Observation 2
$\vec b\times\vec a = -\vec a\times\vec b$	Steps 1, 3	Section 10.6.3, Observation 6
$\hat i\times\hat j = \hat k$ ⁠, $\hat j\times\hat i = -\hat k$ ⁠; the determinant form	Step 4	Section 10.6.3, Observations 4 and 7 (Basics: a determinant measures a signed area)
Diagonals of a parallelogram are $\vec a + \vec b$ and $\vec a - \vec b$ ⁠; area $= \|\vec a\times\vec b\|$	Step 4	Section 10.4 (parallelogram law); Section 10.6.3, Observation 9
The question	Result	Exercise 10.4 Q4 (no answer-key entry) (Lesson: the cross product, exercises)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 10.4 · Q4: (a⃗−b⃗)×(a⃗+b⃗)=2(a⃗×b⃗)(\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)(a−b)×(a+b)=2(a×b)

Basics you need

Solution

Expand, keeping the order

The "square" terms are zero

The cross terms add up

Check with numbers, and what it means

Every fact used, and where it comes from

Exercise 10.4 · Q4: $(\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)$