Chapter 10 · Vector Algebra

Exercise 10.4 · Q8: $\vec a\times\vec b=\vec 0$ with $\vec a\ne\vec 0$ ⁠, $\vec b\ne\vec 0$ ⁠: is the converse true?

Exercise 10.4Zero vector product, parallel vectors · Section 10.6.3

The question

If either $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\vec a\times\vec b = \vec 0$ ⁠. Is the converse true? Justify your answer with an example.

Exam answer

The converse is: if $\vec a\times\vec b = \vec 0$ ⁠, then $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠. It is not true.

Take $\vec a = 2\hat i + 3\hat j + 4\hat k$ and $\vec b = 4\hat i + 6\hat j + 8\hat k$ ⁠. Both are nonzero.

\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{vmatrix} \\ &= \hat i(24 - 24) - \hat j(16 - 16) \\ &\quad + \hat k(12 - 12) \\ &= \vec 0 \end{aligned}

$\therefore\ \vec a\times\vec b = \vec 0$ with $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠, so the converse is not true.

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

The one idea this problem is built on

For nonzero vectors, $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ ⁠. This is 0 not only when a vector is $\vec 0$ but also whenever $\sin\theta = 0$ ⁠, that is, when $\vec a$ and $\vec b$ are parallel ( $\theta = 0$ or $\pi$ ⁠). Any two nonzero parallel vectors break the converse.

Given, to find, and the plan

Given: The statement: if $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\vec a\times\vec b = \vec 0$ ⁠.
To find: Whether the converse holds, with an example.
Plan: (1) See why the given statement is true. (2) Write the converse. (3) Find where it can fail: parallel vectors. (4) Give an explicit example and compute. (5) State exactly when $\vec a\times\vec b = \vec 0$ ⁠.

Basics you need

1The definition of $\vec a\times\vec b$

For nonzero vectors, $\vec a\times\vec b = |\vec a||\vec b|\sin\theta\,\hat n$ ⁠; if either vector is $\vec 0$ ⁠, $\vec a\times\vec b$ is defined to be $\vec 0$ (Definition 3, Section 10.6.3). Lesson: the cross product

2Converse of a statement

The converse of "if P, then Q" is "if Q, then P". A single example where Q holds and P fails shows the converse is false.

3Parallel vectors

$\vec b = k\vec a$ for a nonzero scalar $k$ means $\vec b$ is parallel to $\vec a$ ⁠; then the angle is 0 (for $k > 0$ ⁠) or $\pi$ (for $k < 0$ ⁠) (Section 10.5). Lesson: components

Solution

Why the given statement is true

If $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, the angle $\theta$ is not defined, and the book defines $\vec a\times\vec b = \vec 0$ in this case.

The given statement is part of Definition 3, so there is nothing more to prove. The question is about the opposite direction.

Definition 3, Section 10.6.3. Lesson: the cross product
The converse

"If $\vec a\times\vec b = \vec 0$ ⁠, then $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠."

The converse swaps the "if" part and the "then" part. To show it is false we need two vectors with $\vec a\times\vec b = \vec 0$ that are both nonzero.

Converse of a statement; disproving by a counterexample.
Where it fails: parallel vectors

For nonzero $\vec a$ and $\vec b$ ⁠, $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ with $|\vec a||\vec b| > 0$ ⁠. So $\vec a\times\vec b = \vec 0$ exactly when $\sin\theta = 0$ ⁠, that is, $\theta = 0$ or $\theta = \pi$ ⁠.

On $0 \le \theta \le \pi$ ⁠, $\sin\theta = 0$ only at the two ends, where the vectors point the same way or opposite ways: they are parallel (collinear). So any two nonzero parallel vectors give a zero vector product.

$\vec a\times\vec b = \vec 0$ for nonzero vectors iff they are parallel (Section 10.6.3, Observation 2). Lesson: Examples 22–25
An explicit example

Take $\vec a = 2\hat i + 3\hat j + 4\hat k$ and $\vec b = 4\hat i + 6\hat j + 8\hat k = 2\vec a$ ⁠. Neither is $\vec 0$ ⁠: $|\vec a| = \sqrt{29}$ and $|\vec b| = 2\sqrt{29}$ ⁠. Then
$\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{vmatrix} \\ &= \hat i(24 - 24) - \hat j(16 - 16) \\ &\quad + \hat k(12 - 12) \\ &= \vec 0. \end{aligned}$

$\vec a$ and $\vec b = 2\vec a$ drawn from O along one line with one scale. The angle between them is 0, so $\sin\theta = 0$ ⁠: the parallelogram on $\vec a$ and $\vec b$ is flat and has area $|\vec a\times\vec b| = 0$ ⁠.

The minors are $(3)(8) - (4)(6) = 24 - 24$ ⁠, $(2)(8) - (4)(4) = 16 - 16$ and $(2)(6) - (3)(4) = 12 - 12$ ⁠, all 0. Without the determinant: $\vec a\times(2\vec a) = 2(\vec a\times\vec a) = 2\vec 0 = \vec 0$ ⁠.

The determinant form; $\vec a\times\vec a = \vec 0$ (Section 10.6.3, Observation 2, Property 3). Basics: a determinant measures a signed area
The answer, and the full rule

No, the converse is not true. The complete statement is:
$\vec a\times\vec b = \vec 0 \iff \vec a = \vec 0,\ \vec b = \vec 0 \text{ or } \vec a \parallel \vec b.$
The example above has $\vec a\times\vec b = \vec 0$ while $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠, so the converse fails. Opposite directions work too: $\hat i\times(-3\hat i) = -3(\hat i\times\hat i) = \vec 0$ ⁠.

Section 10.6.3, Definition 3 and Observation 2. Lesson: the cross product, exercises

Result

No. For example, $\vec a = 2\hat i + 3\hat j + 4\hat k$ and $\vec b = 4\hat i + 6\hat j + 8\hat k$ are both nonzero, but $\vec a\times\vec b = \vec 0$ ⁠. Any two nonzero collinear vectors do the same.

(Textbook answer: No; take any two nonzero collinear vectors.)

Check it with numbers

$|\vec a| = \sqrt{29} \approx 5.385165$ and $|\vec b| = 2\sqrt{29} \approx 10.770330$ ⁠, both nonzero.

$\vec a\cdot\vec b = 8 + 18 + 32 = 58$ and $|\vec a||\vec b| = 2 \times 29 = 58$ ⁠, so $\cos\theta = 1$ ⁠, $\theta = 0$ and $\sin\theta = 0$ ⁠, in agreement with $\vec a\times\vec b = \vec 0$ ⁠.

Common mistakes

Answering "yes" because the statement looks symmetric. A statement and its converse are different claims; each needs its own proof or counterexample.
Giving an example that is not checked. The justification needs the computation $\vec a\times\vec b = \vec 0$ and the remark that both vectors are nonzero.
Choosing perpendicular vectors. For $\hat i$ and $\hat j$ ⁠, $\hat i\times\hat j = \hat k \ne \vec 0$ ⁠; perpendicular vectors make the scalar product 0, not the vector product.

What to take away

$\vec a\times\vec b = \vec 0$ means a zero vector or parallel vectors.
To disprove a general statement, one explicit counterexample is enough.
Parallel vectors span a flat parallelogram, so their vector product has length 0.

Every fact used, and where it comes from

Fact	Used in	Source
Definition of $\vec a\times\vec b$ ⁠, including the zero-vector case	Steps 1, 3	Section 10.6.3, Definition 3 (Lesson: the cross product)
$\vec a\times\vec b = \vec 0$ for nonzero vectors iff they are parallel; $\vec a\times\vec a = \vec 0$	Steps 3, 4, 5	Section 10.6.3, Observation 2
$\vec a\times\vec b$ as a determinant in components	Step 4	Section 10.6.3 (Basics: a determinant measures a signed area)
Scalar multiple of a vector is parallel to it	Steps 4, 5	Section 10.5 (Lesson: components)
The question and the answer	Result	Exercise 10.4 Q8; NCERT answer key (Lesson: the cross product, exercises)

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Exercise 10.4 · Q8: a⃗×b⃗=0⃗\vec a\times\vec b=\vec 0a×b=0 with a⃗≠0⃗\vec a\ne\vec 0a=0⁠, b⃗≠0⃗\vec b\ne\vec 0b=0⁠: is the converse true?

Basics you need

Solution

Why the given statement is true

The converse

Where it fails: parallel vectors

An explicit example

The answer, and the full rule

Every fact used, and where it comes from

Exercise 10.4 · Q8: $\vec a\times\vec b=\vec 0$ with $\vec a\ne\vec 0$ ⁠, $\vec b\ne\vec 0$ ⁠: is the converse true?