Chapter 10 · Vector Algebra

Exercise 10.3 · Q14: $\vec a\cdot\vec b=0$ with $\vec a\ne\vec 0$ ⁠, $\vec b\ne\vec 0$ ⁠: an example

Exercise 10.3Perpendicular vectors, scalar product in components · Section 10.6.1

The question

If either vector $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\vec a\cdot\vec b = 0$ ⁠. But the converse need not be true. Justify your answer with an example.

Exam answer

The converse says: if $\vec a\cdot\vec b = 0$ ⁠, then $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠. It is false, for example:

Let $\vec a = \hat i + 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 3\hat j + \hat k$ ⁠.

\begin{aligned} \vec a\cdot\vec b &= 1(3) + 2(-3) + 3(1) \\ &= 3 - 6 + 3 = 0 \end{aligned}

But $|\vec a| = \sqrt{14} \ne 0$ and $|\vec b| = \sqrt{19} \ne 0$ ⁠, so $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠.

Hence $\vec a\cdot\vec b = 0$ does not imply $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠: the converse need not be true.

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

The one idea this problem is built on

For nonzero vectors, $\vec a\cdot\vec b = 0$ means the vectors are perpendicular. So a zero scalar product can come from two perfectly good nonzero vectors at right angles. One such pair is enough to show the converse fails.

Given, to show, and the plan

Given: The statement: if $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠, then $\vec a\cdot\vec b = 0$ ⁠.
To show: The converse, "if $\vec a\cdot\vec b = 0$ then $\vec a = \vec 0$ or $\vec b = \vec 0$ ", is not always true, by an example.
Plan: (1) See why the statement itself is true. (2) Write the converse and what a counterexample needs. (3) Build two nonzero perpendicular vectors. (4) Verify. (5) Note the simplest example.

Basics you need

1Scalar product with the zero vector

The book defines $\vec a\cdot\vec b = 0$ when $\vec a = \vec 0$ or $\vec b = \vec 0$ (Section 10.6.1). Lesson: the dot product

2Converse and counterexample

The converse of "if P then Q" is "if Q then P". One example where Q holds and P fails shows the converse is not always true (school logic).

3Scalar product in components

$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ ⁠, and for nonzero vectors it is 0 exactly when they are perpendicular (Section 10.6.1, Observation 2). Basics: dot products obey the same rule

Solution

Why the statement itself is true

If $\vec a = \vec 0$ ⁠, then $\vec a\cdot\vec b = 0b_1 + 0b_2 + 0b_3 = 0$ ⁠; the same if $\vec b = \vec 0$ ⁠.

With a zero vector the angle $\theta$ is not defined, so the book defines the scalar product to be 0 in that case. The component formula gives the same value.

Definition of the scalar product (Section 10.6.1). Lesson: the dot product
The converse, and what disproves it

Converse: "if $\vec a\cdot\vec b = 0$ ⁠, then $\vec a = \vec 0$ or $\vec b = \vec 0$ ". To show it need not be true, we need one pair with
$\vec a\cdot\vec b = 0, \qquad \vec a \ne \vec 0, \qquad \vec b \ne \vec 0.$
A general claim fails as soon as one case breaks it. Since nonzero vectors with zero scalar product are exactly perpendicular pairs, any two nonzero perpendicular vectors will do.

Section 10.6.1, Observation 2. Lesson: the dot product
Build a perpendicular pair

Choose $\vec a = \hat i + 2\hat j + 3\hat k$ ⁠. A vector $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$ is perpendicular to it when $b_1 + 2b_2 + 3b_3 = 0$ ⁠. Taking $b_1 = 3$ ⁠, $b_3 = 1$ forces $3 + 2b_2 + 3 = 0$ ⁠, so $b_2 = -3$ ⁠:
$\vec b = 3\hat i - 3\hat j + \hat k.$
The condition is one equation in three unknowns, so we may pick two of them freely and solve for the third. Any choice other than $b_1 = b_2 = b_3 = 0$ gives a valid example.

$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ (Section 10.6.1). Basics: dot products obey the same rule
Verify the example
$\begin{aligned} \vec a\cdot\vec b &= 1(3) + 2(-3) + 3(1) \\ &= 3 - 6 + 3 = 0, \end{aligned}$ $\begin{aligned} |\vec a| &= \sqrt{1 + 4 + 9} = \sqrt{14} \ne 0, \\ |\vec b| &= \sqrt{9 + 9 + 1} = \sqrt{19} \ne 0. \end{aligned}$
So $\vec a\cdot\vec b = 0$ although neither vector is $\vec 0$ ⁠: the converse is not true.

The two vectors drawn in their own plane, true to scale (one scale for both directions): $\vec a$ of length $\sqrt{14}$ and $\vec b$ of length $\sqrt{19}$ ⁠, both nonzero, meeting at a right angle, so $\vec a\cdot\vec b = 0$ ⁠.

A vector is $\vec 0$ only when its length is 0. Both lengths are positive, so both vectors are nonzero, and still their scalar product is 0 because they are at $90^\circ$ ⁠: $\cos 90^\circ = 0$ ⁠.

$|\vec a| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ (Section 10.5.1). Basics: use Pythagoras twice
The simplest example

$\hat i\cdot\hat j = 0$ ⁠, yet $\hat i$ and $\hat j$ are unit vectors, far from zero. Any two of $\hat i, \hat j, \hat k$ give a counterexample at once.

The book lists $\hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0$ for these mutually perpendicular unit vectors. The answer key's advice, "take any two non-zero perpendicular vectors", covers both examples.

Section 10.6.1, Observation 5. See also Exercise 10.3 Q12.

Result

For $\vec a = \hat i + 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 3\hat j + \hat k$ ⁠, $\vec a\cdot\vec b = 0$ but $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠. So $\vec a\cdot\vec b = 0$ does not force $\vec a = \vec 0$ or $\vec b = \vec 0$ ⁠: the converse need not be true. Any two nonzero perpendicular vectors give such an example.

(Textbook answer: take any two non-zero perpendicular vectors $\vec a$ and $\vec b$ ⁠.)

Check it with numbers

$|\vec a| = \sqrt{14} \approx 3.741657$ and $|\vec b| = \sqrt{19} \approx 4.358899$ ⁠, both well away from 0. $\cos\theta = \frac{0}{\sqrt{14}\,\sqrt{19}} = 0$ ⁠, so $\theta = 90^\circ$ ⁠.

For contrast, a pair that is not perpendicular: $\vec a\cdot(\hat i + \hat j + \hat k) = 1 + 2 + 3 = 6 \ne 0$ ⁠.

Common mistakes

Using a zero vector in the example. $\vec a = \vec 0$ gives $\vec a\cdot\vec b = 0$ ⁠, but it agrees with the converse instead of breaking it.
Choosing parallel vectors. For parallel nonzero vectors $\vec a\cdot\vec b = \pm|\vec a||\vec b| \ne 0$ ⁠.
Not checking both vectors are nonzero. The example must state $\vec a \ne \vec 0$ and $\vec b \ne \vec 0$ ⁠, for instance through their lengths.

What to take away

$\vec a\cdot\vec b = 0$ means: $\vec a = \vec 0$ ⁠, or $\vec b = \vec 0$ ⁠, or $\vec a \perp \vec b$ ⁠.
Unlike numbers, two nonzero vectors can have product zero.
One explicit counterexample is a complete answer to "need not be true".

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\cdot\vec b = 0$ when $\vec a = \vec 0$ or $\vec b = \vec 0$ (definition)	Step 1	Section 10.6.1 (Lesson: the dot product)
Nonzero vectors: scalar product 0 exactly when perpendicular	Steps 2, 4	Section 10.6.1, Observation 2
$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$	Steps 1, 3, 4	Section 10.6.1 (Basics: dot products obey the same rule)
$\|\vec a\| = \sqrt{a_1^2 + a_2^2 + a_3^2}$	Step 4	Section 10.5.1 (Basics: use Pythagoras twice)
$\hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0$	Step 5	Section 10.6.1, Observation 5
The answer	Result	NCERT answer key, Exercise 10.3 Q14 (Lesson: the dot product exercises)

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Exercise 10.3 · Q14: a⃗⋅b⃗=0\vec a\cdot\vec b=0a⋅b=0 with a⃗≠0⃗\vec a\ne\vec 0a=0⁠, b⃗≠0⃗\vec b\ne\vec 0b=0⁠: an example

Basics you need

Solution

Why the statement itself is true

The converse, and what disproves it

Build a perpendicular pair

Verify the example

The simplest example

Every fact used, and where it comes from

Exercise 10.3 · Q14: $\vec a\cdot\vec b=0$ with $\vec a\ne\vec 0$ ⁠, $\vec b\ne\vec 0$ ⁠: an example