Chapter 10 · Vector Algebra

Exercise 10.3 · Q12: $\vec a\cdot\vec a=0$ and $\vec a\cdot\vec b=0$ ⁠: what about $\vec b$ ⁠?

Exercise 10.3Scalar product of a vector with itself, the zero vector · Section 10.6.1

The question

If $\vec a\cdot\vec a = 0$ and $\vec a\cdot\vec b = 0$ ⁠, then what can be concluded about the vector $\vec b$ ⁠?

Exam answer

$\vec a\cdot\vec a = 0 \Rightarrow |\vec a|^2 = 0 \Rightarrow |\vec a| = 0$ ⁠.

So $\vec a$ is the zero vector: $\vec a = \vec 0$ ⁠.

For $\vec a = \vec 0$ ⁠, $\vec a\cdot\vec b = 0$ for every vector $\vec b$ ⁠, so the second condition puts no restriction on $\vec b$ ⁠.

$\therefore\ \vec b$ can be any vector.

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

The one idea this problem is built on

A vector dotted with itself gives the square of its length: $\vec a\cdot\vec a = |\vec a|^2$ ⁠. So $\vec a\cdot\vec a = 0$ forces $\vec a = \vec 0$ ⁠. And the zero vector has scalar product 0 with every vector, so $\vec a\cdot\vec b = 0$ then tells us nothing about $\vec b$ ⁠.

Given, to find, and the plan

Given: $\vec a\cdot\vec a = 0$ and $\vec a\cdot\vec b = 0$ ⁠.
To find: What these conditions say about $\vec b$ ⁠.
Plan: (1) Turn $\vec a\cdot\vec a$ into $|\vec a|^2$ ⁠. (2) Deduce $\vec a = \vec 0$ ⁠. (3) See that $\vec 0\cdot\vec b = 0$ for every $\vec b$ ⁠. (4) Test with examples.

Basics you need

1A vector dotted with itself

$\vec a\cdot\vec a = |\vec a|^2$ (Section 10.6.1, Observation 3); in components $a_1^2 + a_2^2 + a_3^2$ ⁠. Basics: use Pythagoras twice

2The zero vector

The only vector of magnitude 0 is $\vec 0$ (Section 10.3), and the book defines $\vec a\cdot\vec b = 0$ whenever $\vec a = \vec 0$ or $\vec b = \vec 0$ (Section 10.6.1). Lesson: six kinds of vectors

3What a zero scalar product means

For nonzero vectors only, $\vec a\cdot\vec b = 0$ means $\vec a \perp \vec b$ (Section 10.6.1, Observation 2). Lesson: the dot product

Solution

Read $\vec a\cdot\vec a$ as a length
$\vec a\cdot\vec a = |\vec a|^2 = 0 \;\Rightarrow\; |\vec a| = 0.$
If $\vec a \ne \vec 0$ ⁠, the angle between $\vec a$ and itself is 0 and $\cos 0 = 1$ ⁠, so $\vec a\cdot\vec a = |\vec a||\vec a|\cos 0 = |\vec a|^2$ ⁠. If $\vec a = \vec 0$ ⁠, both sides are 0 (the book defines the scalar product with a zero vector as 0). So $\vec a\cdot\vec a = |\vec a|^2$ for every vector, and a square is 0 only when the number itself is 0.

Section 10.6.1, Observation 3. Basics: use Pythagoras twice
So $\vec a$ is the zero vector

In components, $|\vec a|^2 = a_1^2 + a_2^2 + a_3^2 = 0$ ⁠. A sum of squares of real numbers is 0 only if each one is 0, so $a_1 = a_2 = a_3 = 0$ and
$\vec a = 0\hat i + 0\hat j + 0\hat k = \vec 0.$
Each square is at least 0, so if one of them were positive the sum would be positive. A vector whose length is 0 has its initial and terminal points together: it is the zero vector.

$|\vec a| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ (Section 10.5.1); zero vector (Section 10.3). Lesson: six kinds of vectors
The second condition holds for every $\vec b$

For any $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$ ⁠,
$\vec 0\cdot\vec b = 0\cdot b_1 + 0\cdot b_2 + 0\cdot b_3 = 0.$
So once $\vec a = \vec 0$ ⁠, the condition $\vec a\cdot\vec b = 0$ is true whatever $\vec b$ is. Nothing can be concluded about $\vec b$ ⁠: $\vec b$ can be any vector.

The book defines the scalar product with a zero vector to be 0 (the angle is not defined there), and the component formula agrees. A condition that every vector satisfies gives no information.

Definition of $\vec a\cdot\vec b$ when $\vec a = \vec 0$ (Section 10.6.1). Lesson: the dot product
Test with examples

With $\vec a = \vec 0$ ⁠: $\vec b = \vec 0$ ⁠, $\vec b = \hat i$ and $\vec b = 2\hat i - 3\hat j + 5\hat k$ all give $\vec a\cdot\vec b = 0$ ⁠. They point in different directions and have different lengths, so no property of $\vec b$ is forced.

If instead $\vec a$ were nonzero, $\vec a\cdot\vec b = 0$ would force $\vec b$ to be $\vec 0$ or perpendicular to $\vec a$ ⁠. The first condition rules that case out: here $\vec a$ has no direction for $\vec b$ to be perpendicular to.

Section 10.6.1, Observation 2 (nonzero vectors only). See also Exercise 10.3 Q14.

Result

$\vec a\cdot\vec a = 0$ gives $\vec a = \vec 0$ ⁠, and then $\vec a\cdot\vec b = 0$ for every $\vec b$ ⁠. So nothing can be concluded about $\vec b$ ⁠: it can be any vector.

(Textbook answer: vector $\vec b$ can be any vector.)

Check it with numbers

$\vec a = \vec 0$ ⁠, $\vec b = 2\hat i - 3\hat j + 5\hat k$ ⁠: $\vec a\cdot\vec a = 0$ and $\vec a\cdot\vec b = 0(2) + 0(-3) + 0(5) = 0$ ⁠. Both conditions hold.

Even a very short nonzero vector fails the first condition: $\vec a = 0.01\hat i$ gives $\vec a\cdot\vec a = 0.01^2 = 0.0001$ ⁠, which is not 0. Only $\vec 0$ passes.

Common mistakes

Concluding $\vec b = \vec 0$ ⁠. $\vec b = \hat i$ satisfies both conditions with $\vec a = \vec 0$ ⁠.
Concluding $\vec b \perp \vec a$ ⁠. "Zero scalar product means perpendicular" is only for nonzero vectors; here $\vec a = \vec 0$ ⁠.
Thinking $\vec a\cdot\vec a = 0$ allows a nonzero $\vec a$ ⁠. A nonzero vector cannot be perpendicular to itself: $\vec a\cdot\vec a = |\vec a|^2 > 0$ ⁠.

What to take away

$\vec a\cdot\vec a = |\vec a|^2$ ⁠, so $\vec a\cdot\vec a = 0$ exactly when $\vec a = \vec 0$ ⁠.
$\vec 0\cdot\vec b = 0$ for every $\vec b$ ⁠.
Check that the vectors are nonzero before reading $\vec a\cdot\vec b = 0$ as "perpendicular".

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\cdot\vec a = \|\vec a\|^2$	Step 1	Section 10.6.1, Observation 3 (Basics: use Pythagoras twice)
$\|\vec a\| = \sqrt{a_1^2 + a_2^2 + a_3^2}$	Step 2	Section 10.5.1
Zero vector: initial and terminal points coincide	Step 2	Section 10.3 (Lesson: six kinds of vectors)
$\vec a\cdot\vec b = 0$ when $\vec a = \vec 0$ (definition); component formula	Step 3, Check	Section 10.6.1 (Lesson: the dot product)
Nonzero vectors: scalar product 0 exactly when perpendicular	Step 4	Section 10.6.1, Observation 2
The answer	Result	NCERT answer key, Exercise 10.3 Q12 (Lesson: the dot product exercises)

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Exercise 10.3 · Q12: a⃗⋅a⃗=0\vec a\cdot\vec a=0a⋅a=0 and a⃗⋅b⃗=0\vec a\cdot\vec b=0a⋅b=0⁠: what about b⃗\vec bb⁠?

Basics you need

Solution

Read a⃗⋅a⃗\vec a\cdot\vec aa⋅a as a length

So a⃗\vec aa is the zero vector

The second condition holds for every b⃗\vec bb

Test with examples

Every fact used, and where it comes from

Exercise 10.3 · Q12: $\vec a\cdot\vec a=0$ and $\vec a\cdot\vec b=0$ ⁠: what about $\vec b$ ⁠?

Read $\vec a\cdot\vec a$ as a length

So $\vec a$ is the zero vector

The second condition holds for every $\vec b$