Chapter 10 · Vector Algebra

Exercise 10.2 · Q2: Two different vectors with the same magnitude

Exercise 10.2Components of a vector, equality of vectors · Section 10.5.1

The question

Write two different vectors having same magnitude.

Exam answer

Let $\vec a = 3\hat i + 4\hat j$ and $\vec b = 4\hat i - 3\hat j$ ⁠.

|\vec a| = \sqrt{3^2 + 4^2} = 5

|\vec b| = \sqrt{4^2 + (-3)^2} = 5

So $|\vec a| = |\vec b|$ ⁠. But the $\hat i$ components of $\vec a$ and $\vec b$ are 3 and 4, which are not equal, so $\vec a \ne \vec b$ ⁠.

$\therefore\ 3\hat i + 4\hat j$ and $4\hat i - 3\hat j$ are two different vectors having the same magnitude 5.

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

The one idea this problem is built on

The magnitude of a vector is only its length; two vectors are equal only when they have the same length and the same direction, which in component form means every component matches. So we need two vectors with the same value of $\sqrt{x^2 + y^2 + z^2}$ but with different components. Any two arrows from O to points of one circle (or sphere) about O will do, so there are infinitely many answers.

Given, to find, and the plan

Given: Nothing but the conditions.
To find: Two vectors $\vec a$ and $\vec b$ with $|\vec a| = |\vec b|$ and $\vec a \ne \vec b$ ⁠.
Plan: (1) Pick a vector with a simple length, $3\hat i + 4\hat j$ (length 5). (2) Rearrange and change signs of its components to get another vector of length 5. (3) Check the magnitudes. (4) Show the vectors are different by comparing components.

Basics you need

1Magnitude in component form

$|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

2Equal vectors

Two vectors are equal if they have the same magnitude and direction (Section 10.3); in component form, $\vec a = \vec b$ if and only if $a_1 = b_1$ ⁠, $a_2 = b_2$ and $a_3 = b_3$ (Section 10.5.1, (iii)). Lesson: six kinds of vectors

3A right triangle with whole-number sides

$3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ⁠, so a vector with components 3 and 4 has length exactly 5. Lesson: components

Solution

Choose the first vector

Take $\vec a = 3\hat i + 4\hat j$ (that is, $3\hat i + 4\hat j + 0\hat k$ ⁠). Its magnitude is
$|\vec a| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.$
Any nonzero vector would do. Components 3 and 4 make the square root come out as a whole number, which keeps the comparison clean. A vector in the xy-plane is still a vector in space, with $\hat k$ component 0.

The magnitude formula, from Pythagoras (Section 10.5.1). Basics: use Pythagoras twice
Build a second vector of the same length

The magnitude uses only the squares of the components. Swapping the components and changing a sign keeps the same squares, $16$ and $9$ ⁠. So take $\vec b = 4\hat i - 3\hat j$ ⁠:
$|\vec b| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5.$
The sum of squares $x^2 + y^2 + z^2$ does not change when the components are put in a different order or when any of them changes sign. That gives a quick supply of vectors with a chosen length.

$(-3)^2 = 9$ ⁠: squaring removes the sign. Lesson: components
Show the two vectors are different

Compare the $\hat i$ components: $\vec a$ has 3 and $\vec b$ has 4. Since $3 \ne 4$ ⁠, the vectors are not equal:
$\vec a \ne \vec b, \qquad |\vec a| = |\vec b| = 5.$

Both vectors drawn from O on a unit grid (equal scales). Their tips lie on the dashed circle of radius 5, so they have the same length, but they point in different directions, so they are different vectors.

Equality needs every component to match. One mismatched component is enough to make the vectors different. The picture says the same thing: same length, different directions.

$\vec a = \vec b$ if and only if $a_1 = b_1$ ⁠, $a_2 = b_2$ ⁠, $a_3 = b_3$ (Section 10.5.1, (iii)); compare Example 5, where $\hat i + 2\hat j$ and $2\hat i + \hat j$ have equal magnitudes but are not equal. Lesson: Examples 4–10
Why there are infinitely many answers

Every vector $x\hat i + y\hat j + z\hat k$ with $x^2 + y^2 + z^2 = 25$ has magnitude 5: its tip lies on the sphere of radius 5 about O. Any two different points of that sphere give a valid answer, for example $5\hat k$ and $3\hat i - 4\hat k$ ⁠. The length need not be 5 either: $\hat i + 2\hat j + 2\hat k$ and $2\hat i + \hat j - 2\hat k$ both have magnitude $\sqrt{1 + 4 + 4} = 3$ ⁠.

A sphere has infinitely many points, and the radius can be any positive number, so the question has infinitely many correct answers. Any one valid pair, with the magnitudes worked out and the difference shown, earns full credit.

The points at distance $r$ from O form a sphere of radius $r$ (distance formula, Class XI). Basics: three numbers fix a point in space

Result

For example, $\vec a = 3\hat i + 4\hat j$ and $\vec b = 4\hat i - 3\hat j$ ⁠: $|\vec a| = |\vec b| = 5$ ⁠, but $\vec a \ne \vec b$ ⁠. Any two different vectors whose components have the same sum of squares are also correct.

(Textbook answer: an infinite number of possible answers.)

Check it with numbers

$|\vec a|^2 = 9 + 16 = 25$ and $|\vec b|^2 = 16 + 9 = 25$ ⁠, so both magnitudes are $\sqrt{25} = 5$ ⁠. Directions: $\vec a$ makes $\tan^{-1}\frac43 \approx 53.13^\circ$ with the positive x-axis, and $\vec b$ makes $\tan^{-1}\frac{3}{4} \approx 36.87^\circ$ below it; the directions differ by $53.13^\circ + 36.87^\circ = 90^\circ$ ⁠. The other pair: $1 + 4 + 4 = 9$ and $4 + 1 + 4 = 9$ ⁠, both magnitudes 3.

Common mistakes

Giving a vector and itself. The question asks for two different vectors; show that some component differs.
Thinking the same magnitude makes vectors equal. Equal vectors need the same direction as well.
Not showing the magnitudes. Work out both magnitudes; the claim "same magnitude" must be checked, not just stated.

What to take away

Same magnitude means the same $\sqrt{x^2 + y^2 + z^2}$ ⁠; equal vectors need equal components.
Reordering components or changing their signs keeps the magnitude.
All vectors of magnitude $r$ end on the sphere of radius $r$ about O.

Every fact used, and where it comes from

Fact	Used in	Source
$\|x\hat i + y\hat j + z\hat k\| = \sqrt{x^2 + y^2 + z^2}$	Steps 1, 2, 4	Section 10.5.1 (Basics: Pythagoras twice)
Equal vectors: same magnitude and direction	Step 3	Section 10.3 (Lesson: six kinds of vectors)
$\vec a = \vec b$ if and only if the corresponding components are equal	Step 3	Section 10.5.1, (iii)
Equal magnitudes without equal vectors	Step 3	Example 5, Section 10.5.1 (Lesson: Examples 4–10)
Points at distance $r$ from O form a sphere	Step 4	Distance formula (Class XI)

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Exercise 10.2 · Q2: Two different vectors with the same magnitude

Basics you need

Solution

Choose the first vector

Build a second vector of the same length

Show the two vectors are different

Why there are infinitely many answers

Every fact used, and where it comes from