Chapter 10 · Vector Algebra

Exercise 10.1 · Q4: Coinitial, equal and collinear vectors in a square (Fig 10.6)

Exercise 10.1Types of vectors · Section 10.3

The question

In Fig 10.6 (a square), identify the following vectors.

(i) Coinitial (ii) Equal (iii) Collinear but not equal

Exam answer

Let the square be PQRS with P the top left corner, Q top right, R bottom right and S bottom left (figure).

Then $\vec a = \overrightarrow{PQ}$ ⁠, $\vec b = \overrightarrow{QR}$ ⁠, $\vec c = \overrightarrow{RS}$ and $\vec d = \overrightarrow{PS}$ ⁠.

(i) $\vec a$ and $\vec d$ have the same initial point P, so $\vec a$ and $\vec d$ are coinitial.

(ii) $\vec b$ and $\vec d$ are parallel, both point downwards and have equal magnitudes (sides of a square), so $\vec b = \vec d$ .

(iii) $\vec a$ and $\vec c$ are parallel to the same line with equal magnitudes but opposite directions, so $\vec c = -\vec a$ ⁠: $\vec a$ and $\vec c$ are collinear but not equal.

The NCERT answer key prints $\vec a$ and $\vec b$ for (i), but in Fig 10.6 $\vec b$ starts where $\vec a$ ends; the coinitial pair is $\vec a$ and $\vec d$ ⁠.

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

The one idea this problem is built on

Three definitions from Section 10.3, each applied by looking at the arrows. Coinitial: the vectors start at the same point. Equal: same magnitude and same direction, wherever they start. Collinear: parallel to the same line, whatever their magnitudes and directions. So we read off, for each arrow, where it starts, which way it points and how long it is.

Given, to find, and the plan

Given: Fig 10.6, a square with a vector along each side: $\vec a$ along the top side from left to right, $\vec b$ down the right side, $\vec c$ along the bottom side from right to left, and $\vec d$ down the left side (redrawn in the figure below).
To find: (i) the coinitial vectors, (ii) the equal vectors, (iii) the vectors that are collinear but not equal.
Plan: (1) Name the corners and write each vector as an arrow between corners. (2) Compare initial points for (i). (3) Compare directions and lengths for (ii). (4) Find parallel pairs that are not equal for (iii). (5) Confirm everything with components.

Basics you need

1Initial and terminal points

For $\overrightarrow{AB}$ ⁠, A (where the arrow starts) is the initial point and B (where the arrowhead is) is the terminal point (Section 10.2). Lesson: a vector is a length with a direction

2Coinitial, collinear and equal vectors

Coinitial: two or more vectors with the same initial point. Collinear: parallel to the same line, irrespective of magnitudes and directions. Equal: the same magnitude and direction, regardless of the positions of their initial points (Section 10.3). Lesson: six kinds of vectors worth naming

3Facts about a square

All four sides are equal; opposite sides are parallel; adjacent sides are perpendicular.

4Components

Two vectors are equal exactly when their components are equal, and $\vec b$ is collinear with a nonzero $\vec a$ exactly when $\vec b = \lambda\vec a$ for some nonzero scalar $\lambda$ (Section 10.5.1, Remark (i)). Basics: three numbers fix a point

Part (i): coinitial vectors

Name the corners and the four arrows

Call the corners P (top left), Q (top right), R (bottom right) and S (bottom left). Reading the arrowheads in Fig 10.6:
$\begin{aligned} \vec a &= \overrightarrow{PQ}, & \vec b &= \overrightarrow{QR}, \\ \vec c &= \overrightarrow{RS}, & \vec d &= \overrightarrow{PS}. \end{aligned}$

Fig 10.6 redrawn with its corners named. $\vec a$ (blue) goes from P to Q, $\vec b$ (green) from Q to R, $\vec c$ (rose) from R to S and $\vec d$ (violet) from P to S. The gold dot marks P, the common initial point of $\vec a$ and $\vec d$ ⁠.

The arrowhead marks the terminal point, so the other end is the initial point. In Fig 10.6 the arrowheads of $\vec a$ and $\vec d$ are at the top right and bottom left corners; both arrows leave the top left corner. The corner letters are ours: the book's figure does not name them.

A vector $\overrightarrow{AB}$ starts at its initial point A and ends at its terminal point B (Section 10.2). Lesson: a vector is a length with a direction
Compare the initial points

The initial points are: $\vec a$ ⁠: P, $\vec b$ ⁠: Q, $\vec c$ ⁠: R, $\vec d$ ⁠: P. Only P occurs twice, so $\vec a$ and $\vec d$ are coinitial, and no other pair is.

Coinitial means "same initial point", nothing else: the directions and lengths do not matter. $\vec b$ starts at Q, which is where $\vec a$ ends; $\vec a$ and $\vec b$ are placed head to tail (as in the triangle law of Section 10.4), not tail to tail.

Coinitial vectors: two or more vectors having the same initial point (Section 10.3). Lesson: six kinds of vectors

Part (ii): equal vectors

Compare directions and lengths

$\vec b$ and $\vec d$ lie on the two vertical sides, so they are parallel, and both arrowheads are at the bottom: they point the same way (downwards). Their lengths are sides of a square, so $|\vec b| = |\vec d|$ ⁠. Hence $\vec b = \vec d$ .

Equal vectors need the same magnitude and the same direction, but they may start at different points. Sliding $\vec d$ from the left side to the right side (a parallel displacement, allowed for free vectors) lands it exactly on $\vec b$ ⁠.

Equal vectors: same magnitude and direction regardless of the positions of their initial points; vectors may be moved parallel to themselves (Remark on free vectors, Section 10.3). Lesson: six kinds of vectors
Rule out the other pairs

$\vec a$ and $\vec c$ have equal lengths and are parallel, but $\vec a$ points right and $\vec c$ points left, so $\vec a \ne \vec c$ ⁠. Every other pair is perpendicular (one horizontal, one vertical), so their directions differ. Thus $\vec b$ and $\vec d$ are the only equal vectors.

Equal length alone is not enough: all four vectors have the same length, yet only one pair is equal. The direction decides.

Adjacent sides of a square are perpendicular; opposite sides are parallel.

Part (iii): collinear but not equal

Find the parallel pairs

Two vectors along a square's sides are collinear exactly when they lie on opposite sides. The pairs are $\vec a, \vec c$ (horizontal) and $\vec b, \vec d$ (vertical).

Collinear vectors are parallel to the same line, "irrespective of their magnitudes and directions": opposite directions are allowed. Vectors on adjacent sides are perpendicular, so they are not collinear.

Collinear vectors: two or more vectors parallel to the same line (Section 10.3). Lesson: six kinds of vectors
Remove the equal pair

$\vec b = \vec d$ (Part (ii)), so that pair is excluded. The pair $\vec a, \vec c$ has equal magnitudes and opposite directions, so
$\vec c = -\vec a \ne \vec a.$
Hence $\vec a$ and $\vec c$ are collinear but not equal.

$\vec c$ has the same magnitude as $\vec a$ but the opposite direction, which is the definition of the negative of $\vec a$ ⁠. A nonzero vector is never equal to its negative.

Negative of a vector: same magnitude, opposite direction (Section 10.3). Lesson: six kinds of vectors
Confirm with components

Put the square on axes with S at the origin and side 1: $S(0, 0)$ ⁠, $R(1, 0)$ ⁠, $Q(1, 1)$ ⁠, $P(0, 1)$ ⁠. Then (end minus start)
$\begin{aligned} \vec a &= \overrightarrow{PQ} = \hat i, & \vec b &= \overrightarrow{QR} = -\hat j, \\ \vec c &= \overrightarrow{RS} = -\hat i, & \vec d &= \overrightarrow{PS} = -\hat j. \end{aligned}$
So $\vec b = \vec d$ (equal components), $\vec c = (-1)\vec a$ (collinear, $\lambda = -1$ ⁠) with $\vec c \ne \vec a$ ⁠, and no $\lambda$ gives $-\hat j = \lambda\hat i$ (adjacent sides are not collinear).

The component test turns the picture into arithmetic: equality means every component matches, collinearity means one vector is a scalar multiple of the other. Both tests agree with what we read from the figure.

The vector joining two points is "end minus start" (Section 10.5.2); equality of vectors (Section 10.5.1, (iii)); collinearity $\vec b = \lambda\vec a$ (Remark (i), Section 10.5.1). Lesson: components

Result

(i) $\vec a$ and $\vec d$ are coinitial (both start at the top left corner). (ii) $\vec b$ and $\vec d$ are equal. (iii) $\vec a$ and $\vec c$ are collinear but not equal ( $\vec c = -\vec a$ ⁠).

(Textbook answer: (i) vectors $\vec a$ and $\vec b$ are coinitial; (ii) vectors $\vec b$ and $\vec d$ are equal; (iii) vectors $\vec a$ and $\vec c$ are collinear but not equal.)

For (i) the key's pair does not match Fig 10.6: $\vec a$ starts at the top left corner and $\vec b$ at the top right corner, where $\vec a$ ends. The vectors that share an initial point are $\vec a$ and $\vec d$ ⁠. Parts (ii) and (iii) agree with the key.

Check it with numbers

With the coordinates of Part (iii), Step 3: $\vec b - \vec d = (-\hat j) - (-\hat j) = \vec 0$ ⁠, so $\vec b = \vec d$ ⁠. $\vec a + \vec c = \hat i - \hat i = \vec 0$ ⁠, so $\vec c = -\vec a$ ⁠, and $\vec a - \vec c = 2\hat i \ne \vec 0$ ⁠, so $\vec a \ne \vec c$ ⁠. All four magnitudes are 1. Initial points: $\vec a$ and $\vec d$ both start at $(0, 1)$ ⁠; $\vec b$ starts at $(1, 1)$ and $\vec c$ at $(1, 0)$ ⁠.

Common mistakes

Calling head-to-tail vectors coinitial. $\vec b$ begins where $\vec a$ ends; coinitial vectors begin at the same point.
Calling $\vec a$ and $\vec c$ equal because they have the same length. They point in opposite directions, so $\vec c = -\vec a$ ⁠.
Thinking collinear vectors must point the same way. Collinear allows opposite directions; that is why $\vec a$ and $\vec c$ are collinear.
Thinking equal vectors must start at the same point. $\vec b$ and $\vec d$ are on different sides and still equal.

What to take away

Coinitial: compare starting points only. Equal: compare length and direction only. Collinear: compare lines only.
Equal vectors are always collinear, but collinear vectors need not be equal (they may differ in length or direction).
Components make all three tests mechanical: $\vec b = \vec d$ means equal components; $\vec c = \lambda\vec a$ means collinear.

Every fact used, and where it comes from

Fact	Used in	Source
Initial and terminal points of a vector	Part (i), Steps 1–2	Section 10.2
Coinitial, collinear and equal vectors; negative of a vector; free vectors	Parts (i) to (iii)	Section 10.3 (Lesson: six kinds of vectors)
Triangle law: vectors placed head to tail	Part (i), Step 2	Section 10.4 (Lesson: adding arrows)
Vector joining two points; equality of components	Part (iii), Step 3	Sections 10.5.2 and 10.5.1
$\vec b$ collinear with $\vec a$ if and only if $\vec b = \lambda\vec a$	Part (iii), Step 3	Remark (i), Section 10.5.1
The vectors and their directions	Whole solution	Fig 10.6, Exercise 10.1

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 10.1 · Q4: Coinitial, equal and collinear vectors in a square (Fig 10.6)

Basics you need

Part (i): coinitial vectors

Name the corners and the four arrows

Compare the initial points

Part (ii): equal vectors

Compare directions and lengths

Rule out the other pairs

Part (iii): collinear but not equal

Find the parallel pairs

Remove the equal pair

Confirm with components

Every fact used, and where it comes from