Chapter 10 · Vector Algebra

Exercise 10.2 · Q3: Two different vectors with the same direction

Exercise 10.2Multiplication of a vector by a scalar · Sections 10.5 and 10.5.1

The question

Write two different vectors having same direction.

Exam answer

Let $\vec a = \hat i + 2\hat j + 2\hat k$ and $\vec b = 2\hat i + 4\hat j + 4\hat k$ ⁠.

Then $\vec b = 2\vec a$ with $2 > 0$ ⁠, so $\vec b$ has the same direction as $\vec a$ ⁠.

|\vec a| = \sqrt{1 + 4 + 4} = 3

|\vec b| = \sqrt{4 + 16 + 16} = 6

As $|\vec a| \ne |\vec b|$ ⁠, $\vec a \ne \vec b$ ⁠.

$\therefore\ \hat i + 2\hat j + 2\hat k$ and $2\hat i + 4\hat j + 4\hat k$ are two different vectors having the same direction.

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

The one idea this problem is built on

Multiplying a vector by a positive scalar $\lambda$ keeps its direction and multiplies its length by $\lambda$ (Section 10.5). So $\vec a$ and $2\vec a$ point the same way, but $2\vec a$ is twice as long; with different lengths they cannot be equal. Same direction, different vectors.

Given, to find, and the plan

Given: Nothing but the conditions.
To find: Two vectors $\vec a \ne \vec b$ with the same direction.
Plan: (1) Pick a nonzero vector $\vec a = \hat i + 2\hat j + 2\hat k$ ⁠. (2) Take $\vec b = 2\vec a$ ⁠. (3) Show the magnitudes differ, so $\vec a \ne \vec b$ ⁠. (4) Confirm the directions agree through the unit vectors (direction cosines).

Basics you need

1Multiplying a vector by a scalar

$\lambda\vec a$ has the same direction as $\vec a$ if $\lambda > 0$ and the opposite direction if $\lambda < 0$ ⁠; its magnitude is $|\lambda\vec a| = |\lambda|\,|\vec a|$ ⁠. In components, $\lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k$ (Sections 10.5 and 10.5.1, (iv)). Lesson: components

2Magnitude

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

3Unit vector and direction cosines

The unit vector $\hat a = \frac{1}{|\vec a|}\vec a$ points in the direction of $\vec a$ ⁠; its components are the direction cosines $l, m, n$ of $\vec a$ (Remark (iii), Section 10.5.1). Two nonzero vectors have the same direction exactly when their unit vectors are equal. Lesson: six kinds of vectors

Solution

Choose the first vector

Take $\vec a = \hat i + 2\hat j + 2\hat k$ ⁠. Its magnitude is
$|\vec a| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt9 = 3.$
Any nonzero vector works (the zero vector has no definite direction, so it is no use here). Components 1, 2, 2 give the whole-number length 3, which keeps every later number simple.

The magnitude formula (Section 10.5.1). Basics: use Pythagoras twice
Multiply by a positive scalar

Take $\lambda = 2$ ⁠:
$\vec b = 2\vec a = 2\hat i + 4\hat j + 4\hat k.$
Since $\lambda = 2 > 0$ ⁠, $\vec b$ has the same direction as $\vec a$ ⁠.

Doubling a vector stretches the arrow along its own line without turning it. A negative multiplier such as $-2$ would also stay on the same line but would reverse the arrow, giving the opposite direction, so the multiplier must be positive.

"The vector $\lambda\vec a$ has the direction same (or opposite) to that of vector $\vec a$ according as the value of $\lambda$ is positive (or negative)" (Section 10.5). Lesson: components
Show the vectors are different

The magnitude of $\vec b$ is
$\begin{aligned} |\vec b| &= \sqrt{2^2 + 4^2 + 4^2} \\ &= \sqrt{36} = 6 = 2|\vec a|. \end{aligned}$
As $6 \ne 3$ ⁠, the vectors have different magnitudes, so $\vec a \ne \vec b$ (also, their $\hat i$ components 1 and 2 differ).

$\vec a$ and $\vec b = 2\vec a$ drawn side by side, parallel to one dashed line and to the same scale (40 px per unit). Both arrows point the same way; $\vec b$ is twice as long, so the two vectors are not equal.

Equal vectors must have the same magnitude and direction. Here the directions agree but the magnitudes do not, so the vectors are different, exactly as the question requires.

$|\lambda\vec a| = |\lambda|\,|\vec a|$ (Section 10.5); equal vectors (Section 10.3). Lesson: six kinds of vectors
Confirm the directions agree

Divide each vector by its magnitude:
$\begin{aligned} \hat a &= \tfrac13(\hat i + 2\hat j + 2\hat k) \\ &= \tfrac13\hat i + \tfrac23\hat j + \tfrac23\hat k, \\ \hat b &= \tfrac16(2\hat i + 4\hat j + 4\hat k) \\ &= \tfrac13\hat i + \tfrac23\hat j + \tfrac23\hat k. \end{aligned}$
The unit vectors are equal, so both vectors have direction cosines $\frac13, \frac23, \frac23$ ⁠: they make the same angles with the three axes.

A unit vector keeps only the direction of a vector and throws away its length. Equal unit vectors therefore mean equal directions. This check does not rely on spotting $\vec b = 2\vec a$ ⁠.

The components of the unit vector are the direction cosines $\cos\alpha, \cos\beta, \cos\gamma$ (Remark (iii), Section 10.5.1). Lesson: components
Why there are infinitely many answers

For any nonzero $\vec a$ ⁠, every vector $\lambda\vec a$ with $\lambda > 0$ and $\lambda \ne 1$ has the same direction as $\vec a$ and a different length. For example, $\hat i + \hat j$ and $3\hat i + 3\hat j$ ⁠, or $\hat k$ and $5\hat k$ ⁠.

There are infinitely many choices of $\vec a$ and of $\lambda$ ⁠, so the question has infinitely many correct answers. Any one valid pair, shown to have the same direction and to be different, is a complete answer.

Collinear vectors in the same direction are positive multiples of each other (Section 10.5). Lesson: components

Result

For example, $\vec a = \hat i + 2\hat j + 2\hat k$ and $\vec b = 2\hat i + 4\hat j + 4\hat k$ ⁠: $\vec b = 2\vec a$ with $2 > 0$ ⁠, so they have the same direction (direction cosines $\frac13, \frac23, \frac23$ ⁠), but $|\vec a| = 3$ and $|\vec b| = 6$ ⁠, so $\vec a \ne \vec b$ ⁠. Any $\vec a$ and $\lambda\vec a$ with $\lambda > 0$ ⁠, $\lambda \ne 1$ ⁠, is also correct.

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

Check it with numbers

Ratios of components: $\frac21 = \frac42 = \frac42 = 2$ ⁠, one positive ratio, so $\vec b = 2\vec a$ ⁠. Direction cosines: $\frac13 \approx 0.3333$ and $\frac23 \approx 0.6667$ ⁠, with $\frac19 + \frac49 + \frac49 = 1$ ⁠. The angles with the axes are $\cos^{-1}\frac13 \approx 70.53^\circ$ ⁠, $\cos^{-1}\frac23 \approx 48.19^\circ$ and $48.19^\circ$ ⁠, the same for both vectors.

Common mistakes

Using a negative multiplier. $\vec a$ and $-2\vec a$ are collinear but point in opposite directions.
Using $\lambda = 1$ ⁠. Then $\vec b = \vec a$ ⁠, the same vector, not two different ones.
Changing the components unevenly. $\hat i + 2\hat j + 2\hat k$ and $2\hat i + 4\hat j + 3\hat k$ do not point the same way: the ratios $2, 2, \frac32$ are not all equal.

What to take away

Same direction means a positive scalar multiple: $\vec b = \lambda\vec a$ ⁠, $\lambda > 0$ ⁠.
Two vectors with the same direction are equal only if their magnitudes are also equal.
Equal unit vectors (equal direction cosines) is a sure test of equal direction.

Every fact used, and where it comes from

Fact	Used in	Source
$\lambda\vec a$ ⁠: same direction for $\lambda > 0$ ⁠, opposite for $\lambda < 0$ ⁠; $\|\lambda\vec a\| = \|\lambda\|\|\vec a\|$	Steps 2, 3, 5	Section 10.5
$\lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k$	Step 2	Section 10.5.1, (iv)
$\|x\hat i + y\hat j + z\hat k\| = \sqrt{x^2 + y^2 + z^2}$	Steps 1, 3	Section 10.5.1 (Basics: Pythagoras twice)
Equal vectors: same magnitude and direction	Step 3	Section 10.3 (Lesson: six kinds of vectors)
$\hat a = \frac{1}{\|\vec a\|}\vec a$ ⁠; its components are the direction cosines	Step 4	Section 10.5; Remark (iii), Section 10.5.1

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Exercise 10.2 · Q3: Two different vectors with the same direction

Basics you need

Solution

Choose the first vector

Multiply by a positive scalar

Show the vectors are different

Confirm the directions agree

Why there are infinitely many answers

Every fact used, and where it comes from