Chapter 10 · Vector Algebra

Exercise 10.3 · Q8: Equal magnitudes, angle $60^\circ$ ⁠, scalar product $\frac12$

Exercise 10.3Scalar product and magnitudes · Section 10.6.1

The question

Find the magnitude of two vectors $\vec a$ and $\vec b$ ⁠, having the same magnitude and such that the angle between them is $60^\circ$ and their scalar product is $\frac12$ ⁠.

Exam answer

Let $|\vec a| = |\vec b| = x$ ⁠. Given $\theta = 60^\circ$ and $\vec a\cdot\vec b = \frac12$ ⁠.

\begin{aligned} \vec a\cdot\vec b &= |\vec a||\vec b|\cos 60^\circ \\ \Rightarrow \frac12 &= x \times x \times \frac12 \\ \Rightarrow x^2 &= 1 \end{aligned}

As a magnitude is non-negative, $x = 1$ ⁠.

$\therefore\ |\vec a| = 1$ and $|\vec b| = 1$ ⁠.

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

The one idea this problem is built on

The definition $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ links four numbers. Here three facts are given (equal magnitudes, the angle, the scalar product), so the definition becomes an equation for the one unknown, the common magnitude:

\frac12 = x \times x \times \cos 60^\circ.

Given, to find, and the plan

Given: $|\vec a| = |\vec b|$ ⁠, the angle between $\vec a$ and $\vec b$ is $60^\circ$ ⁠, and $\vec a\cdot\vec b = \frac12$ ⁠.
To find: $|\vec a|$ and $|\vec b|$ ⁠.
Plan: (1) Call the common magnitude $x$ ⁠. (2) Write the definition with $\cos 60^\circ = \frac12$ ⁠. (3) Solve $x^2 = 1$ and keep the root a length can have. (4) Check with a concrete pair.

Basics you need

1The scalar product

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ ⁠, with $\theta$ the angle between $\vec a$ and $\vec b$ ⁠, $0 \le \theta \le \pi$ (Section 10.6.1, Definition 2). Lesson: the dot product

2A standard cosine

$\cos 60^\circ = \frac12$ ⁠. Basics: cos θ is a shadow

3Magnitudes are non-negative

$|\vec a| \ge 0$ for every vector (Section 10.2, Note), so from $x^2 = 1$ only $x = 1$ is possible. Lesson: Examples 13–21

Solution

Name the common magnitude

Let $|\vec a| = |\vec b| = x$ ⁠, where $x \ge 0$ is to be found.

The question says the two vectors have the same magnitude, so one letter is enough for both. A single unknown means a single equation will do.

The magnitude of a vector is its length (Section 10.2). Lesson: a vector is a length with a direction
Write the definition
$\begin{aligned} \vec a\cdot\vec b &= |\vec a||\vec b|\cos 60^\circ \\ \Rightarrow \frac12 &= x \times x \times \frac12. \end{aligned}$
Every number in the definition is known except $x$ ⁠: the scalar product is $\frac12$ ⁠, both lengths are $x$ ⁠, and $\cos 60^\circ = \frac12$ ⁠.

Section 10.6.1, Definition 2; $\cos 60^\circ = \frac12$ ⁠. Basics: cos θ is a shadow
Solve for $x$
$\frac{x^2}{2} = \frac12 \;\Rightarrow\; x^2 = 1 \;\Rightarrow\; x = 1.$
So $|\vec a| = |\vec b| = 1$ : both are unit vectors.

Drawn to scale, one scale on both axes: $\vec a$ and $\vec b$ ⁠, both of length 1, with $60^\circ$ between them. The perpendicular from the tip of $\vec b$ meets the line of $\vec a$ at its midpoint, so the shadow of $\vec b$ on $\vec a$ is $\cos 60^\circ = \frac12$ ⁠, and $\vec a\cdot\vec b = |\vec a| \times \frac12 = \frac12$ ⁠.

Multiplying both sides by 2 gives $x^2 = 1$ ⁠, whose roots are $x = 1$ and $x = -1$ ⁠. A magnitude is a length and is never negative, so $x = -1$ is rejected.

$|\vec a| \ge 0$ (Section 10.2, Note). Lesson: Examples 13–21
Check with a concrete pair

Take $\vec a = \hat i$ and $\vec b = \frac12\hat i + \frac{\sqrt3}{2}\hat j$ ⁠. Then $|\vec a| = 1$ ⁠, $|\vec b| = \sqrt{\frac14 + \frac34} = 1$ ⁠, and
$\vec a\cdot\vec b = (1)\Big(\frac12\Big) + (0)\Big(\frac{\sqrt3}{2}\Big) = \frac12.$
The vector $\vec b$ makes $\tan^{-1}\frac{\sqrt3/2}{1/2} = \tan^{-1}\sqrt3 = 60^\circ$ with the x-axis, which is the direction of $\vec a$ ⁠. So this pair has equal magnitudes 1, angle $60^\circ$ and scalar product $\frac12$ ⁠: all three conditions hold.

$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ (Section 10.6.1). Lesson: the dot product

Result

$|\vec a| = 1$ and $|\vec b| = 1$ ⁠.

(Textbook answer: $|\vec a| = 1$ ⁠, $|\vec b| = 1$ ⁠.)

Check it with numbers

$\cos 60^\circ = 0.5$ ⁠, so $|\vec a||\vec b|\cos 60^\circ = 1 \times 1 \times 0.5 = 0.5 = \frac12$ ⁠.

In the concrete pair, $\vec b \approx 0.5\,\hat i + 0.866025\,\hat j$ ⁠, and $\vec a\cdot\vec b = 1 \times 0.5 + 0 \times 0.866025 = 0.5$ ⁠. If the common length were 2 instead, the scalar product would be $2 \times 2 \times 0.5 = 2$ ⁠, not $\frac12$ ⁠: only length 1 works.

Common mistakes

Using $\cos 60^\circ$ as $\frac{\sqrt3}{2}$ ⁠. That is $\sin 60^\circ$ (or $\cos 30^\circ$ ⁠); with it, $x^2 = \frac{1}{\sqrt3}$ ⁠, a wrong answer.
Writing $x$ instead of $x^2$ ⁠. Both magnitudes are $x$ ⁠, so their product is $x^2$ ⁠; $\frac{x}{2} = \frac12$ happens to give $x = 1$ here, but by a wrong route.
Keeping $x = -1$ ⁠. A magnitude cannot be negative.

What to take away

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ can be solved for whichever quantity is unknown.
Equal magnitudes: call the common value $x$ ⁠, and the product of the lengths is $x^2$ ⁠.
Keep only the non-negative root for a length.

Every fact used, and where it comes from

Fact	Used in	Source
Magnitude of a vector; $\|\vec a\| \ge 0$	Steps 1, 3	Section 10.2, Note (Lesson: a vector is a length with a direction)
$\vec a\cdot\vec b = \|\vec a\|\|\vec b\|\cos\theta$	Step 2	Section 10.6.1, Definition 2 (Lesson: the dot product)
$\cos 60^\circ = \frac12$	Step 2	Trigonometry (Basics: cos θ is a shadow)
Scalar product in components	Step 4	Section 10.6.1
The answer	Result	NCERT answer key, Exercise 10.3 Q8 (Lesson: the dot product)

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 10.3 · Q8: Equal magnitudes, angle 60∘60^\circ60∘⁠, scalar product 12\frac1221​

Basics you need

Solution

Name the common magnitude

Write the definition

Solve for xxx

Check with a concrete pair

Every fact used, and where it comes from

Exercise 10.3 · Q8: Equal magnitudes, angle $60^\circ$ ⁠, scalar product $\frac12$

Solve for $x$