Chapter 10 · Vector Algebra

Exercise 10.3 · Q1: Angle when $|\vec a|=\sqrt3$ ⁠, $|\vec b|=2$ ⁠, $\vec a\cdot\vec b=\sqrt6$

Exercise 10.3Angle between two vectors · Section 10.6.1

The question

Find the angle between two vectors $\vec a$ and $\vec b$ with magnitudes $\sqrt3$ and 2, respectively having $\vec a\cdot\vec b = \sqrt6$ ⁠.

Exam answer

Given $|\vec a| = \sqrt3$ ⁠, $|\vec b| = 2$ and $\vec a\cdot\vec b = \sqrt6$ ⁠. Let $\theta$ be the angle between $\vec a$ and $\vec b$ ⁠. Then

\begin{aligned} \cos\theta &= \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} \\ &= \frac{\sqrt6}{\sqrt3 \times 2} = \frac{\sqrt2}{2} \\ &= \frac{1}{\sqrt2} \end{aligned}

Since $0 \le \theta \le \pi$ ⁠,

\theta = \cos^{-1}\Big(\frac{1}{\sqrt2}\Big)

$\therefore\ \theta = \dfrac{\pi}{4}$ ⁠.

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

The one idea this problem is built on

The scalar product is defined by $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ ⁠, where $\theta$ is the angle between the vectors and $0 \le \theta \le \pi$ ⁠. With the scalar product and both magnitudes known, the definition can be solved for $\cos\theta$ :

\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}.

On $[0, \pi]$ each value of $\cos\theta$ belongs to exactly one angle, so the answer is a single angle.

Given, to find, and the plan

Given: $|\vec a| = \sqrt3$ ⁠, $|\vec b| = 2$ ⁠, $\vec a\cdot\vec b = \sqrt6$ ⁠.
To find: The angle $\theta$ between $\vec a$ and $\vec b$ ⁠.
Plan: (1) Write $\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ ⁠. (2) Substitute and simplify the surds. (3) Find the one angle in $[0, \pi]$ with that cosine. (4) Check by putting $\theta$ back into the definition.

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, Fig 10.19). Lesson: the dot product

2The angle formula

For nonzero vectors, $\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ ⁠, or $\theta = \cos^{-1}\left(\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right)$ (Section 10.6.1, Observation 6). Lesson: Examples 13–21

3A standard cosine

$\cos\frac{\pi}{4} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}$ ⁠, and $\cos$ takes each value in $[-1, 1]$ exactly once on $[0, \pi]$ ⁠. Basics: cos θ is a shadow

Solution

Write the angle formula

Both vectors are nonzero (their magnitudes are $\sqrt3$ and 2), so the angle between them is defined and
$\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}.$
Dividing the definition $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ by $|\vec a||\vec b|$ leaves $\cos\theta$ alone. The division is allowed because $|\vec a||\vec b| = 2\sqrt3 \ne 0$ ⁠; for a zero vector the angle would not be defined.

Definition 2 and Observation 6 (Section 10.6.1). Lesson: the dot product
Substitute and simplify
$\cos\theta = \frac{\sqrt6}{\sqrt3 \times 2} = \frac{\sqrt2}{2} = \frac{1}{\sqrt2}.$
Since $\sqrt6 = \sqrt2 \times \sqrt3$ ⁠, the $\sqrt3$ on top cancels the $\sqrt3$ underneath and leaves $\frac{\sqrt2}{2}$ ⁠. Writing $\frac{\sqrt2}{2} = \frac{\sqrt2}{\sqrt2 \times \sqrt2} = \frac{1}{\sqrt2}$ shows the familiar value of $\cos 45^\circ$ ⁠. Only the three given numbers are needed; the components of the vectors are never used.

$\sqrt{mn} = \sqrt m\,\sqrt n$ for $m, n \ge 0$ ⁠; $\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ (Section 10.6.1, Observation 6). Lesson: Examples 13–21
Find the angle in $[0, \pi]$
$\theta = \cos^{-1}\Big(\frac{1}{\sqrt2}\Big) = \frac{\pi}{4}.$
So $\theta = \frac{\pi}{4}$ , that is $45^\circ$ ⁠.

Drawn to scale, one scale on both axes: $\vec a$ of length $\sqrt3$ ⁠, $\vec b$ of length 2 at $\theta = 45^\circ$ to it. The perpendicular from the tip of $\vec b$ meets the line of $\vec a$ at distance $2\cos 45^\circ = \sqrt2$ from O, so the shadow of $\vec b$ on $\vec a$ has length $\sqrt2$ ⁠, and $\vec a\cdot\vec b = |\vec a| \times \sqrt2 = \sqrt3 \times \sqrt2 = \sqrt6$ ⁠.

The angle between two vectors is taken in $[0, \pi]$ ⁠, which is also the principal range of $\cos^{-1}$ ⁠. On that interval cosine decreases from 1 to $-1$ ⁠, so it takes the value $\frac{1}{\sqrt2}$ exactly once, at $\frac{\pi}{4}$ ⁠. Angles such as $-\frac{\pi}{4}$ or $\frac{7\pi}{4}$ have the same cosine but are not in $[0, \pi]$ ⁠.

$0 \le \theta \le \pi$ (Section 10.6.1, Definition 2); $\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠. Basics: cos θ is a shadow
Check in the definition
$\begin{aligned} |\vec a||\vec b|\cos\frac{\pi}{4} &= \sqrt3 \times 2 \times \frac{1}{\sqrt2} \\ &= \sqrt3 \times \sqrt2 = \sqrt6. \end{aligned}$
Putting the angle back into the definition must give the scalar product we were told, and it does: $\frac{2}{\sqrt2} = \sqrt2$ ⁠, and $\sqrt3 \times \sqrt2 = \sqrt6$ ⁠. The figure shows the same fact: the shadow of $\vec b$ on the line of $\vec a$ is $|\vec b|\cos\theta = \sqrt2$ ⁠, and the scalar product is $|\vec a|$ times that shadow.

$\vec a\cdot\vec b = |\vec a|\,(|\vec b|\cos\theta)$ ⁠: the length of $\vec a$ times the projection of $\vec b$ on $\vec a$ (Sections 10.6.1 and 10.6.2). Lesson: the dot product

Result

The angle between $\vec a$ and $\vec b$ is $\theta = \frac{\pi}{4}$ ( $45^\circ$ ⁠).

(Textbook answer: $\frac{\pi}{4}$ ⁠.)

Check it with numbers

$\sqrt6 \approx 2.449490$ and $\sqrt3 \times 2 \approx 3.464102$ ⁠, so $\cos\theta \approx \frac{2.449490}{3.464102} \approx 0.707107$ ⁠. Also $\frac{\pi}{4} \approx 0.785398$ radians, and $\cos(0.785398) \approx 0.707107$ ⁠: the same number.

A concrete pair with these data: $\vec a = \sqrt3\,\hat i$ and $\vec b = \sqrt2\,\hat i + \sqrt2\,\hat j$ ⁠. Then $|\vec a| = \sqrt3$ ⁠, $|\vec b| = \sqrt{2 + 2} = 2$ and $\vec a\cdot\vec b = \sqrt3 \times \sqrt2 = \sqrt6$ ⁠, while $\vec b$ makes $\tan^{-1}\frac{\sqrt2}{\sqrt2} = \tan^{-1}1 = 45^\circ$ with the x-axis, the direction of $\vec a$ ⁠.

Common mistakes

Dividing by one magnitude only. $\frac{\sqrt6}{\sqrt3} = \sqrt2$ is bigger than 1 and is not a cosine at all; the denominator is $|\vec a||\vec b| = 2\sqrt3$ ⁠.
Simplifying the surds wrongly. $\frac{\sqrt6}{2\sqrt3}$ is $\frac{\sqrt2}{2}$ ⁠, not $\frac{\sqrt3}{2}$ ⁠; the slip $\frac{\sqrt3}{2}$ would give $\frac{\pi}{6}$ ⁠.
Giving an angle outside $[0, \pi]$ ⁠. $\frac{7\pi}{4}$ also has cosine $\frac{1}{\sqrt2}$ ⁠, but the angle between two vectors is never more than $\pi$ ⁠.

What to take away

$\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ needs only the scalar product and the two lengths.
Simplify surds with $\sqrt{mn} = \sqrt m\,\sqrt n$ before reading off the angle.
The angle between two vectors lies in $[0, \pi]$ ⁠, so $\theta = \cos^{-1}(\ldots)$ gives the one answer.

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\cdot\vec b = \|\vec a\|\|\vec b\|\cos\theta$ ⁠, $0 \le \theta \le \pi$	Steps 1, 3, 4	Section 10.6.1, Definition 2, Fig 10.19 (Lesson: the dot product)
$\cos\theta = \frac{\vec a\cdot\vec b}{\|\vec a\|\|\vec b\|}$	Steps 1, 2	Section 10.6.1, Observation 6
$\sqrt6 = \sqrt2\,\sqrt3$	Steps 2, 4	Algebra of surds
$\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠; principal range of $\cos^{-1}$ is $[0, \pi]$	Step 3	Trigonometry (Basics: cos θ is a shadow)
Projection of $\vec b$ on $\vec a$ is $\|\vec b\|\cos\theta$	Step 4	Section 10.6.2
The answer	Result	NCERT answer key, Exercise 10.3 Q1 (Lesson: the dot product)

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Exercise 10.3 · Q1: Angle when ∣a⃗∣=3|\vec a|=\sqrt3∣a∣=3​⁠, ∣b⃗∣=2|\vec b|=2∣b∣=2⁠, a⃗⋅b⃗=6\vec a\cdot\vec b=\sqrt6a⋅b=6​

Basics you need

Solution

Write the angle formula

Substitute and simplify

Find the angle in [0,π][0, \pi][0,π]

Check in the definition

Every fact used, and where it comes from

Exercise 10.3 · Q1: Angle when $|\vec a|=\sqrt3$ ⁠, $|\vec b|=2$ ⁠, $\vec a\cdot\vec b=\sqrt6$

Find the angle in $[0, \pi]$