Chapter 10 · Vector Algebra

Exercise 10.4 · Q3: Unit vector at $\frac\pi3$ to $\hat i$ and $\frac\pi4$ to $\hat j$ ⁠: its angle with $\hat k$ and its components

Exercise 10.4Direction cosines of a unit vector · Sections 10.2 and 10.6.2

The question

If a unit vector $\vec a$ makes angles $\frac{\pi}{3}$ with $\hat i$ ⁠, $\frac{\pi}{4}$ with $\hat j$ and an acute angle $\theta$ with $\hat k$ ⁠, then find $\theta$ and hence, the components of $\vec a$ ⁠.

Exam answer

Let $\vec a = l\hat i + m\hat j + n\hat k$ ⁠. As $\vec a$ is a unit vector, $l, m, n$ are its direction cosines:

$l = \cos\frac{\pi}{3} = \frac12$ ⁠, $m = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠, $n = \cos\theta$ ⁠.

Since $l^2 + m^2 + n^2 = 1$ ⁠,

\begin{aligned} &\tfrac14 + \tfrac12 + \cos^2\theta = 1 \\ &\Rightarrow \cos^2\theta = \tfrac14 \\ &\Rightarrow \cos\theta = \tfrac12 \quad (\theta \text{ acute}) \end{aligned}

So $\theta = \frac{\pi}{3}$ ⁠, and the components of $\vec a$ are $\frac12, \frac{1}{\sqrt2}, \frac12$ ⁠.

$\therefore\ \theta = \dfrac{\pi}{3}$ and $\vec a = \dfrac12\hat i + \dfrac{1}{\sqrt2}\hat j + \dfrac12\hat k$ ⁠.

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

The one idea this problem is built on

The components of a unit vector are the cosines of the angles it makes with $\hat i$ ⁠, $\hat j$ and $\hat k$ ⁠, its direction cosines $l, m, n$ ⁠. They always satisfy

l^2 + m^2 + n^2 = 1,

so two of the angles fix the cosine of the third up to its sign, and "acute" fixes the sign.

Given, to find, and the plan

Given: $|\vec a| = 1$ ⁠; the angle of $\vec a$ with $\hat i$ is $\frac{\pi}{3}$ ⁠, with $\hat j$ is $\frac{\pi}{4}$ ⁠, and with $\hat k$ is an acute angle $\theta$ ⁠.
To find: $\theta$ ⁠, and the components of $\vec a$ ⁠.
Plan: (1) Write the direction cosines. (2) Use $l^2 + m^2 + n^2 = 1$ to find $\cos^2\theta$ ⁠. (3) Use "acute" to choose the sign, and find $\theta$ ⁠. (4) Write $\vec a$ from its direction cosines and check it.

Basics you need

1Direction cosines

If a vector makes angles $\alpha, \beta, \gamma$ with the positive x-, y- and z-axes, then $l = \cos\alpha$ ⁠, $m = \cos\beta$ ⁠, $n = \cos\gamma$ are its direction cosines, and $l^2 + m^2 + n^2 = 1$ (Section 10.2 and its Note). Lesson: components

2A unit vector in terms of its direction cosines

A unit vector is $\cos\alpha\,\hat i + \cos\beta\,\hat j + \cos\gamma\,\hat k$ (Section 10.6.2, Remark). For example, $\vec a\cdot\hat i = |\vec a||\hat i|\cos\alpha = \cos\alpha$ ⁠, and $\vec a\cdot\hat i$ is the $\hat i$ component. Lesson: the dot product

3Cosines of special angles

$\cos\frac{\pi}{3} = \frac12$ ⁠, $\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠; on $0 \le \theta \le \pi$ ⁠, the cosine is positive exactly for acute $\theta$ ⁠. Basics: cos θ is a shadow, sin θ is a height

Solution

The direction cosines of $\vec a$
$\begin{aligned} l &= \cos\frac{\pi}{3} = \frac12, \\ m &= \cos\frac{\pi}{4} = \frac{1}{\sqrt2}, \\ n &= \cos\theta. \end{aligned}$
The angles that $\vec a$ makes with $\hat i$ ⁠, $\hat j$ ⁠, $\hat k$ are its angles with the positive x-, y- and z-axes, because $\hat i, \hat j, \hat k$ point along those axes. Their cosines are the direction cosines.

Direction cosines (Section 10.2). Lesson: components
Use $l^2 + m^2 + n^2 = 1$
$\begin{aligned} \left(\tfrac12\right)^2 + \left(\tfrac{1}{\sqrt2}\right)^2 + \cos^2\theta &= 1 \\ \tfrac14 + \tfrac12 + \cos^2\theta &= 1 \\ \cos^2\theta &= 1 - \tfrac34 = \tfrac14. \end{aligned}$
The identity holds because $l, m, n$ are the components of the unit vector $\hat a$ ⁠, and its length is 1: $l^2 + m^2 + n^2 = |\hat a|^2 = 1$ ⁠. Here $\vec a$ itself is the unit vector.

$l^2 + m^2 + n^2 = 1$ (Section 10.2, Note). Basics: use Pythagoras twice
Choose the sign, and find $\theta$

$\cos^2\theta = \frac14$ gives $\cos\theta = \frac12$ or $\cos\theta = -\frac12$ ⁠. Since $\theta$ is acute, $\cos\theta > 0$ ⁠, so
$\cos\theta = \frac12, \qquad \theta = \frac{\pi}{3}.$
An angle between two vectors lies in $[0, \pi]$ ⁠, where the cosine takes each value exactly once. The value $-\frac12$ belongs to $\theta = \frac{2\pi}{3}$ ⁠, which is obtuse and is ruled out. In $\left(0, \frac{\pi}{2}\right)$ the only angle with cosine $\frac12$ is $\frac{\pi}{3}$ ⁠.

$\cos\frac{\pi}{3} = \frac12$ ⁠; cosine is positive on acute angles. Basics: cos θ is a shadow, sin θ is a height
The components of $\vec a$ ⁠, and a check
$\vec a = l\hat i + m\hat j + n\hat k = \frac12\hat i + \frac{1}{\sqrt2}\hat j + \frac12\hat k.$
Its scalar components are $\frac12, \frac{1}{\sqrt2}, \frac12$ ⁠. Check: $|\vec a|^2 = \frac14 + \frac12 + \frac14 = 1$ ⁠, and $\vec a\cdot\hat j = \frac{1}{\sqrt2} = \cos\frac{\pi}{4}$ ⁠.

For a unit vector the components are the direction cosines themselves; for a vector of length $r$ they would be $lr, mr, nr$ ⁠. The check confirms the length is 1 and the angle with $\hat j$ is $\frac{\pi}{4}$ ⁠, as given.

Unit vector $= \cos\alpha\,\hat i + \cos\beta\,\hat j + \cos\gamma\,\hat k$ (Section 10.6.2, Remark). Lesson: the cross product, exercises

Result

$\theta = \frac{\pi}{3}$ ⁠, and $\vec a = \frac12\hat i + \frac{1}{\sqrt2}\hat j + \frac12\hat k$ ⁠: its components are $\frac12, \frac{1}{\sqrt2}, \frac12$ ⁠.

(Textbook answer: $\frac{\pi}{3}$ ⁠; $\frac12, \frac{1}{\sqrt2}, \frac12$ ⁠.)

Check it with numbers

$\frac{1}{\sqrt2} \approx 0.707107$ ⁠, and $\cos 45^\circ \approx 0.707107$ ⁠, the same number; $\cos 60^\circ = \frac12$ ⁠.

The angle with $\hat k$ ⁠: $\vec a\cdot\hat k = \frac12$ ⁠, so $\theta = \cos^{-1}\frac12 = 60^\circ$ ⁠, acute as required. With $\cos\theta = -\frac12$ the angle would be $120^\circ$ ⁠, not acute.

Common mistakes

Adding the angles instead of the squared cosines. Nothing makes $\frac{\pi}{3} + \frac{\pi}{4} + \theta$ a fixed number; the rule is $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ ⁠.
Keeping both signs. $\cos\theta = -\frac12$ gives $\theta = \frac{2\pi}{3}$ ⁠, which is not acute; the question rules it out.
Writing $\cos\frac{\pi}{4} = \frac12$ ⁠. $\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠, whose square is $\frac12$ ⁠.

What to take away

A unit vector is $l\hat i + m\hat j + n\hat k$ ⁠, with its direction cosines as components.
$l^2 + m^2 + n^2 = 1$ finds a missing direction cosine up to sign.
Words such as "acute" choose the sign: acute means a positive cosine.

Every fact used, and where it comes from

Fact	Used in	Source
Direction cosines $l, m, n$ and $l^2 + m^2 + n^2 = 1$	Steps 1, 2	Section 10.2 and its Note (Lesson: components)
$\cos\frac{\pi}{3} = \frac12$ ⁠, $\cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠; the angle between vectors lies in $[0, \pi]$	Steps 1, 3	Trigonometry; Section 10.6.1 (Basics: cos θ is a shadow)
A unit vector is $\cos\alpha\,\hat i + \cos\beta\,\hat j + \cos\gamma\,\hat k$	Step 4	Section 10.6.2, Remark
$\|x\hat i + y\hat j + z\hat k\| = \sqrt{x^2 + y^2 + z^2}$	Steps 2, 4	Section 10.5.1
The question and the answer	Result	Exercise 10.4 Q3; NCERT answer key (Lesson: the cross product, exercises)

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 10.4 · Q3: Unit vector at π3\frac\pi33π​ to i^\hat ii^ and π4\frac\pi44π​ to j^\hat jj^​⁠: its angle with k^\hat kk^ and its components

Basics you need

Solution

The direction cosines of a⃗\vec aa

Use l2+m2+n2=1l^2 + m^2 + n^2 = 1l2+m2+n2=1

Choose the sign, and find θ\thetaθ

The components of a⃗\vec aa⁠, and a check

Every fact used, and where it comes from

Exercise 10.4 · Q3: Unit vector at $\frac\pi3$ to $\hat i$ and $\frac\pi4$ to $\hat j$ ⁠: its angle with $\hat k$ and its components

The direction cosines of $\vec a$

Use $l^2 + m^2 + n^2 = 1$

Choose the sign, and find $\theta$

The components of $\vec a$ ⁠, and a check