Chapter 10 · Vector Algebra

Exercise 10.2 · Q7: Unit vector along $\hat i+\hat j+2\hat k$

Exercise 10.2Unit vector in the direction of a vector · Sections 10.5 and 10.5.1

The question

Find the unit vector in the direction of the vector $\vec a = \hat i + \hat j + 2\hat k$ ⁠.

Exam answer

The unit vector in the direction of $\vec a$ is $\hat a = \frac{1}{|\vec a|}\vec a$ ⁠.

\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + 2^2} \\ &= \sqrt6 \end{aligned}

\therefore\ \hat a = \frac{1}{\sqrt6}(\hat i + \hat j + 2\hat k)

$\hat a = \dfrac{1}{\sqrt6}\hat i + \dfrac{1}{\sqrt6}\hat j + \dfrac{2}{\sqrt6}\hat k$ ⁠.

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

The one idea this problem is built on

Dividing a nonzero vector by its own length keeps its direction and makes its length 1: $\hat a = \frac{1}{|\vec a|}\vec a$ ⁠. So we find $|\vec a|$ and divide every component by it.

Given, to find, and the plan

Given: $\vec a = \hat i + \hat j + 2\hat k$ ⁠, that is, the components are 1, 1 and 2.
To find: The unit vector $\hat a$ in the direction of $\vec a$ ⁠.
Plan: (1) Compute $|\vec a|$ ⁠. (2) Form $\hat a = \frac{1}{|\vec a|}\vec a$ ⁠. (3) Write it component by component. (4) Check that $|\hat a| = 1$ and that $\hat a$ points the same way as $\vec a$ ⁠.

Basics you need

1Magnitude from components

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

2Multiplying by a scalar

$\lambda\vec a$ has magnitude $|\lambda|\,|\vec a|$ and the same direction as $\vec a$ when $\lambda > 0$ (Section 10.5); in components, $\lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k$ (Section 10.5.1, (iv)). Lesson: components

3The unit vector

For $\vec a \ne \vec 0$ ⁠, $\hat a = \frac{1}{|\vec a|}\vec a$ is the unit vector in the direction of $\vec a$ (Section 10.5). Lesson: six kinds of vectors worth naming

Solution

Find the magnitude
$\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + 2^2} \\ &= \sqrt{1 + 1 + 4} = \sqrt6. \end{aligned}$
The magnitude of $x\hat i + y\hat j + z\hat k$ is the distance from O to the point $(x, y, z)$ ⁠, here A(1, 1, 2). The $\hat i$ and $\hat j$ terms have no number in front, so their components are 1. Since $|\vec a| = \sqrt6 \ne 0$ ⁠, $\vec a$ is not the zero vector, so a unit vector along it exists.

$|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice
Divide by the magnitude
$\hat a = \frac{1}{|\vec a|}\vec a = \frac{1}{\sqrt6}(\hat i + \hat j + 2\hat k).$
With $\lambda = \frac{1}{\sqrt6} > 0$ ⁠, the vector $\lambda\vec a$ points the same way as $\vec a$ and has length $\frac{1}{\sqrt6}\cdot\sqrt6 = 1$ ⁠. Those are exactly the two things a "unit vector in the direction of $\vec a$ " must have.

$\hat a = \frac{1}{|\vec a|}\vec a$ (Section 10.5). Lesson: components
Write it component by component
$\hat a = \frac{1}{\sqrt6}\hat i + \frac{1}{\sqrt6}\hat j + \frac{2}{\sqrt6}\hat k.$
This is the required unit vector.

Multiplying a vector by a scalar multiplies each component by that scalar: $1 \cdot \frac{1}{\sqrt6}$ ⁠, $1 \cdot \frac{1}{\sqrt6}$ and $2 \cdot \frac{1}{\sqrt6}$ ⁠. After rationalising, the same vector is $\frac{\sqrt6}{6}\hat i + \frac{\sqrt6}{6}\hat j + \frac{\sqrt6}{3}\hat k$ ⁠.

$\lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k$ (Section 10.5.1, (iv)). Lesson: components
Check the length and the direction
$|\hat a|^2 = \frac16 + \frac16 + \frac46 = 1.$
So $|\hat a| = 1$ ⁠. And $\hat a = \frac{1}{\sqrt6}\vec a$ is a positive multiple of $\vec a$ ⁠, so it points the same way.

The vector $\vec a = \overrightarrow{OA}$ to A(1, 1, 2) is the diagonal of the box with edges 1, 1 and 2 along the axes (the components, dashed). The unit vector $\hat a$ lies along it, with the same direction and $\frac{1}{\sqrt6}$ of the length. Drawn in a true (orthographic) view with one scale on all three axes.

These checks are the two halves of the definition: length 1, and the direction of $\vec a$ ⁠. Dividing by 6 instead of $\sqrt6$ would fail the first check, since the squares would add to $\frac{6}{36} = \frac16$ ⁠, not 1.

A unit vector has magnitude 1 (Section 10.3); $|\lambda\vec a| = |\lambda||\vec a|$ (Section 10.5). This is the same method as Example 6.

Result

$\hat a = \frac{1}{\sqrt6}(\hat i + \hat j + 2\hat k) = \frac{1}{\sqrt6}\hat i + \frac{1}{\sqrt6}\hat j + \frac{2}{\sqrt6}\hat k$ ⁠.

(Textbook answer: $\frac{1}{\sqrt6}\hat i + \frac{1}{\sqrt6}\hat j + \frac{2}{\sqrt6}\hat k$ ⁠.)

Check it with numbers

$\sqrt6 \approx 2.449490$ ⁠, so $\hat a \approx 0.408248\hat i + 0.408248\hat j + 0.816497\hat k$ ⁠: each component is $\frac{1}{\sqrt6} \approx 0.408248$ times the matching component 1, 1, 2 of $\vec a$ ⁠.

The angles with the axes: $\cos^{-1}(0.408248) \approx 65.905^\circ$ for OX and OY, and $\cos^{-1}(0.816497) \approx 35.264^\circ$ for OZ, so $\vec a$ leans most towards the z-axis, as the figure shows.

Common mistakes

Dividing by $|\vec a|^2 = 6$ instead of $|\vec a| = \sqrt6$ ⁠. The result then has length $\frac{1}{\sqrt6}$ ⁠, not 1.
Not squaring the 2. $|\vec a| = \sqrt{1 + 1 + 4}$ ⁠, not $\sqrt{1 + 1 + 2} = 2$ ⁠.
Dividing only one component. Every component is divided by $\sqrt6$ ⁠, including the 1s.

What to take away

$\hat a = \frac{1}{|\vec a|}\vec a$ ⁠: find the length, then divide every component by it.
The squares of the components of a unit vector add to 1; that is a quick check.
A positive multiple keeps the direction, so the components keep their ratios 1 : 1 : 2.

Every fact used, and where it comes from

Fact	Used in	Source
$\|x\hat i + y\hat j + z\hat k\| = \sqrt{x^2 + y^2 + z^2}$	Steps 1, 4	Section 10.5.1 (Basics: use Pythagoras twice)
$\hat a = \frac{1}{\|\vec a\|}\vec a$ for $\vec a \ne \vec 0$ ⁠; $\|\lambda\vec a\| = \|\lambda\|\|\vec a\|$	Steps 2, 4	Section 10.5
$\lambda\vec a$ by components	Step 3	Section 10.5.1, (iv) (Lesson: components)
Unit vector: magnitude 1	Step 4	Section 10.3
The answer	Result	NCERT answer key, Exercise 10.2 Q7 (Lesson: components, lengths and unit vectors)

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Exercise 10.2 · Q7: Unit vector along i^+j^+2k^\hat i+\hat j+2\hat ki^+j^​+2k^

Basics you need

Solution

Find the magnitude

Divide by the magnitude

Write it component by component

Check the length and the direction

Every fact used, and where it comes from

Exercise 10.2 · Q7: Unit vector along $\hat i+\hat j+2\hat k$