Chapter 10 · Vector Algebra

Example 16: Projection of $2\hat i+3\hat j+2\hat k$ on $\hat i+2\hat j+\hat k$

ExamplesProjection of a vector on a vector · Section 10.6.2

The question

Find the projection of the vector $\vec a = 2\hat i + 3\hat j + 2\hat k$ on the vector $\vec b = \hat i + 2\hat j + \hat k$ ⁠.

Exam answer

The projection of $\vec a$ on $\vec b$ is $\frac{1}{|\vec b|}(\vec a\cdot\vec b)$ ⁠.

\begin{aligned} \vec a\cdot\vec b &= (2)(1) + (3)(2) + (2)(1) \\ &= 2 + 6 + 2 = 10 \end{aligned}

|\vec b| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt6

\begin{aligned} \text{Projection} &= \frac{1}{\sqrt6}(10) = \frac{10}{\sqrt6} \\ &= \frac{10\sqrt6}{6} \end{aligned}

$\therefore$ the projection of $\vec a$ on $\vec b$ is $\dfrac53\sqrt6$ ⁠.

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

The one idea this problem is built on

The projection of $\vec a$ on $\vec b$ is the signed length of the shadow of $\vec a$ on the line of $\vec b$ ⁠: $|\vec a|\cos\theta$ ⁠. Since $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ ⁠, it equals

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

So we need only the scalar product and the length of the vector we project on.

Given, to find, and the plan

Given: $\vec a = 2\hat i + 3\hat j + 2\hat k$ ⁠, $\vec b = \hat i + 2\hat j + \hat k$ ⁠.
To find: The projection of $\vec a$ on $\vec b$ ⁠.
Plan: (1) Compute $\vec a\cdot\vec b$ ⁠. (2) Compute $|\vec b|$ ⁠. (3) Divide and rationalise. (4) Check that the shadow is not longer than $\vec a$ ⁠.

Basics you need

1Projection on a vector

The projection of $\vec a$ on $\vec b$ is $\vec a\cdot\hat b = \vec a\cdot\frac{\vec b}{|\vec b|} = \frac{1}{|\vec b|}(\vec a\cdot\vec b)$ (Section 10.6.2, Observation 2, Fig 10.20). Lesson: the dot product

2Scalar product in components

$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ (Section 10.6.1). Lesson: Examples 13–21

3Cosine as a shadow

A segment of length $r$ at angle $\theta$ to a line casts a shadow of signed length $r\cos\theta$ on it. Basics: cos θ is a shadow

Solution

The scalar product
$\begin{aligned} \vec a\cdot\vec b &= (2)(1) + (3)(2) + (2)(1) \\ &= 2 + 6 + 2 = 10. \end{aligned}$
Multiply the matching components of $\vec a$ and $\vec b$ and add, exactly as the book writes $2 \times 1 + 3 \times 2 + 2 \times 1$ ⁠.

$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ (Section 10.6.1). Lesson: the dot product
The length of $\vec b$
$|\vec b| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt6.$
We project on $\vec b$ ⁠, so we divide by the length of $\vec b$ ⁠, the vector that fixes the direction of the line. The length of $\vec a$ is not used.

$|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice
Divide, and rationalise
$\frac{1}{|\vec b|}(\vec a\cdot\vec b) = \frac{10}{\sqrt6} = \frac{10\sqrt6}{6} = \frac53\sqrt6.$
So the projection of $\vec a$ on $\vec b$ is $\frac{10}{\sqrt6} = \frac53\sqrt6$ .

$\vec a$ and $\vec b$ drawn in their own plane, true to scale. The perpendicular from the tip of $\vec a$ to the line $l$ of $\vec b$ marks the shadow of $\vec a$ ⁠: its length is the projection $\frac{10}{\sqrt6} \approx 4.08$ ⁠, a little less than $|\vec a| = \sqrt{17} \approx 4.12$ ⁠, because the angle between the vectors is small.

Multiplying top and bottom by $\sqrt6$ gives $\frac{10\sqrt6}{6}$ ⁠, and cancelling 2 gives $\frac{5\sqrt6}{3}$ ⁠, the book's form. The answer is positive because $\vec a\cdot\vec b > 0$ ⁠: the shadow points the same way as $\vec b$ ⁠, and the angle between the vectors is acute.

Projection of $\vec a$ on $\vec b$ (Section 10.6.2, Observation 2). Lesson: Examples 13–21
Check against the length of $\vec a$

$|\vec a| = \sqrt{4 + 9 + 4} = \sqrt{17}$ ⁠. The projection is $|\vec a|\cos\theta$ and $|\cos\theta| \le 1$ ⁠, so it can never be longer than $\sqrt{17}$ ⁠. Indeed $\left(\frac{10}{\sqrt6}\right)^2 = \frac{100}{6} = \frac{50}{3}$ ⁠, which is less than 17.

This guards against the most common slip, dividing by $|\vec a|$ instead of $|\vec b|$ ⁠: that would give $\frac{10}{\sqrt{17}}$ ⁠, the projection of $\vec b$ on $\vec a$ ⁠, which must be at most $|\vec b| = \sqrt6$ ⁠.

$|\cos\theta| \le 1$ ⁠. Basics: cos θ is a shadow

Result

The projection of $\vec a$ on $\vec b$ is $\frac{10}{\sqrt6} = \frac53\sqrt6$ ⁠.

(Textbook answer: $\frac{1}{|\vec b|}(\vec a\cdot\vec b)$ $= \frac{(2 \times 1 + 3 \times 2 + 2 \times 1)}{\sqrt{(1)^2 + (2)^2 + (1)^2}}$ $= \frac{10}{\sqrt6} = \frac53\sqrt6$ ⁠.)

Check it with numbers

$\frac{10}{\sqrt6} \approx 4.082483$ and $\frac53\sqrt6 \approx 4.082483$ (the same number), while $|\vec a| = \sqrt{17} \approx 4.123106$ ⁠.

The angle: $\cos\theta = \frac{10}{\sqrt{17}\,\sqrt6} = \frac{10}{\sqrt{102}} \approx 0.990148$ ⁠, so $\theta \approx 8.05^\circ$ ⁠, and $\sqrt{17}\cos\theta = \frac{10}{\sqrt6}$ exactly.

The projection vector (the shadow as a vector along $\vec b$ ⁠) is $(\vec a\cdot\hat b)\,\hat b = \frac{10}{6}(\hat i + 2\hat j + \hat k)$ $= \frac53(\hat i + 2\hat j + \hat k)$ ⁠, whose length is $\frac53\sqrt6$ again.

Common mistakes

Dividing by $|\vec a|$ ⁠. $\frac{10}{\sqrt{17}}$ is the projection of $\vec b$ on $\vec a$ ⁠; projecting on $\vec b$ divides by $|\vec b|$ ⁠.
Forgetting to divide at all. $\vec a\cdot\vec b = 10$ is not the projection; it is $|\vec b|$ times the projection.
Mixing up the projection and the projection vector. The question asks for the number $\frac53\sqrt6$ ⁠; the vector $\frac53(\hat i + 2\hat j + \hat k)$ is the projection vector.

What to take away

Projection of $\vec a$ on $\vec b$ ⁠: $\frac{\vec a\cdot\vec b}{|\vec b|}$ ⁠, dividing by the length of the vector projected on.
It is the signed length of the shadow, $|\vec a|\cos\theta$ ⁠, so its size is at most $|\vec a|$ ⁠.
Its sign tells whether the angle is acute (positive) or obtuse (negative).

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$	Step 1	Section 10.6.1 (Lesson: the dot product)
$\|x\hat i + y\hat j + z\hat k\| = \sqrt{x^2 + y^2 + z^2}$	Steps 2, 4	Section 10.5.1
Projection of $\vec a$ on $\vec b$ is $\vec a\cdot\hat b = \frac{1}{\|\vec b\|}(\vec a\cdot\vec b)$	Step 3	Section 10.6.2, Observation 2, Fig 10.20
$\|\cos\theta\| \le 1$	Step 4	Trigonometry (Basics: cos θ is a shadow)
Projection vector	Check	Section 10.6.2
The book's working and answer	Result	Example 16, Section 10.6.2 (Lesson: Examples 13–21)

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Example 16: Projection of 2i^+3j^+2k^2\hat i+3\hat j+2\hat k2i^+3j^​+2k^ on i^+2j^+k^\hat i+2\hat j+\hat ki^+2j^​+k^

Basics you need

Solution

The scalar product

The length of b⃗\vec bb

Divide, and rationalise

Check against the length of a⃗\vec aa

Every fact used, and where it comes from

Example 16: Projection of $2\hat i+3\hat j+2\hat k$ on $\hat i+2\hat j+\hat k$

The length of $\vec b$

Check against the length of $\vec a$