Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q1: Three lines with given direction cosines are mutually perpendicular

Exercise 11.2Perpendicular lines: l₁l₂ + m₁m₂ + n₁n₂ = 0 · Section 11.4

The question

Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ ⁠; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ ⁠; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Exam answer

Let the direction cosines of the three lines be

\begin{aligned} l_1, m_1, n_1 &= \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} \\ l_2, m_2, n_2 &= \frac{4}{13}, \frac{12}{13}, \frac{3}{13} \\ l_3, m_3, n_3 &= \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \end{aligned}

Two lines are perpendicular if $l_1l_2 + m_1m_2 + n_1n_2 = 0$ ⁠.

Lines 1 and 2:

\begin{aligned} &l_1l_2 + m_1m_2 + n_1n_2 \\ &= \frac{12(4) + (-3)(12) + (-4)(3)}{169} \\ &= \frac{48 - 36 - 12}{169} = 0 \end{aligned}

Lines 2 and 3:

\begin{aligned} &l_2l_3 + m_2m_3 + n_2n_3 \\ &= \frac{4(3) + 12(-4) + 3(12)}{169} \\ &= \frac{12 - 48 + 36}{169} = 0 \end{aligned}

Lines 1 and 3:

\begin{aligned} &l_1l_3 + m_1m_3 + n_1n_3 \\ &= \frac{12(3) + (-3)(-4) + (-4)(12)}{169} \\ &= \frac{36 + 12 - 48}{169} = 0 \end{aligned}

So each pair of lines is perpendicular.

$\therefore$ The three lines are mutually perpendicular. Hence proved.

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

The one idea this problem is built on

The angle $\theta$ between two lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ has $\cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|$ ⁠. The lines are perpendicular exactly when this sum is 0. "Mutually perpendicular" means every pair, so there are three sums to check.

Given, to find, and the plan

Given: Three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ ⁠; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ ⁠; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ ⁠.
To show: Each pair of the lines is perpendicular.
Plan: (1) Name the direction cosines and check they are genuine. (2) State the test. (3) to (5) Test the three pairs. (6) Conclude.

Basics you need

1Angle between two lines

For lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ ⁠: $\cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|$ (Section 11.4, equation (3)). Lesson: the angle between two lines

2Perpendicular lines

$\theta = 90^\circ$ exactly when $\cos\theta = 0$ ⁠, that is when $l_1l_2 + m_1m_2 + n_1n_2 = 0$ (Section 11.4, condition (i)). Basics: the dot product

3Direction cosines

Genuine direction cosines satisfy $l^2 + m^2 + n^2 = 1$ ⁠. Basics: direction cosines

Solution

Name the direction cosines
$\begin{aligned} l_1, m_1, n_1 &= \tfrac{12}{13}, \tfrac{-3}{13}, \tfrac{-4}{13}, \\ l_2, m_2, n_2 &= \tfrac{4}{13}, \tfrac{12}{13}, \tfrac{3}{13}, \\ l_3, m_3, n_3 &= \tfrac{3}{13}, \tfrac{-4}{13}, \tfrac{12}{13}. \end{aligned}$
Each set has squares adding up to 1: $144 + 9 + 16 = 169$ ⁠, $16 + 144 + 9 = 169$ and $9 + 16 + 144 = 169$ ⁠, each over $169$ ⁠.

Subscripts keep track of which line each number belongs to. The squares check confirms the given numbers really are direction cosines, so the formula $\cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|$ (which assumes this) applies.

$l^2 + m^2 + n^2 = 1$ (Section 11.2). Basics: direction cosines
The test for perpendicular lines

Two lines are perpendicular when
$l_1l_2 + m_1m_2 + n_1n_2 = 0.$
$l_1\hat i + m_1\hat j + n_1\hat k$ and $l_2\hat i + m_2\hat j + n_2\hat k$ are unit vectors along the two lines, and their dot product is $\cos\theta$ ⁠. A right angle has cosine 0. Since every factor here has denominator 13, every product has denominator $13 \times 13 = 169$ ⁠, and only the numerators need adding.

$\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3$ ⁠. Basics: the dot product
Lines 1 and 2
$\begin{aligned} &l_1l_2 + m_1m_2 + n_1n_2 \\ &= \frac{12(4) + (-3)(12) + (-4)(3)}{169} \\ &= \frac{48 - 36 - 12}{169} = 0. \end{aligned}$
Multiply matching direction cosines ( $x$ with $x$ ⁠, $y$ with $y$ ⁠, $z$ with $z$ ⁠) and add. A negative times a positive is negative: $(-3)(12) = -36$ and $(-4)(3) = -12$ ⁠. The sum is 0, so lines 1 and 2 are perpendicular.

Section 11.4, condition (i). Lesson: the angle between two lines
Lines 2 and 3
$\begin{aligned} &l_2l_3 + m_2m_3 + n_2n_3 \\ &= \frac{4(3) + 12(-4) + 3(12)}{169} \\ &= \frac{12 - 48 + 36}{169} = 0. \end{aligned}$
The same test for the second pair: $12 + 36 = 48$ cancels $-48$ ⁠. So lines 2 and 3 are perpendicular.

Perpendicular: $l_1l_2 + m_1m_2 + n_1n_2 = 0$ ⁠. Exercise 11.2 player
Lines 1 and 3
$\begin{aligned} &l_1l_3 + m_1m_3 + n_1n_3 \\ &= \frac{12(3) + (-3)(-4) + (-4)(12)}{169} \\ &= \frac{36 + 12 - 48}{169} = 0. \end{aligned}$
A negative times a negative is positive: $(-3)(-4) = 12$ ⁠. The sum is again 0, so lines 1 and 3 are perpendicular. It is easy to forget this third pair; "mutually" needs all three.

Signs of products. Basics: the dot product
Conclude

Every pair of the three lines is perpendicular, so the three lines are mutually perpendicular.

"Mutually perpendicular" means each line is perpendicular to each of the other two. Three lines give exactly three pairs, $(1, 2)$ ⁠, $(2, 3)$ and $(1, 3)$ ⁠, and all three have been checked. Like the $x$ ⁠, $y$ and $z$ -axes, the three unit vectors along these lines form a set of mutually perpendicular directions.

Every pair checked. Lesson: perpendicular and parallel lines

Result

For each pair, $l_1l_2 + m_1m_2 + n_1n_2 = 0$ (the numerators are $48 - 36 - 12$ ⁠, $12 - 48 + 36$ and $36 + 12 - 48$ ⁠), so the three lines are mutually perpendicular.

(The answer key prints no answer for this "show that" question.)

Check it with numbers

As decimals the direction cosines are $0.923077, -0.230769, -0.307692$ ⁠; $0.307692, 0.923077, 0.230769$ ⁠; $0.230769, -0.307692, 0.923077$ ⁠.

Lines 1 and 2, product by product: $\frac{48}{169} \approx 0.284024$ ⁠, $-\frac{36}{169} \approx -0.213018$ and $-\frac{12}{169} \approx -0.071006$ ⁠, and $0.284024 - 0.213018$ $-\ 0.071006 = 0.000000$ ⁠. (Multiplying the rounded six-place decimals instead gives about $0.284023 - 0.213018 - 0.071006$ ⁠, which is 0 up to rounding in the last place.)

Common mistakes

Checking only one or two pairs. Three lines have three pairs; all must give 0.
Sign slips in the products, such as $(-3)(-4) = -12$ ⁠. A negative times a negative is positive.
Multiplying across the wrong positions. Pair $x$ with $x$ ⁠, $y$ with $y$ ⁠, $z$ with $z$ ⁠.

What to take away

Perpendicular lines: $l_1l_2 + m_1m_2 + n_1n_2 = 0$ (or $a_1a_2 + b_1b_2 + c_1c_2 = 0$ with direction ratios).
"Mutually perpendicular" for three lines means three separate checks.
With a common denominator, add the numerators only.

Every fact used, and where it comes from

Fact	Used in	Source
$l^2 + m^2 + n^2 = 1$	Step 1	Section 11.2 (Basics)
$\cos\theta = \|l_1l_2 + m_1m_2 + n_1n_2\|$	Step 2	Section 11.4, equation (3) (Lesson)
Perpendicular lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$	Steps 3 to 5	Section 11.4, condition (i) (Exercise 11.2 player)
Dot product in components	Step 2	Section 10.6.1 (Basics)

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 11.2 · Q1: Three lines with given direction cosines are mutually perpendicular

Basics you need

Solution

Name the direction cosines

The test for perpendicular lines

Lines 1 and 2

Lines 2 and 3

Lines 1 and 3

Conclude

Every fact used, and where it comes from