Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q11: $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular

Exercise 11.2Perpendicular lines · Section 11.4

The question

Show that the lines $\frac{x - 5}{7} = \frac{y + 2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.

Exam answer

The direction ratios of the first line are $7, -5, 1$ and of the second line $1, 2, 3$ ⁠.

Two lines are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$ ⁠.

\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 7(1) + (-5)(2) + 1(3) \\ &= 7 - 10 + 3 = 0 \end{aligned}

$\therefore$ The two lines are perpendicular to each other. Hence proved.

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

The one idea this problem is built on

Two lines are perpendicular when their directions are: the angle between the lines is the angle between their direction vectors, and that angle is $90^\circ$ exactly when the dot product of the direction ratios, $a_1a_2 + b_1b_2 + c_1c_2$ ⁠, is 0.

Given, to find, and the plan

Given: The lines $\frac{x - 5}{7} = \frac{y + 2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ ⁠.
To show: The lines are perpendicular to each other.
Plan: (1) Read off the direction ratios. (2) Compute $a_1a_2 + b_1b_2 + c_1c_2$ ⁠. (3) Conclude. (4) See that the lines do not meet, and why that does not matter.

Basics you need

1Reading the Cartesian form

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ passes through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ (Section 11.3.1). Basics: a line is a point plus a direction

2Perpendicular lines

Lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$ (Section 11.4, condition (i)). Lesson: perpendicular and parallel lines

3Why the dot product

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ ⁠, and $\cos 90^\circ = 0$ ⁠. Basics: the dot product

Solution

Read off the direction ratios

First line: $\frac{x - 5}{7} = \frac{y - (-2)}{-5} = \frac{z - 0}{1}$ ⁠, so $a_1, b_1, c_1 = 7, -5, 1$ ⁠. Second line: $\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{3}$ ⁠, so $a_2, b_2, c_2 = 1, 2, 3$ ⁠.

Both lines are in standard form, so the denominators are the direction ratios, with their signs. The first line passes through $(5, -2, 0)$ ⁠; the second passes through the origin.

Cartesian equation (3), Section 11.3.1. Basics: point plus direction
Compute $a_1a_2 + b_1b_2 + c_1c_2$
$\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 7(1) + (-5)(2) + 1(3) \\ &= 7 - 10 + 3 = 0. \end{aligned}$
Multiply matching direction ratios and add. The negative middle product $-10$ exactly cancels $7 + 3$ ⁠.

Dot product in components (Section 10.6.1). Basics: the dot product
Conclude

Since $a_1a_2 + b_1b_2 + c_1c_2 = 0$ ⁠, $\cos\theta = 0$ and $\theta = 90^\circ$ ⁠: the lines are perpendicular to each other.

In the angle formula of Section 11.4, $\cos\theta$ has top $|a_1a_2 + b_1b_2 + c_1c_2| = 0$ and bottom $\sqrt{75}\,\sqrt{14}$ ⁠, the product of the two lengths, which is not 0. So $\cos\theta = 0$ ⁠.

Section 11.4, condition (i). Lesson: perpendicular lines
The lines do not meet

A common point would satisfy $5 + 7t = s$ ⁠, $-2 - 5t = 2s$ and $t = 3s$ for some $t$ ⁠, $s$ ⁠. The first and third give $5 + 21s = s$ ⁠, so $s = -\frac14$ and $t = -\frac34$ ⁠. Then $-2 - 5t = \frac74$ ⁠, but $2s = -\frac12$ ⁠. So there is no common point.

Points of the lines are $(5 + 7t, -2 - 5t, t)$ and $(s, 2s, 3s)$ ⁠. The two lines are perpendicular but skew: in space, "perpendicular lines" means perpendicular directions, and the lines need not intersect. The question asks only about the directions, so Step 3 already proves it.

Skew lines: neither parallel nor intersecting (Section 11.5). Lesson: skew lines

Result

$7(1) + (-5)(2) + 1(3) = 0$ ⁠, so the two lines are perpendicular to each other. (They do not meet: they are perpendicular skew lines.)

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

Check it with numbers

The direction lengths are $\sqrt{49 + 25 + 1} = \sqrt{75} \approx 8.660254$ and $\sqrt{1 + 4 + 9} = \sqrt{14} \approx 3.741657$ ⁠. Unit vectors along the lines are about $(0.808290, -0.577350, 0.115470)$ and $(0.267261, 0.534522, 0.801784)$ ⁠, and

$0.216024 - 0.308606$ $+\ 0.092582 = 0.000000$

(each product of the six-place decimals, rounded to six places), which is the cosine of the angle between the lines: 0.

Common mistakes

Using the numerators $5, -2, 0$ as direction ratios. They give a point of the first line.
Dropping the minus sign of $-5$ : that gives $7 + 10 + 3 = 20 \ne 0$ ⁠.
Thinking perpendicular lines must meet. In space they need not; only the directions matter.

What to take away

Perpendicular lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ for their direction ratios.
Read the direction ratios from the denominators of the standard form, signs included.
Lines in space can be perpendicular without meeting (compare Exercise 11.2 Q2).

Every fact used, and where it comes from

Fact	Used in	Source
Denominators of the standard Cartesian form are direction ratios	Step 1	Section 11.3.1, equation (3) (Basics)
Perpendicular lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$	Steps 2, 3	Section 11.4, condition (i) (Lesson)
Angle between two lines from direction ratios	Step 3	Section 11.4, equation (1) (Basics)
Skew lines	Step 4	Section 11.5 (Lesson)

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Exercise 11.2 · Q11: x−57=y+2−5=z1\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}7x−5​=−5y+2​=1z​ and x1=y2=z3\frac{x}{1} = \frac{y}{2} = \frac{z}{3}1x​=2y​=3z​ are perpendicular

Basics you need

Solution

Read off the direction ratios

Compute a1a2+b1b2+c1c2a_1a_2 + b_1b_2 + c_1c_2a1​a2​+b1​b2​+c1​c2​

Conclude

The lines do not meet

Every fact used, and where it comes from

Exercise 11.2 · Q11: $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular

Compute $a_1a_2 + b_1b_2 + c_1c_2$