Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q10: Values of $p$ that make two lines perpendicular

Exercise 11.2Perpendicular lines · Section 11.4

The question

Find the values of $p$ so that the lines $\frac{1 - x}{3} = \frac{7y - 14}{2p} = \frac{z - 3}{2}$ and $\frac{7 - 7x}{3p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles.

Exam answer

Writing each fraction in the form $\frac{x - x_1}{a}$ ⁠:

\begin{aligned} \frac{1 - x}{3} &= \frac{x - 1}{-3} \\ \frac{7y - 14}{2p} &= \frac{y - 2}{2p/7} \\ \frac{7 - 7x}{3p} &= \frac{x - 1}{-3p/7} \\ \frac{6 - z}{5} &= \frac{z - 6}{-5} \end{aligned}

So the lines are

\frac{x - 1}{-3} = \frac{y - 2}{2p/7} = \frac{z - 3}{2}

\frac{x - 1}{-3p/7} = \frac{y - 5}{1} = \frac{z - 6}{-5}

Their direction ratios are $-3, \frac{2p}{7}, 2$ and $-\frac{3p}{7}, 1, -5$ ⁠.

The lines are perpendicular, so $a_1a_2 + b_1b_2 + c_1c_2 = 0$ ⁠:

\begin{aligned} &(-3)\left(-\frac{3p}{7}\right) + \frac{2p}{7}(1) \\ &\quad + 2(-5) = 0 \\ \Rightarrow\ &\frac{9p}{7} + \frac{2p}{7} - 10 = 0 \\ \Rightarrow\ &\frac{11p}{7} = 10 \end{aligned}

$\therefore\ p = \dfrac{70}{11}$ ⁠.

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

The one idea this problem is built on

Two lines are at right angles exactly when $a_1a_2 + b_1b_2 + c_1c_2 = 0$ for their direction ratios. But the direction ratios can only be read off once each line is in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ ⁠, with $x$ ⁠, $y$ ⁠, $z$ each having coefficient $+1$ ⁠. Neither line here is in that form yet, so we rewrite them first.

Given, to find, and the plan

Given: $\frac{1 - x}{3} = \frac{7y - 14}{2p} = \frac{z - 3}{2}$ and $\frac{7 - 7x}{3p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ ⁠, with $p \ne 0$ (it is in denominators).
To find: Every value of $p$ for which the lines are perpendicular.
Plan: (1) Put the first line in standard form. (2) Put the second line in standard form. (3) Apply the perpendicularity condition and solve for $p$ ⁠. (4) Check by substituting back.

Basics you need

1Standard Cartesian form

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ ⁠: point $(x_1, y_1, z_1)$ ⁠, direction ratios $a, b, c$ (Section 11.3.1). The coefficients of $x$ ⁠, $y$ ⁠, $z$ must be $+1$ ⁠. 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

3Moving a factor into the denominator

$\frac{kA}{B} = \frac{A}{B/k}$ for $k \ne 0$ ⁠, and $\frac{-A}{B} = \frac{A}{-B}$ ⁠. These keep the value of each fraction, so the line does not change.

Solution

Standard form of the first line
$\begin{aligned} \frac{1 - x}{3} &= \frac{-(x - 1)}{3} = \frac{x - 1}{-3}, \\ \frac{7y - 14}{2p} &= \frac{7(y - 2)}{2p} = \frac{y - 2}{2p/7}. \end{aligned}$
So the first line is $\frac{x - 1}{-3} = \frac{y - 2}{2p/7} = \frac{z - 3}{2}$ ⁠, with direction ratios $-3, \frac{2p}{7}, 2$ ⁠.

In $1 - x$ the coefficient of $x$ is $-1$ ⁠; taking the minus sign into the denominator makes it $+1$ ⁠. In $7y - 14$ the coefficient of $y$ is 7; take out the factor 7 and divide the denominator by it. The value of each fraction is unchanged, so this is the same line, now in a form whose denominators are direction ratios. The line passes through $(1, 2, 3)$ ⁠.

Cartesian equation (3), Section 11.3.1. Basics: point plus direction
Standard form of the second line
$\begin{aligned} \frac{7 - 7x}{3p} &= \frac{-7(x - 1)}{3p} = \frac{x - 1}{-3p/7}, \\ \frac{6 - z}{5} &= \frac{-(z - 6)}{5} = \frac{z - 6}{-5}. \end{aligned}$
So the second line is $\frac{x - 1}{-3p/7} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ ⁠, with direction ratios $-\frac{3p}{7}, 1, -5$ ⁠.

Here the coefficient of $x$ is $-7$ ⁠, so we divide the denominator by $-7$ ⁠: $\frac{3p}{-7} = -\frac{3p}{7}$ ⁠. The middle fraction is already standard. In $6 - z$ the coefficient of $z$ is $-1$ ⁠, so the denominator 5 becomes $-5$ ⁠. The line passes through $(1, 5, 6)$ ⁠.

Cartesian equation (3), Section 11.3.1. Basics: point plus direction
Apply the condition for right angles
$\begin{aligned} &(-3)\left(-\frac{3p}{7}\right) + \frac{2p}{7}(1) + 2(-5) = 0 \\ \Rightarrow\ &\frac{9p}{7} + \frac{2p}{7} - 10 = 0 \\ \Rightarrow\ &\frac{11p}{7} = 10 \ \Rightarrow\ p = \frac{70}{11}. \end{aligned}$
Multiply matching direction ratios and add, then set the sum to 0. The equation is linear in $p$ ⁠, so it has exactly one solution: although the question says "values", there is only one value of $p$ ⁠. It is not 0, so the denominators $2p$ and $3p$ of the given equations make sense.

Section 11.4, condition (i) for perpendicular lines. Lesson: perpendicular lines
Check by substituting back

With $p = \frac{70}{11}$ ⁠: $\frac{2p}{7} = \frac{20}{11}$ and $\frac{3p}{7} = \frac{30}{11}$ ⁠. The direction ratios become $-3, \frac{20}{11}, 2$ and $-\frac{30}{11}, 1, -5$ ⁠; multiplying each set by 11 gives $-33, 20, 22$ and $-30, 11, -55$ ⁠. Then
$\begin{aligned} &(-33)(-30) + 20(11) + 22(-55) \\ &= 990 + 220 - 1210 = 0. \end{aligned}$
Multiplying a set of direction ratios by the same nonzero number gives direction ratios of the same line, so whole numbers can be used for the check. The sum is 0, so the lines are at right angles.

Direction ratios are fixed only up to a nonzero multiple (Section 11.2). Basics: direction cosines and ratios

Result

The lines are at right angles for exactly one value, $p = \frac{70}{11}$ ⁠.

(Textbook answer: $p = \frac{70}{11}$ ⁠.)

Check it with numbers

$p = \frac{70}{11} \approx 6.363636$ ⁠. Then $\frac{2p}{7} \approx 1.818182$ and $-\frac{3p}{7} \approx -2.727273$ ⁠, and

$(-3)(-2.727273) + 1.818182 + 2(-5)$ $= 8.181819 + 1.818182 - 10 = 0.000001$ ⁠,

which is 0 up to the rounding of the decimals.

Common mistakes

Reading the denominators directly as $3, 2p, 2$ and $3p, 1, 5$ ⁠. That gives $9p + 2p + 10 = 0$ ⁠, $p = -\frac{10}{11}$ ⁠, which is wrong: the numerators $1 - x$ ⁠, $7y - 14$ ⁠, $7 - 7x$ ⁠, $6 - z$ are not in standard form.
Fixing the sign but not the factor 7 (or the other way round): every coefficient of $x$ ⁠, $y$ ⁠, $z$ must become $+1$ ⁠.
Sign slip in $-\frac{3p}{7}$ : dividing $3p$ by $-7$ gives a negative direction ratio.

What to take away

Before reading direction ratios, make the coefficients of $x$ ⁠, $y$ ⁠, $z$ equal to $+1$ ⁠: divide the denominator by the coefficient.
Perpendicular lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ ⁠; this gives an equation for the unknown.
Check by substituting the value back, scaling the direction ratios to whole numbers if that helps.

Every fact used, and where it comes from

Fact	Used in	Source
Standard Cartesian form; its denominators are direction ratios	Steps 1, 2	Section 11.3.1, equation (3) (Basics)
Perpendicular lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$	Steps 3, 4	Section 11.4, condition (i) (Lesson)
Direction ratios are proportional to the direction cosines; any nonzero multiple serves	Step 4	Section 11.2 (Basics)

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Exercise 11.2 · Q10: Values of ppp that make two lines perpendicular

Basics you need

Solution

Standard form of the first line

Standard form of the second line

Apply the condition for right angles

Check by substituting back

Every fact used, and where it comes from

Exercise 11.2 · Q10: Values of $p$ that make two lines perpendicular