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Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q2: Line through (1,−1,2)(1, -1, 2) and (3,4,−2)(3, 4, -2) is perpendicular to the line through (0,3,2)(0, 3, 2) and (3,5,6)(3, 5, 6)

Exercise 11.2Perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0 · Section 11.4
The question

Show that the line through the points (1,−1,2)(1, -1, 2)⁠, (3,4,−2)(3, 4, -2) is perpendicular to the line through the points (0,3,2)(0, 3, 2) and (3,5,6)(3, 5, 6)⁠.

Exam answer

Let A(1,−1,2)A(1, -1, 2)⁠, B(3,4,−2)B(3, 4, -2)⁠, C(0,3,2)C(0, 3, 2) and D(3,5,6)D(3, 5, 6)⁠.

Direction ratios of ABAB⁠:

a1,b1,c1=3−1, 4+1, −2−2=2, 5, −4\begin{aligned} a_1, b_1, c_1 &= 3 - 1,\ 4 + 1,\ -2 - 2 \\ &= 2,\ 5,\ -4 \end{aligned}

Direction ratios of CDCD⁠:

a2,b2,c2=3−0, 5−3, 6−2=3, 2, 4\begin{aligned} a_2, b_2, c_2 &= 3 - 0,\ 5 - 3,\ 6 - 2 \\ &= 3,\ 2,\ 4 \end{aligned}

Two lines are perpendicular if a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0⁠.

a1a2+b1b2+c1c2=2(3)+5(2)+(−4)(4)=6+10−16=0\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(3) + 5(2) + (-4)(4) \\ &= 6 + 10 - 16 = 0 \end{aligned}

∴\therefore ABAB is perpendicular to CDCD⁠. Hence proved.

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

The one idea this problem is built on

Two points give the direction ratios of the line through them, by subtraction. Two lines are perpendicular exactly when their direction ratios satisfy a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0⁠: this is the dot product of the two direction vectors.

Given, to find, and the plan
Given
The line ABAB through A(1,−1,2)A(1, -1, 2)⁠, B(3,4,−2)B(3, 4, -2) and the line CDCD through C(0,3,2)C(0, 3, 2)⁠, D(3,5,6)D(3, 5, 6)⁠.
To show
AB⊥CDAB \perp CD⁠.
Plan
(1) Direction ratios of ABAB⁠. (2) Direction ratios of CDCD⁠. (3) Apply the perpendicularity test. (4) A closer look: the lines do not meet, and that is fine.

Basics you need

1Direction ratios of the line through two points

x2−x1,y2−y1,z2−z1x_2 - x_1, y_2 - y_1, z_2 - z_1 (Note in Section 11.2.1). Lesson: direction cosines and direction ratios

2Perpendicular lines

Lines with direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2 are perpendicular when a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0 (Section 11.4, condition (i)). Lesson: the angle between two lines

3The dot product

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=a1b1+a2b2+a3b3\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3⁠, which is 0 for perpendicular nonzero vectors. Basics: the dot product

Solution

  1. Direction ratios of ABAB

    a1,b1,c1=3−1, 4−(−1), −2−2=2, 5, −4.\begin{aligned} a_1, b_1, c_1 &= 3 - 1,\ 4 - (-1),\ -2 - 2 \\ &= 2,\ 5,\ -4. \end{aligned}
    Second point minus first point, coordinate by coordinate. 4−(−1)=4+1=54 - (-1) = 4 + 1 = 5⁠. These are the components of AB→\overrightarrow{AB}⁠, a vector along the first line.
    Direction ratios from two points (Section 11.2.1). Lesson: the line through two points

  2. Direction ratios of CDCD

    a2,b2,c2=3−0, 5−3, 6−2=3, 2, 4.\begin{aligned} a_2, b_2, c_2 &= 3 - 0,\ 5 - 3,\ 6 - 2 \\ &= 3,\ 2,\ 4. \end{aligned}
    The same rule for the second line, with CC first and DD second. These are the components of CD→\overrightarrow{CD}⁠.
    Direction ratios from two points. Basics: points in space

  3. Apply the test

    a1a2+b1b2+c1c2=2(3)+5(2)+(−4)(4)=6+10−16=0.\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(3) + 5(2) + (-4)(4) \\ &= 6 + 10 - 16 = 0. \end{aligned}

    So the lines ABAB and CDCD are perpendicular.

    For the angle θ\theta between the lines, cos⁡θ\cos\theta is ∣a1a2+b1b2+c1c2∣|a_1a_2 + b_1b_2 + c_1c_2| divided by the lengths a12+b12+c12\sqrt{a_1^2 + b_1^2 + c_1^2} and a22+b22+c22\sqrt{a_2^2 + b_2^2 + c_2^2}⁠. The numerator is 0, so cos⁡θ=0\cos\theta = 0 and θ=90∘\theta = 90^\circ⁠. Direction ratios are enough here; there is no need to turn them into direction cosines, because dividing by the lengths cannot make 0 into anything else.
    Section 11.4, equation (1) and condition (i). Exercise 11.2 player

  4. A closer look: the lines do not meet

    Points of the two lines are (1+2t,−1+5t,2−4t)(1 + 2t, -1 + 5t, 2 - 4t) and (3s,3+2s,2+4s)(3s, 3 + 2s, 2 + 4s)⁠. Equal zz gives t=−st = -s⁠; equal xx then gives 1−2s=3s1 - 2s = 3s⁠, so s=15s = \frac15⁠, t=−15t = -\frac15⁠. But then the yy-coordinates are −2-2 and 175\frac{17}{5}⁠, which differ.

    So the two lines never meet: they are skew lines. That does not change the answer. In space, the angle between two lines is defined through lines parallel to them drawn from the origin (Note in Section 11.4), so it only depends on the directions. "Perpendicular" means the directions are at 90∘90^\circ⁠, whether or not the lines meet, just as an east-west beam in a ceiling is perpendicular to a north-south beam in the floor.
    A line is a point plus a direction. Basics: a line is a point plus a direction
Result

The direction ratios 2,5,−42, 5, -4 and 3,2,43, 2, 4 give 6+10−16=06 + 10 - 16 = 0⁠, so the line through (1,−1,2)(1, -1, 2) and (3,4,−2)(3, 4, -2) is perpendicular to the line through (0,3,2)(0, 3, 2) and (3,5,6)(3, 5, 6)⁠.

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

Check it with numbers

∣AB→∣=45≈6.708204|\overrightarrow{AB}| = \sqrt{45} \approx 6.708204 and ∣CD→∣=29≈5.385165|\overrightarrow{CD}| = \sqrt{29} \approx 5.385165⁠, so cos⁡θ=036.124784=0\cos\theta = \frac{0}{36.124784} = 0 and θ=90∘\theta = 90^\circ⁠.

Reversing one direction, BA→\overrightarrow{BA} has −2,−5,4-2, -5, 4⁠, and −6−10+16=0-6 - 10 + 16 = 0 still: the test does not depend on which way each line is taken.

Common mistakes
  1. Using the points instead of the direction ratios, for example 1(0)+(−1)(3)+2(2)1(0) + (-1)(3) + 2(2)⁠. The test uses the differences of the points.
  2. Sign slips: 4−(−1)=54 - (-1) = 5⁠, not 3; (−4)(4)=−16(-4)(4) = -16⁠.
  3. Thinking perpendicular lines must meet. In space they need not; only the directions matter.
What to take away
  1. Perpendicular lines: a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0⁠, with direction ratios from the given points.
  2. Direction ratios work directly; no need to divide by lengths for this test.
  3. Perpendicular lines in space may be skew.

Every fact used, and where it comes from

FactUsed inSource
Direction ratios of PQPQ⁠: x2−x1,y2−y1,z2−z1x_2 - x_1, y_2 - y_1, z_2 - z_1Steps 1, 2Note in Section 11.2.1 (Lesson)
cos⁡θ\cos\theta from direction ratios; perpendicular when a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0Step 3Section 11.4, equation (1) and condition (i) (Lesson)
Lines not through the origin: use parallel lines through the originStep 4Note in Section 11.4
Skew lines: neither parallel nor meetingStep 4Section 11.5 (Lesson)
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