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Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q9: Angle between two pairs of lines in Cartesian form

Exercise 11.2Angle between two lines, Cartesian form · Section 11.4
The question

Find the angle between the following pair of lines:

(i) x−22=y−15=z+3−3\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{-3} and x+2−1=y−48=z−54\frac{x + 2}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}

(ii) x2=y2=z1\frac{x}{2} = \frac{y}{2} = \frac{z}{1} and x−54=y−21=z−38\frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8}

Exam answer

If θ\theta is the angle between lines with direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2⁠, then

cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12×1a22+b22+c22\begin{aligned} \cos\theta &= \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}} \\ &\quad \times \frac{1}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \end{aligned}

(i) The direction ratios are 2,5,−32, 5, -3 and −1,8,4-1, 8, 4⁠.

a1a2+b1b2+c1c2=2(−1)+5(8)+(−3)(4)=−2+40−12=26\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(-1) + 5(8) + (-3)(4) \\ &= -2 + 40 - 12 = 26 \end{aligned} 4+25+9=381+64+16=81=9\begin{aligned} &\sqrt{4 + 25 + 9} = \sqrt{38} \\ &\sqrt{1 + 64 + 16} = \sqrt{81} = 9 \end{aligned} cos⁡θ=∣26938∣=26938\cos\theta = \left|\frac{26}{9\sqrt{38}}\right| = \frac{26}{9\sqrt{38}}

∴ θ=cos⁡−1(26938)\therefore\ \theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)⁠.

(ii) The direction ratios are 2,2,12, 2, 1 and 4,1,84, 1, 8⁠.

a1a2+b1b2+c1c2=2(4)+2(1)+1(8)=8+2+8=18\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(4) + 2(1) + 1(8) \\ &= 8 + 2 + 8 = 18 \end{aligned} 4+4+1=9=316+1+64=81=9\begin{aligned} &\sqrt{4 + 4 + 1} = \sqrt9 = 3 \\ &\sqrt{16 + 1 + 64} = \sqrt{81} = 9 \end{aligned} cos⁡θ=∣183×9∣=1827=23\cos\theta = \left|\frac{18}{3 \times 9}\right| = \frac{18}{27} = \frac23

∴ θ=cos⁡−1(23)\therefore\ \theta = \cos^{-1}\left(\dfrac23\right)⁠.

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

The one idea this problem is built on

In the Cartesian form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠, the denominators are the direction ratios, and the angle between two lines depends only on their directions. So the angle formula uses the two sets of denominators and nothing else.

Given, to find, and the plan
Given
Two pairs of lines in Cartesian form: (i) denominators 2,5,−32, 5, -3 and −1,8,4-1, 8, 4⁠; (ii) denominators 2,2,12, 2, 1 and 4,1,84, 1, 8⁠.
To find
The angle θ\theta between the lines of each pair.
Plan
For each part: (1) read off the direction ratios; (2) compute a1a2+b1b2+c1c2a_1a_2 + b_1b_2 + c_1c_2⁠; (3) compute the two lengths; (4) divide and name θ\theta⁠. Then check with sin⁡θ\sin\theta⁠.

Basics you need

1Reading the Cartesian form

The line x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} passes through (x1,y1,z1)(x_1, y_1, z_1) and has direction ratios a,b,ca, b, c (Section 11.3.1). Basics: a line is a point plus a direction

2Angle between two lines, Cartesian form

For direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2⁠, the acute angle θ\theta between the lines has

cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12×1a22+b22+c22\begin{aligned} \cos\theta &= \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}} \\ &\quad \times \frac{1}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \end{aligned}

(Section 11.4). Lesson: the angle between two lines

3The dot product behind it

The top is the dot product of a1i^+b1j^+c1k^a_1\hat i + b_1\hat j + c_1\hat k and a2i^+b2j^+c2k^a_2\hat i + b_2\hat j + c_2\hat k⁠; the bottom is the product of their lengths. Basics: the dot product

Part (i): direction ratios 2,5,−32, 5, -3 and −1,8,4-1, 8, 4

  1. Read off the direction ratios

    First line: a1,b1,c1=2,5,−3a_1, b_1, c_1 = 2, 5, -3⁠. Second line: a2,b2,c2=−1,8,4a_2, b_2, c_2 = -1, 8, 4⁠.

    Both lines are already in the form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠, with coefficient 1 in front of xx⁠, yy⁠, zz⁠, so the denominators are the direction ratios, signs included. The numerators only give the points (2,1,−3)(2, 1, -3) and (−2,4,5)(-2, 4, 5)⁠.
    Cartesian equation (3) of Section 11.3.1. Basics: point plus direction

  2. The top

    a1a2+b1b2+c1c2=2(−1)+5(8)+(−3)(4)=−2+40−12=26.\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(-1) + 5(8) + (-3)(4) \\ &= -2 + 40 - 12 = 26. \end{aligned}
    Multiply matching direction ratios and add. Two of the three products are negative here, so add carefully: 40−2−12=2640 - 2 - 12 = 26⁠.
    a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠. Basics: the dot product

  3. The bottom

    22+52+(−3)2=4+25+9=38,(−1)2+82+42=1+64+16=81=9.\begin{aligned} &\sqrt{2^2 + 5^2 + (-3)^2} \\ &= \sqrt{4 + 25 + 9} = \sqrt{38}, \\ &\sqrt{(-1)^2 + 8^2 + 4^2} \\ &= \sqrt{1 + 64 + 16} = \sqrt{81} = 9. \end{aligned}
    These are the lengths of the two direction vectors. 38=2×1938 = 2 \times 19 has no square factor, so 38\sqrt{38} does not simplify; 81=9281 = 9^2⁠.
    Length of a vector. Basics: Pythagoras in 3D

  4. Divide and name the angle

    cos⁡θ=∣26938∣=26938,θ=cos⁡−1(26938).\begin{aligned} \cos\theta &= \left|\frac{26}{9\sqrt{38}}\right| = \frac{26}{9\sqrt{38}}, \\ \theta &= \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right). \end{aligned}
    The fraction is positive and less than 1 (262=676<81×38=307826^2 = 676 < 81 \times 38 = 3078⁠), so θ\theta is an acute angle. Rationalising gives the equal form 2638342=1338171\frac{26\sqrt{38}}{342} = \frac{13\sqrt{38}}{171}⁠; the answer key keeps 26938\frac{26}{9\sqrt{38}}⁠.
    Section 11.4, Cartesian form; principal value of cos⁡−1\cos^{-1}⁠. Lesson: compare the directions

Part (ii): direction ratios 2,2,12, 2, 1 and 4,1,84, 1, 8

  1. Read off the direction ratios

    x2=y2=z1\frac{x}{2} = \frac{y}{2} = \frac{z}{1} is x−02=y−02=z−01\frac{x - 0}{2} = \frac{y - 0}{2} = \frac{z - 0}{1}⁠, so a1,b1,c1=2,2,1a_1, b_1, c_1 = 2, 2, 1⁠. The second line gives a2,b2,c2=4,1,8a_2, b_2, c_2 = 4, 1, 8⁠.

    A numerator with no number subtracted means the coordinate of the point is 0: the first line passes through the origin. The denominators are still the direction ratios.
    Cartesian equation (3) of Section 11.3.1. Basics: point plus direction

  2. The top and the bottom

    a1a2+b1b2+c1c2=2(4)+2(1)+1(8)=18,4+4+1=3,16+1+64=9.\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(4) + 2(1) + 1(8) = 18, \\ &\sqrt{4 + 4 + 1} = 3, \\ &\sqrt{16 + 1 + 64} = 9. \end{aligned}
    8+2+8=188 + 2 + 8 = 18⁠. Both lengths are whole numbers because 99 and 8181 are perfect squares.
    Dot product and length in components. Basics: the dot product

  3. Divide and name the angle

    cos⁡θ=∣183×9∣=1827=23,θ=cos⁡−1(23).\begin{aligned} \cos\theta &= \left|\frac{18}{3 \times 9}\right| = \frac{18}{27} = \frac23, \\ \theta &= \cos^{-1}\left(\frac23\right). \end{aligned}
    X Y Z θ along (4, 1, 8) along (2, 2, 1) O
    Oblique projection, not to scale, so the angle drawn is not the true size. Unit arrows from OO along the direction ratios 2,2,12, 2, 1 and 4,1,84, 1, 8⁠; the angle θ\theta between the lines is the angle between these directions.
    Cancel the common factor 9: 1827=23\frac{18}{27} = \frac{2}{3}⁠. It is positive and less than 1, so θ\theta is acute.
    Section 11.4, Cartesian form. Lesson: compare the directions

  4. Check both parts with sin⁡θ\sin\theta

    By equation (2) of Section 11.4, sin⁡θ\sin\theta uses a1b2−a2b1a_1b_2 - a_2b_1⁠, b1c2−b2c1b_1c_2 - b_2c_1⁠, c1a2−c2a1c_1a_2 - c_2a_1⁠. In part (i) these are 2121⁠, 4444⁠, −5-5⁠; in part (ii) they are −6-6⁠, 1515⁠, −12-12⁠. So

    (i)  sin⁡2θ=441+1936+2538×81=24023078,(ii)  sin⁡2θ=36+225+1449×81=405729=59,\begin{aligned} \text{(i)}\ \ \sin^2\theta &= \frac{441 + 1936 + 25}{38 \times 81} \\ &= \frac{2402}{3078}, \\ \text{(ii)}\ \ \sin^2\theta &= \frac{36 + 225 + 144}{9 \times 81} \\ &= \frac{405}{729} = \frac59, \end{aligned}

    and 6763078+24023078=1\frac{676}{3078} + \frac{2402}{3078} = 1⁠, 49+59=1\frac49 + \frac59 = 1⁠.

    The three brackets are the components of the cross product of the direction vectors, so sin⁡θ\sin\theta measures the same angle. In part (i), a1b2−a2b1=2(8)−(−1)(5)=21a_1b_2 - a_2b_1 = 2(8) - (-1)(5) = 21⁠, b1c2−b2c1=5(4)−8(−3)=44b_1c_2 - b_2c_1 = 5(4) - 8(-3) = 44 and c1a2−c2a1=(−3)(−1)−4(2)=−5c_1a_2 - c_2a_1 = (-3)(-1) - 4(2) = -5⁠. The squares of cos⁡θ\cos\theta and sin⁡θ\sin\theta add up to 1 in both parts, as they must.
    Section 11.4, equation (2); ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta⁠. Basics: the cross product
Result

(i) θ=cos⁡−1(26938)\theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)⁠. (ii) θ=cos⁡−1(23)\theta = \cos^{-1}\left(\frac23\right)⁠.

(Textbook answer: (i) θ=cos⁡−1(26938)\theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)⁠; (ii) θ=cos⁡−1(23)\theta = \cos^{-1}\left(\frac23\right)⁠.)

Check it with numbers

(i) 38≈6.164414\sqrt{38} \approx 6.164414⁠, so cos⁡θ=26938≈0.468640\cos\theta = \frac{26}{9\sqrt{38}} \approx 0.468640 and θ≈62.05∘\theta \approx 62.05^\circ⁠.

(ii) cos⁡θ=23≈0.666667\cos\theta = \frac23 \approx 0.666667⁠, so θ≈48.19∘\theta \approx 48.19^\circ⁠.

Common mistakes
  1. Using the numerators. 2,1,−32, 1, -3 and −2,4,5-2, 4, 5 in part (i) are coordinates of points, not directions.
  2. Dropping the minus signs of the denominators −3-3 and −1-1 in part (i): that gives 4+40+12=564 + 40 + 12 = 56 instead of 26.
  3. Reading x2=y2=z1\frac{x}{2} = \frac{y}{2} = \frac{z}{1} as having no direction ratios. It is a line through the origin with direction ratios 2,2,12, 2, 1⁠.
What to take away
  1. In Cartesian form the denominators, with their signs, are the direction ratios.
  2. For cos⁡θ\cos\theta⁠, the top is ∣a1a2+b1b2+c1c2∣|a_1a_2 + b_1b_2 + c_1c_2| and the bottom is the product of the two lengths a2+b2+c2\sqrt{a^2 + b^2 + c^2}⁠; the modulus gives the acute angle.
  3. The same method in vector form is Exercise 11.2 Q8.

Every fact used, and where it comes from

FactUsed inSource
The denominators of the Cartesian form are direction ratiosPart (i), Step 1; Part (ii), Step 1Section 11.3.1, equation (3) (Basics)
Angle between lines from their direction ratios, θ\theta acutePart (i), Steps 2 to 4; Part (ii), Steps 2, 3Section 11.4, Cartesian form (Lesson)
sin⁡θ\sin\theta by equation (2)Part (ii), Step 4Section 11.4, equation (2) (Basics)
Dot product in componentsPart (i), Step 2; Part (ii), Step 2Section 10.6.1 (Basics)
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