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Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q8: Angle between two pairs of lines in vector form

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

Find the angle between the following pairs of lines:

(i) r⃗=2i^−5j^+k^+λ(3i^+2j^+6k^)\vec r = 2\hat i - 5\hat j + \hat k + \lambda(3\hat i + 2\hat j + 6\hat k) and

r⃗=7i^−6k^+μ(i^+2j^+2k^)\vec r = 7\hat i - 6\hat k + \mu(\hat i + 2\hat j + 2\hat k)

(ii) r⃗=3i^+j^−2k^+λ(i^−j^−2k^)\vec r = 3\hat i + \hat j - 2\hat k + \lambda(\hat i - \hat j - 2\hat k) and

r⃗=2i^−j^−56k^+μ(3i^−5j^−4k^)\vec r = 2\hat i - \hat j - 56\hat k + \mu(3\hat i - 5\hat j - 4\hat k)

Exam answer

If θ\theta is the angle between r⃗=a⃗1+λb⃗1\vec r = \vec a_1 + \lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r = \vec a_2 + \mu\vec b_2⁠, then

cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣ ∣b⃗2∣∣\cos\theta = \left|\frac{\vec b_1\cdot\vec b_2}{|\vec b_1|\,|\vec b_2|}\right|

(i) Here b⃗1=3i^+2j^+6k^\vec b_1 = 3\hat i + 2\hat j + 6\hat k and b⃗2=i^+2j^+2k^\vec b_2 = \hat i + 2\hat j + 2\hat k⁠.

b⃗1⋅b⃗2=3(1)+2(2)+6(2)=3+4+12=19\begin{aligned} \vec b_1\cdot\vec b_2 &= 3(1) + 2(2) + 6(2) \\ &= 3 + 4 + 12 = 19 \end{aligned} ∣b⃗1∣=9+4+36=7∣b⃗2∣=1+4+4=3\begin{aligned} |\vec b_1| &= \sqrt{9 + 4 + 36} = 7 \\ |\vec b_2| &= \sqrt{1 + 4 + 4} = 3 \end{aligned} cos⁡θ=∣197×3∣=1921\cos\theta = \left|\frac{19}{7 \times 3}\right| = \frac{19}{21}

∴ θ=cos⁡−1(1921)\therefore\ \theta = \cos^{-1}\left(\frac{19}{21}\right)⁠.

(ii) Here b⃗1=i^−j^−2k^\vec b_1 = \hat i - \hat j - 2\hat k and b⃗2=3i^−5j^−4k^\vec b_2 = 3\hat i - 5\hat j - 4\hat k⁠.

b⃗1⋅b⃗2=1(3)+(−1)(−5)+(−2)(−4)=3+5+8=16\begin{aligned} \vec b_1\cdot\vec b_2 &= 1(3) + (-1)(-5) \\ &\quad + (-2)(-4) \\ &= 3 + 5 + 8 = 16 \end{aligned} ∣b⃗1∣=1+1+4=6∣b⃗2∣=9+25+16=50=52\begin{aligned} |\vec b_1| &= \sqrt{1 + 1 + 4} = \sqrt6 \\ |\vec b_2| &= \sqrt{9 + 25 + 16} \\ &= \sqrt{50} = 5\sqrt2 \end{aligned} cos⁡θ=∣166×52∣=16512=16103=853\begin{aligned} \cos\theta &= \left|\frac{16}{\sqrt6 \times 5\sqrt2}\right| \\ &= \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt3} \\ &= \frac{8}{5\sqrt3} \end{aligned}

∴ θ=cos⁡−1(853)\therefore\ \theta = \cos^{-1}\left(\dfrac{8}{5\sqrt3}\right)⁠.

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

The one idea this problem is built on

The angle between two lines is the angle between their directions. In r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b the direction is b⃗\vec b⁠, the vector multiplied by the parameter; the point a⃗\vec a plays no part. So for each pair, read off b⃗1\vec b_1 and b⃗2\vec b_2 and use cos⁡θ=∣b⃗1⋅b⃗2∣∣b⃗1∣ ∣b⃗2∣\cos\theta = \frac{|\vec b_1\cdot\vec b_2|}{|\vec b_1|\,|\vec b_2|}⁠, where the modulus makes θ\theta the acute angle.

Given, to find, and the plan
Given
Two pairs of lines in vector form: (i) directions 3i^+2j^+6k^3\hat i + 2\hat j + 6\hat k and i^+2j^+2k^\hat i + 2\hat j + 2\hat k⁠; (ii) directions i^−j^−2k^\hat i - \hat j - 2\hat k and 3i^−5j^−4k^3\hat i - 5\hat j - 4\hat k⁠.
To find
The angle θ\theta between the lines of each pair.
Plan
For each part: (1) read off b⃗1\vec b_1⁠, b⃗2\vec b_2⁠; (2) compute b⃗1⋅b⃗2\vec b_1\cdot\vec b_2⁠; (3) compute the two lengths; (4) divide, simplify and name θ\theta⁠. Then check with sin⁡θ\sin\theta⁠.

Basics you need

1Reading a vector equation

In r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, the vector multiplied by the parameter is the direction b⃗\vec b⁠; the other one, a⃗\vec a⁠, is a point of the line (Section 11.3.1). Basics: go to A, then walk λ copies of b

2Angle between two lines

If θ\theta is the acute angle between r⃗=a⃗1+λb⃗1\vec r = \vec a_1 + \lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r = \vec a_2 + \mu\vec b_2⁠, then cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣∣b⃗2∣∣\cos\theta = \left|\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}\right| (Section 11.4). Lesson: the angle between two lines

3The dot product and length

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 and ∣a⃗∣=a12+a22+a32|\vec a| = \sqrt{a_1^2 + a_2^2 + a_3^2} (Section 10.6.1). Basics: the dot product

4Simplifying surds

50=52\sqrt{50} = 5\sqrt2 and 6 2=12=23\sqrt6\,\sqrt2 = \sqrt{12} = 2\sqrt3⁠. Basics: Pythagoras in 3D

Part (i): directions 3i^+2j^+6k^3\hat i + 2\hat j + 6\hat k and i^+2j^+2k^\hat i + 2\hat j + 2\hat k

  1. Read off the directions

    b⃗1=3i^+2j^+6k^\vec b_1 = 3\hat i + 2\hat j + 6\hat k (multiplied by λ\lambda⁠) and b⃗2=i^+2j^+2k^\vec b_2 = \hat i + 2\hat j + 2\hat k (multiplied by μ\mu⁠).

    The points 2i^−5j^+k^2\hat i - 5\hat j + \hat k and 7i^−6k^7\hat i - 6\hat k only say where the lines are, not which way they run, so they do not affect the angle. In the second point there is no j^\hat j term: its yy-coordinate is 0.
    Vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, Section 11.3.1. Basics: a⃗ + λb⃗

  2. The dot product

    b⃗1⋅b⃗2=3(1)+2(2)+6(2)=3+4+12=19.\begin{aligned} \vec b_1\cdot\vec b_2 &= 3(1) + 2(2) + 6(2) \\ &= 3 + 4 + 12 = 19. \end{aligned}
    Multiply matching components and add. The result is positive, so the angle between the two arrows b⃗1\vec b_1 and b⃗2\vec b_2 is already acute.
    Dot product in components (Section 10.6.1). Basics: the dot product

  3. The two lengths

    ∣b⃗1∣=9+4+36=49=7,∣b⃗2∣=1+4+4=9=3.\begin{aligned} |\vec b_1| &= \sqrt{9 + 4 + 36} = \sqrt{49} = 7, \\ |\vec b_2| &= \sqrt{1 + 4 + 4} = \sqrt9 = 3. \end{aligned}
    Both lengths come out whole numbers because 4949 and 99 are perfect squares, so no surds appear in this part.
    Length of a vector. Basics: Pythagoras in 3D

  4. Divide and name the angle

    cos⁡θ=∣197×3∣=1921,θ=cos⁡−1(1921).\begin{aligned} \cos\theta &= \left|\frac{19}{7 \times 3}\right| = \frac{19}{21}, \\ \theta &= \cos^{-1}\left(\frac{19}{21}\right). \end{aligned}
    0<1921<10 < \frac{19}{21} < 1⁠, so θ\theta lies between 0∘0^\circ and 90∘90^\circ⁠: the acute angle between the lines. These are the same two directions as in Example 7 (in the other order), so the angle is the same.
    Section 11.4, angle between two lines. Lesson: compare the directions

Part (ii): directions i^−j^−2k^\hat i - \hat j - 2\hat k and 3i^−5j^−4k^3\hat i - 5\hat j - 4\hat k

  1. Read off the directions

    b⃗1=i^−j^−2k^\vec b_1 = \hat i - \hat j - 2\hat k and b⃗2=3i^−5j^−4k^\vec b_2 = 3\hat i - 5\hat j - 4\hat k⁠.

    Again only the vectors multiplied by λ\lambda and μ\mu count. The book prints the second point as 2i^−j^−56k^2\hat i - \hat j - 56\hat k⁠; the large number −56-56 is just the zz-coordinate of a point of the line and has no effect on the angle.
    Vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, Section 11.3.1. Basics: a⃗ + λb⃗

  2. The dot product

    b⃗1⋅b⃗2=1(3)+(−1)(−5)+(−2)(−4)=3+5+8=16.\begin{aligned} \vec b_1\cdot\vec b_2 &= 1(3) + (-1)(-5) + (-2)(-4) \\ &= 3 + 5 + 8 = 16. \end{aligned}
    Each of the last two products is a negative times a negative, so both are positive: (−1)(−5)=5(-1)(-5) = 5 and (−2)(−4)=8(-2)(-4) = 8⁠.
    Dot product in components (Section 10.6.1). Basics: the dot product

  3. The two lengths

    ∣b⃗1∣=1+1+4=6,∣b⃗2∣=9+25+16=50=52.\begin{aligned} |\vec b_1| &= \sqrt{1 + 1 + 4} = \sqrt6, \\ |\vec b_2| &= \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt2. \end{aligned}
    Squares are never negative: (−1)2=1(-1)^2 = 1⁠, (−2)2=4(-2)^2 = 4⁠, (−5)2=25(-5)^2 = 25⁠, (−4)2=16(-4)^2 = 16⁠. Then 50=25×2=52\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2⁠.
    Length of a vector. Basics: Pythagoras in 3D

  4. Divide, simplify and name the angle

    cos⁡θ=∣166×52∣=16512=16103=853=8315.\begin{aligned} \cos\theta &= \left|\frac{16}{\sqrt6 \times 5\sqrt2}\right| = \frac{16}{5\sqrt{12}} \\ &= \frac{16}{10\sqrt3} = \frac{8}{5\sqrt3} = \frac{8\sqrt3}{15}. \end{aligned}

    So θ=cos⁡−1(853)\theta = \cos^{-1}\left(\frac{8}{5\sqrt3}\right)⁠.

    θ along (1, −1, −2) along (3, −5, −4) O
    Oblique projection, not to scale, so the angle drawn is not the true size. Unit arrows from OO along b⃗1=i^−j^−2k^\vec b_1 = \hat i - \hat j - 2\hat k and b⃗2=3i^−5j^−4k^\vec b_2 = 3\hat i - 5\hat j - 4\hat k⁠; the angle between the lines is the angle θ\theta between these directions.
    6 2=12=23\sqrt6\,\sqrt2 = \sqrt{12} = 2\sqrt3⁠, so the bottom is 5×23=1035 \times 2\sqrt3 = 10\sqrt3⁠; cancelling 2 gives 853\frac{8}{5\sqrt3}⁠, the form of the answer key, and multiplying top and bottom by 3\sqrt3 gives the equal form 8315\frac{8\sqrt3}{15}⁠. It is less than 1 because 64×3=192<225=15264 \times 3 = 192 < 225 = 15^2⁠. The dot product, 16, and the lengths, 6\sqrt6 and 525\sqrt2⁠, are the same numbers as in Example 8, so the angle is the same there too.
    Section 11.4; principal value of cos⁡−1\cos^{-1}⁠. Lesson: compare the directions

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

    ∣b⃗1×b⃗2∣=∣b⃗1∣ ∣b⃗2∣sin⁡θ|\vec b_1\times\vec b_2| = |\vec b_1|\,|\vec b_2|\sin\theta⁠. In part (i), (3i^+2j^+6k^)×(i^+2j^+2k^)=−8i^+4k^(3\hat i + 2\hat j + 6\hat k)\times(\hat i + 2\hat j + 2\hat k) = -8\hat i + 4\hat k⁠, of length 80\sqrt{80}⁠; in part (ii), (i^−j^−2k^)×(3i^−5j^−4k^)=−6i^−2j^−2k^(\hat i - \hat j - 2\hat k)\times(3\hat i - 5\hat j - 4\hat k) = -6\hat i - 2\hat j - 2\hat k⁠, of length 44\sqrt{44}⁠. So

    (i)  cos⁡2θ+sin⁡2θ=361441+80441=1,(ii)  cos⁡2θ+sin⁡2θ=6475+44300=1.\begin{aligned} \text{(i)}\ \ \cos^2\theta + \sin^2\theta &= \frac{361}{441} + \frac{80}{441} = 1, \\ \text{(ii)}\ \ \cos^2\theta + \sin^2\theta &= \frac{64}{75} + \frac{44}{300} = 1. \end{aligned}
    The cross product measures the same angle with sin⁡\sin instead of cos⁡\cos⁠, so the two must satisfy cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1⁠. In part (ii), sin⁡2θ=446×50=44300=1175\sin^2\theta = \frac{44}{6 \times 50} = \frac{44}{300} = \frac{11}{75} and 6475+1175=1\frac{64}{75} + \frac{11}{75} = 1⁠.
    ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta (Section 10.6.3). Basics: the cross product
Result

(i) θ=cos⁡−1(1921)\theta = \cos^{-1}\left(\frac{19}{21}\right)⁠. (ii) θ=cos⁡−1(853)\theta = \cos^{-1}\left(\frac{8}{5\sqrt3}\right)⁠, that is cos⁡−1(8315)\cos^{-1}\left(\frac{8\sqrt3}{15}\right)⁠.

(Textbook answer: (i) θ=cos⁡−1(1921)\theta = \cos^{-1}\left(\frac{19}{21}\right)⁠; (ii) θ=cos⁡−1(853)\theta = \cos^{-1}\left(\frac{8}{5\sqrt3}\right)⁠.)

Check it with numbers

(i) 1921≈0.904762\frac{19}{21} \approx 0.904762⁠, so θ≈25.21∘\theta \approx 25.21^\circ⁠.

(ii) 6≈2.449490\sqrt6 \approx 2.449490 and 52≈7.0710685\sqrt2 \approx 7.071068⁠, so 166×52≈0.923760\frac{16}{\sqrt6 \times 5\sqrt2} \approx 0.923760⁠, and 853≈0.923760\frac{8}{5\sqrt3} \approx 0.923760 too. So θ≈22.52∘\theta \approx 22.52^\circ⁠.

Common mistakes
  1. Using the points instead of the directions, for example 2i^−5j^+k^2\hat i - 5\hat j + \hat k and 7i^−6k^7\hat i - 6\hat k in part (i). The angle depends only on the vectors multiplied by λ\lambda and μ\mu⁠.
  2. Sign slips in the dot product of part (ii): (−1)(−5)=+5(-1)(-5) = +5 and (−2)(−4)=+8(-2)(-4) = +8⁠, so the total is 16, not −10-10 or 0.
  3. Simplifying 6×52\sqrt6 \times 5\sqrt2 wrongly. It is 512=1035\sqrt{12} = 10\sqrt3⁠, not 585\sqrt8 or 56\sqrt{56}⁠.
What to take away
  1. Angle between lines in vector form: cos⁡θ=∣b⃗1⋅b⃗2∣∣b⃗1∣ ∣b⃗2∣\cos\theta = \frac{|\vec b_1\cdot\vec b_2|}{|\vec b_1|\,|\vec b_2|}⁠, using only the direction vectors.
  2. The modulus gives the acute angle; if the dot product is negative, drop its sign.
  3. 853\frac{8}{5\sqrt3} and 8315\frac{8\sqrt3}{15} are the same number; either form is accepted.

Every fact used, and where it comes from

FactUsed inSource
Reading b⃗\vec b from r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec bPart (i), Step 1; Part (ii), Step 1Section 11.3.1 (Basics)
Dot product and length in componentsPart (i), Steps 2, 3; Part (ii), Steps 2, 3Section 10.6.1 (Basics)
cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣∣b⃗2∣∣\cos\theta = \left|\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}\right|⁠, θ\theta acutePart (i), Step 4; Part (ii), Step 4Section 11.4 (Lesson)
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta⁠; cross product as a determinantPart (ii), Step 5Section 10.6.3 (Basics)
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