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Chapter 11 · Three Dimensional Geometry

Example 7: Angle between lines along i^+2j^+2k^\hat i + 2\hat j + 2\hat k and 3i^+2j^+6k^3\hat i + 2\hat j + 6\hat k

ExamplesAngle between two lines, vector form · Section 11.4
The question

Find the angle between the pair of lines given by

r⃗=3i^+2j^−4k^+λ(i^+2j^+2k^)\vec r = 3\hat i + 2\hat j - 4\hat k + \lambda(\hat i + 2\hat j + 2\hat k)

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

Exam answer

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

The angle θ\theta between the lines is given by

cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣ ∣b⃗2∣∣\cos\theta = \left|\frac{\vec b_1\cdot\vec b_2}{|\vec b_1|\,|\vec b_2|}\right| b⃗1⋅b⃗2=1(3)+2(2)+2(6)=3+4+12=19\begin{aligned} \vec b_1\cdot\vec b_2 &= 1(3) + 2(2) + 2(6) \\ &= 3 + 4 + 12 = 19 \end{aligned} ∣b⃗1∣=1+4+4=3∣b⃗2∣=9+4+36=7\begin{aligned} |\vec b_1| &= \sqrt{1 + 4 + 4} = 3 \\ |\vec b_2| &= \sqrt{9 + 4 + 36} = 7 \end{aligned} cos⁡θ=∣193×7∣=1921\cos\theta = \left|\frac{19}{3 \times 7}\right| = \frac{19}{21}

∴ θ=cos⁡−1(1921)\therefore\ \theta = \cos^{-1}\left(\dfrac{19}{21}\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. The points the lines pass through do not matter. So read off the direction vectors b⃗1\vec b_1 and b⃗2\vec b_2 and use the dot product: 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 lines 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 with b⃗1=i^+2j^+2k^\vec b_1 = \hat i + 2\hat j + 2\hat k and b⃗2=3i^+2j^+6k^\vec b_2 = 3\hat i + 2\hat j + 6\hat k⁠.
To find
The angle θ\theta between the lines.
Plan
(1) Read off b⃗1\vec b_1⁠, b⃗2\vec b_2⁠. (2) Write the formula for cos⁡θ\cos\theta⁠. (3) Compute the dot product. (4) Compute the two lengths. (5) Divide and name θ\theta⁠. (6) 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

2The 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 (Section 10.6.1). Basics: the dot product

3Angle 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

4Length of a vector

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}⁠. Basics: Pythagoras in 3D

Solution

  1. Read off the directions

    Comparing with 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:

    b⃗1=i^+2j^+2k^,b⃗2=3i^+2j^+6k^.\begin{aligned} \vec b_1 &= \hat i + 2\hat j + 2\hat k, \\ \vec b_2 &= 3\hat i + 2\hat j + 6\hat k. \end{aligned}
    The vectors multiplied by λ\lambda and μ\mu are the directions. The other parts, 3i^+2j^−4k^3\hat i + 2\hat j - 4\hat k and 5i^−2j^5\hat i - 2\hat j⁠, only say where each line is; moving a line parallel to itself does not change its angle with another line (the Note in Section 11.4: we may use parallel lines through the origin instead).
    Direction of r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b is b⃗\vec b⁠. Basics: a⃗ + λb⃗

  2. The formula

    cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣ ∣b⃗2∣∣.\cos\theta = \left|\frac{\vec b_1\cdot\vec b_2}{|\vec b_1|\,|\vec b_2|}\right|.
    X Y Z θ b₂ along (3, 2, 6) b₁ along (1, 2, 2) O
    Oblique projection, not to scale, so the angle drawn is not the true size. Unit arrows from OO along b⃗1\vec b_1 and b⃗2\vec b_2⁠: the angle θ\theta between the lines is the angle between these directions.
    The dot product gives cos⁡\cos of the angle between the two vectors. Two lines make two angles, θ\theta and 180∘−θ180^\circ - \theta⁠; the modulus picks the acute one, which is the angle between the lines as the book defines it.
    Section 11.4, 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. Lesson: compare the directions

  3. The dot product

    b⃗1⋅b⃗2=(1)(3)+(2)(2)+(2)(6)=3+4+12=19.\begin{aligned} \vec b_1\cdot\vec b_2 &= (1)(3) + (2)(2) + (2)(6) \\ &= 3 + 4 + 12 = 19. \end{aligned}
    Multiply matching components and add, because i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1 and the dot product of two different unit vectors among i^,j^,k^\hat i, \hat j, \hat k is 0.
    a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠. Basics: the dot product

  4. The two lengths

    ∣b⃗1∣=1+4+4=9=3,∣b⃗2∣=9+4+36=49=7.\begin{aligned} |\vec b_1| &= \sqrt{1 + 4 + 4} = \sqrt9 = 3, \\ |\vec b_2| &= \sqrt{9 + 4 + 36} = \sqrt{49} = 7. \end{aligned}
    Square each component, add, take the square root. Both sums are perfect squares here, which is why the answer comes out as a simple fraction.
    Length of a vector in space. Basics: Pythagoras twice

  5. Divide and name the angle

    cos⁡θ=∣193×7∣=1921,θ=cos⁡−1(1921).\begin{aligned} \cos\theta &= \left|\frac{19}{3 \times 7}\right| = \frac{19}{21}, \\ \theta &= \cos^{-1}\Big(\frac{19}{21}\Big). \end{aligned}
    0<1921<10 < \frac{19}{21} < 1⁠, so cos⁡−11921\cos^{-1}\frac{19}{21} is a genuine angle between 00 and π2\frac{\pi}{2}⁠: the acute angle we want. It is not a standard angle, so it is left in the form cos⁡−11921\cos^{-1}\frac{19}{21}⁠, as in the book.
    Principal value of cos⁡−1\cos^{-1} lies in [0,π][0, \pi]⁠; a positive argument gives an acute angle. Lesson: Examples 7–8

  6. Check with sin⁡θ\sin\theta

    b⃗1×b⃗2=∣i^j^k^122326∣=8i^+0j^−4k^,sin⁡θ=∣b⃗1×b⃗2∣∣b⃗1∣ ∣b⃗2∣=8021=4521.\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{vmatrix} = 8\hat i + 0\hat j - 4\hat k, \\ \sin\theta &= \frac{|\vec b_1\times\vec b_2|}{|\vec b_1|\,|\vec b_2|} = \frac{\sqrt{80}}{21} = \frac{4\sqrt5}{21}. \end{aligned}

    Then cos⁡2θ+sin⁡2θ=361441+80441=1\cos^2\theta + \sin^2\theta = \frac{361}{441} + \frac{80}{441} = 1⁠.

    The cross product gives ∣b⃗1∣∣b⃗2∣sin⁡θ|\vec b_1||\vec b_2|\sin\theta⁠, so it measures the same angle a second way; this is the vector form of equation (2) of Section 11.4. The components are 2⋅6−2⋅2=82 \cdot 6 - 2 \cdot 2 = 8⁠, −(1⋅6−2⋅3)=0-(1 \cdot 6 - 2 \cdot 3) = 0 and 1⋅2−2⋅3=−41 \cdot 2 - 2 \cdot 3 = -4⁠. The two results fit together exactly.
    ∣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

The angle between the lines is θ=cos⁡−1(1921)\theta = \cos^{-1}\left(\frac{19}{21}\right)⁠.

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

Check it with numbers

1921≈0.904762\frac{19}{21} \approx 0.904762⁠, so θ≈25.21∘\theta \approx 25.21^\circ⁠, an acute angle. Also 4521≈0.425918\frac{4\sqrt5}{21} \approx 0.425918⁠, and the squares 0.9047622≈0.8185940.904762^2 \approx 0.818594 and 0.4259182≈0.1814060.425918^2 \approx 0.181406 add up to 1.

Common mistakes
  1. Using the position vectors. 3i^+2j^−4k^3\hat i + 2\hat j - 4\hat k and 5i^−2j^5\hat i - 2\hat j are points, not directions; they play no part in the angle.
  2. Forgetting a square root. ∣b⃗1∣=9=3|\vec b_1| = \sqrt9 = 3⁠, not 9.
  3. Dropping the modulus. Without it, a negative dot product would give the obtuse angle, which is not the angle between the lines.
What to take away
  1. Angle between lines = angle between their direction vectors; ignore the points.
  2. cos⁡θ=∣b⃗1⋅b⃗2∣∣b⃗1∣∣b⃗2∣\cos\theta = \frac{|\vec b_1\cdot\vec b_2|}{|\vec b_1||\vec b_2|} gives the acute angle.
  3. sin⁡θ=∣b⃗1×b⃗2∣∣b⃗1∣∣b⃗2∣\sin\theta = \frac{|\vec b_1\times\vec b_2|}{|\vec b_1||\vec b_2|} is a good check.

Every fact used, and where it comes from

FactUsed inSource
In r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, b⃗\vec b is the directionStep 1Section 11.3.1 (Basics)
Lines not through the origin may be replaced by parallel lines through itStep 1Note in Section 11.4
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 acuteSteps 2, 5Section 11.4 and Fig 11.4; the text there calls the figure Fig 11.6, a slip for Fig 11.4 (Lesson)
a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3Step 3Section 10.6.1 (Basics)
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta⁠; determinant form of a⃗×b⃗\vec a\times\vec bStep 6Section 10.6.3 (Basics)
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