Example 8: Angle between two lines in Cartesian form, direction ratios $3, 5, 4$ and $1, 1, 2$

Basics you need

Solution

1. Read off the direction ratios

   First line: $a_1, b_1, c_1 = 3, 5, 4$⁠. Second line: $a_2, b_2, c_2 = 1, 1, 2$⁠.

   Each line is already in the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$⁠, with coefficient 1 in front of $x$⁠, $y$ and $z$ in every numerator, so the denominators can be read off directly. The numerators only give the points $(-3, 1, -3)$ and $(-1, 4, 5)$ on the lines, which do not affect the angle.
   Cartesian equation (3) of Section 11.3.1. Basics: point plus direction

2. The formula

   $$\begin{aligned} \cos\theta &= \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}} \\ &\quad \times \frac{1}{\sqrt{a_2^2 + b_2^2 + c_2^2}}. \end{aligned}$$
   This is the dot product formula $\cos\theta = \frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}$ written with components, for $\vec b_1 = 3\hat i + 5\hat j + 4\hat k$ and $\vec b_2 = \hat i + \hat j + 2\hat k$⁠. The modulus gives the acute angle.
   Section 11.4, Cartesian form of the angle between two lines. Lesson: compare the directions

3. The top

   $$\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 3(1) + 5(1) + 4(2) \\ &= 3 + 5 + 8 = 16. \end{aligned}$$
   Multiply matching direction ratios and add. The book writes this as $3.1 + 5.1 + 4.2$⁠, where the dot means "times".
   $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$⁠. Basics: the dot product

4. The bottom

   $$\begin{aligned} \sqrt{3^2 + 5^2 + 4^2} &= \sqrt{9 + 25 + 16} \\ &= \sqrt{50} = 5\sqrt2, \\ \sqrt{1^2 + 1^2 + 2^2} &= \sqrt{1 + 1 + 4} = \sqrt6. \end{aligned}$$
   These are the lengths of the two direction vectors. $\sqrt{50}$ simplifies because $50 = 25 \times 2$ and $\sqrt{25} = 5$⁠; $\sqrt6$ does not simplify.
   Length of a vector. Basics: Pythagoras in 3D

5. Simplify and name the angle

   $$\begin{aligned} \cos\theta &= \left|\frac{16}{5\sqrt2\,\sqrt6}\right| = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt3} \\ &= \frac{8}{5\sqrt3} = \frac{8\sqrt3}{15}. \end{aligned}$$

   So $\theta = \cos^{-1}\left(\frac{8\sqrt3}{15}\right)$⁠.

   $\sqrt2\,\sqrt6 = \sqrt{12} = 2\sqrt3$⁠, so the bottom is $5 \times 2\sqrt3 = 10\sqrt3$⁠. Cancelling 2 gives $\frac{8}{5\sqrt3}$⁠, and multiplying top and bottom by $\sqrt3$ gives $\frac{8\sqrt3}{15}$⁠. This number is positive and less than 1 (since $64 \times 3 = 192 < 225 = 15^2$⁠), so $\theta$ is a genuine acute angle.
   Principal value of $\cos^{-1}$⁠; positive argument, acute angle. Lesson: Examples 7–8

6. Check with $\sin\theta$

   By equation (2) of Section 11.4, with $a_1b_2 - a_2b_1 = 3 - 5 = -2$⁠, $b_1c_2 - b_2c_1 = 10 - 4 = 6$ and $c_1a_2 - c_2a_1 = 4 - 6 = -2$⁠:

   $$\sin\theta = \frac{\sqrt{4 + 36 + 4}}{\sqrt{50}\,\sqrt6} = \frac{\sqrt{44}}{10\sqrt3} = \frac{\sqrt{33}}{15}.$$

   Then $\cos^2\theta + \sin^2\theta = \frac{192}{225} + \frac{33}{225} = 1$⁠.

   The three brackets are the components of the cross product of the direction vectors, so this measures the same angle with $\sin$ instead of $\cos$⁠. $\sqrt{44} = 2\sqrt{11}$⁠, and $\frac{2\sqrt{11}}{10\sqrt3} = \frac{\sqrt{11}}{5\sqrt3} = \frac{\sqrt{33}}{15}$⁠. The two answers agree.
   $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$⁠. Basics: the cross product

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