Question

Find the exact angular measurement $\theta$ between the following pair of spatial lines in three-dimensional space: \[ \frac{x-2}{3} = \frac{y+5}{2} = \frac{1-z}{-6} \quad \text{and} \quad \frac{x-7}{1} = \frac{y}{2} = \frac{6-z}{-2} \]

Hint:

Always look out for terms like $1-z$ or $3-x$ in line equations. Forgetting to rearrange them to standard form will change the sign of your direction numbers and result in an incorrect angle calculation.

Answer 1

Concept: The angle $\theta$ between two lines in a three-dimensional coordinate system is equivalent to the angle between their respective direction vectors, $\vec{b_1}$ and $\vec{b_2}$. Before extracting these directional components, both equations must be written in the standard symmetrical form: \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \] where the variables $x, y, z$ must have a positive coefficient of $+1$. Once the direction vectors $\vec{b_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ and $\vec{b_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$ are determined, the angle is computed using the formula: \[ \cos\theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|} \]

Step 1: Converting Line 1 into standard form and extracting its direction vector.
The given equation for Line 1 is: \[ \frac{x-2}{3} = \frac{y+5}{2} = \frac{1-z}{-6} \] Notice that the third term, $\frac{1-z}{-6}$, is not in standard form because the variable $z$ has a negative sign. Let us multiply both the numerator and the denominator of this term by $-1$: \[ \frac{1-z}{-6} = \frac{-(z-1)}{-6} = \frac{z-1}{6} \] Now, rewriting Line 1 in standard form: \[ \frac{x-2}{3} = \frac{y+5}{2} = \frac{z-1}{6} \] The direction vector $\vec{b_1}$ is formed from the denominators: \[ \vec{b_1} = 3\hat{i} + 2\hat{j} + 6\hat{k} \]

Step 2: Converting Line 2 into standard form and extracting its direction vector.
The given equation for Line 2 is: \[ \frac{x-7}{1} = \frac{y}{2} = \frac{6-z}{-2} \] Similarly, the third term $\frac{6-z}{-2}$ has a negative $z$ variable. Multiplying its numerator and denominator by $-1$: \[ \frac{6-z}{-2} = \frac{-(z-6)}{-2} = \frac{z-6}{2} \] Rewriting Line 2 in standard form: \[ \frac{x-7}{1} = \frac{y}{2} = \frac{z-6}{2} \] The direction vector $\vec{b_2}$ is formed from the denominators: \[ \vec{b_2} = 1\hat{i} + 2\hat{j} + 2\hat{k} \]

Step 3: Computing the dot product and the magnitudes of the direction vectors.
Let us calculate the vector dot product $\vec{b_1} \cdot \vec{b_2}$: \[ \vec{b_1} \cdot \vec{b_2} = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19 \] Next, calculate the absolute magnitudes of each direction vector: \[ |\vec{b_1}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \] \[ |\vec{b_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4 = } \sqrt{9} = 3 \]

Step 4: Substituting values into the cosine formula to solve for $\theta$.
Substitute the derived values into the angle formula: \[ \cos\theta = \frac{19}{7 \times 3} = \frac{19}{21} \] Taking the inverse cosine on both sides to isolate $\theta$: \[ \theta = \cos^{-1}\left(\frac{19}{21}\right) \]