<p>The shortest distance between the <em>z</em>-axis and the line \(x + y + 2z - 3 = 0 = 2x + 3y + 4z - 4\) is</p>
Step-by-Step Solution
Key Concept: The shortest distance from a line to the z-axis equals the distance from any point on the z-axis to the given line. Convert the line to parametric form, then use the skew lines distance formula or find the perpendicular distance from a point on the z-axis to the line.
Step 1: Find the direction vector and a point on the given line by solving the system of planes: x + y + 2z - 3 = 0 and 2x + 3y + 4z - 4 = 0. From the first equation: x + y = 3 - 2z. From the second: 2x + 3y = 4 - 4z. Solving: x = 5 - 2z, y = -2 + z. So the line is: (5 - 2z, -2 + z, z) or parametrically r_1 = (5, -2, 0) + t(-2, 1, 1) . Step 2: The z-axis has direction vector d_2 = (0, 0, 1) and passes through origin (0, 0, 0) . Step 3: Use the skew lines distance formula: d = |((a_2 - a_1) · (d_1 × d_2))| / |d_1 × d_2|, where a_1 = (5, -2, 0), a_2 = (0, 0, 0), d_1 = (-2, 1, 1), d_2 = (0, 0, 1). Step 4: Calculate d_1 × d_2 = | i j k ; -2 1 1; 0 0 1| = (1, 2, 0). Step 5: (a_2 - a_1) = (-5, 2, 0). So (a_2 - a_1) · (d_1 × d_2) = (-5)(1) + (2)(2) + (0)(0) = -5 + 4 = -1. Step 6: |d_1 × d_2| = √(1 + 4 + 0) = √5. ∴ Distance = |-1|/√5 = 1/√5 = √5/5
Correct Answer: A