Question

The shortest distance between the z-axis and the line x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4, is ______.

Options

• 1

• 2

• 4

• 3

Answer 1

The shortest distance between the z-axis and the line x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4, is 2.

Explanation:

The equation of any plane passing through a given line is

(x + y + 2z – 3) + λ (2x + 3y + 4z – 4) = 0

`\implies` (1 + 2λ) x + (1 + 3λ) y + (2 + 4λ) z – (3 + 4λ) = 0

If this plane is parallel to the z-axis then normal to the plane will be perpendicular to the z-axis.

∴ (1 + 2λ) (0) + (1 + 3λ) (0) + (2 + 4λ) (1) = 0

λ = `-1/2`

Thus, Required plane is

`(x + y + 2z - 3) - 1/2(2x + 3y + 4z)` = 0

`\implies` y + 2 = 0

∴ Shortest distance from z-axis = `2/sqrt((1)^2` = 2