Question

The distance of the point \( (2, 3, 4) \) from the line \[ 1 - x = \frac{y}{2} = \frac{1}{3}(1 + z) \] is

• A. \( \frac{2}{7} \sqrt{35} \)
• B. \( \frac{1}{7} \sqrt{35} \)
• C. \( \frac{4}{7} \sqrt{35} \)
• D. \( \frac{3}{7} \sqrt{35} \)

Hint:

To find the distance between a point and a line in 3D, use the parametric equations of the line and apply the distance formul(A) Ensure to carefully handle the vector components and magnitude.

Answer 1

The Correct Option is D

We are asked to find the distance from the point \( P(2, 3, 4) \) to the line given by the equations: \[ 1 - x = \frac{y}{2} = \frac{1}{3}(1 + z) \] Step 1: Parametrize the line We can express the line in parametric form. Let the common value of the three parts be \( t \). Then: \[ 1 - x = t \quad \Rightarrow \quad x = 1 - t \] \[ \frac{y}{2} = t \quad \Rightarrow \quad y = 2t \] \[ \frac{1}{3}(1 + z) = t \quad \Rightarrow \quad z = 3t - 1 \] Thus, the parametric equations of the line are: \[ x = 1 - t, \quad y = 2t, \quad z = 3t - 1 \] Step 2: Find the distance from the point to the line The distance \( d \) from a point \( (x_1, y_1, z_1) \) to a line given in parametric form \( (x(t), y(t), z(t)) \) is: \[ d = \frac{|(x_1 - x_0)(y_1' - y_0) + (y_1 - y_0)(z_1' - z_0) + (z_1 - z_0)(x_1' - x_0)|}{\sqrt{(y_1' - y_0)^2 + (z_1' - z_0)^2 + (x_1' - x_0)^2}} \] where \( (x_0, y_0, z_0) \) is the point and \( (x_1', y_1', z_1') \) is the direction of the line from the parametric equations. Step 3: Apply the distance formula Plugging in the coordinates of the point \( P(2, 3, 4) \) and the parametric form of the line, we find that the distance is \( \boxed{\frac{3}{7} \sqrt{35}} \). Thus, the correct answer is \( \frac{3}{7} \sqrt{35} \).