Question

The distance of the line \( \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4} \) from the point \( (1, 4, 0) \) along the line \( \frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \) is:

• A. \( \sqrt{7} \)
• B. \( \sqrt{14} \)
• C. \( \sqrt{15} \)
• D. \( \sqrt{13} \)

Hint:

When finding the distance between a point and a line in 3D, use the cross product of the vector from the point to any point on the line with the direction vector of the line, and divide by the magnitude of the direction vector of the line.

Answer 1

The Correct Option is B

The given lines are expressed in symmetric form: For the first line, \[ \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}, \] which represents the parametric equations: \[ x = 2t + 2, \quad y = 3t + 6, \quad z = 4t + 3. \] For the second line, \[ \frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}, \] which represents the parametric equations: \[ x = t, \quad y = 2t + 2, \quad z = 3t - 3. \] Step 1: The direction vector of the first line is \( \vec{d_1} = \langle 2, 3, 4 \rangle \) and the direction vector of the second line is \( \vec{d_2} = \langle 1, 2, 3 \rangle \). Step 2: The position vector of the point \( P(1, 4, 0) \) is \( \vec{OP} = \langle 1, 4, 0 \rangle \). Step 3: The vector connecting the point \( P(1, 4, 0) \) to any point on the first line can be represented as \( \vec{OP'} = \langle 1 - 2t - 2, 4 - 3t - 6, 0 - 4t - 3 \rangle \), which simplifies to: \[ \vec{OP'} = \langle -2t - 1, -3t - 2, -4t - 3 \rangle. \] Step 4: The distance between the point \( P(1, 4, 0) \) and the line is given by the formula: \[ \text{Distance} = \frac{| \vec{OP'} \times \vec{d_1} |}{| \vec{d_1} |}. \] Here, we calculate the cross product \( \vec{OP'} \times \vec{d_1} \) and its magnitude. After performing the cross product and simplifying, we find that the magnitude of the distance is \( \sqrt{14} \).