Question

Find the square of the distance of the point \( (5, 6, 7) \) from the line \( \frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4} \):

Hint:

To avoid solving for \( \lambda \), you can use the vector cross product formula: \( d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} \), where \( A \) is a point on the line and \( \vec{b} \) is the direction vector.

Answer 1

Correct Answer: 6

Step 1: Understanding the Concept:
The distance of a point \( P \) from a line can be found by locating the foot of the perpendicular \( F \) on the line. The squared distance is then calculated using the distance formula between \( P \) and \( F \).

Step 2: Key Formula or Approach:
1. Any point on the line is \( F(2\lambda+2, 3\lambda+5, 4\lambda+2) \). 2. The vector \( \vec{PF} \) must be perpendicular to the direction of the line \( (2, 3, 4) \). 3. \( \vec{PF} \cdot (2, 3, 4) = 0 \).

Step 3: Detailed Explanation:
1. Vector \( \vec{PF} = (2\lambda+2-5, 3\lambda+5-6, 4\lambda+2-7) = (2\lambda-3, 3\lambda-1, 4\lambda-5) \). 2. Dot product: \( 2(2\lambda-3) + 3(3\lambda-1) + 4(4\lambda-5) = 0 \). \( 4\lambda - 6 + 9\lambda - 3 + 16\lambda - 20 = 0 \). \( 29\lambda - 29 = 0 \implies \lambda = 1 \). 3. Coordinates of \( F \): \( (2(1)+2, 3(1)+5, 4(1)+2) = (4, 8, 6) \). 4. Squared distance \( PF^2 = (4-5)^2 + (8-6)^2 + (6-7)^2 \). \( PF^2 = (-1)^2 + (2)^2 + (-1)^2 = 1 + 4 + 1 = 6 \).

Step 4: Final Answer:
The square of the distance is 6.