Question

The square of the distance of the point \(\left( \frac{15}{7}, \frac{32}{7}, 7 \right)\) from the line \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) in the direction of the vector \(\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\) is:

• A. \(41\)
• B. \(44\)
• C. \(54\)
• D. \(66\)

Hint:

To find the square of the distance between two points, use the distance formula and then square the result.

Answer 1

The Correct Option is D

Step 1: Equation of the Line and Point

The given line is:
\( L: \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \)
The point \( P = \left( \frac{15}{7}, \frac{32}{7}, 7 \right) \).

Step 2: Finding the Coordinates of Point Q

Let \( Q \) be the point on the line with parametric coordinates:
\( Q = \left( \lambda + \frac{15}{7}, 4\lambda + \frac{32}{7}, 7\lambda + 7 \right) \)

Step 3: Condition for Point \( Q \) on Line \( L \)

Using the equation:
\( \frac{\lambda + \frac{15}{7} + 1}{3} = \frac{4\lambda + \frac{32}{7} + 3}{5} = \frac{7\lambda + 7 + 5}{7} \)

Equating terms:
\( 7\lambda + 22 = 21\lambda + 36 \)
\( \Rightarrow \lambda = -1 \)

Step 4: Coordinates of Point \( Q \)

Substituting \( \lambda = -1 \):
\( Q = \left( \frac{8}{7}, 4, 0 \right) \)

Step 5: Distance \( PQ \)

The distance between points \( P \) and \( Q \) is:
\( PQ = \sqrt{\left( \frac{15}{7} - \frac{8}{7} \right)^2 + \left( \frac{32}{7} - 4 \right)^2 + (7 - 0)^2} \)
\( PQ = \sqrt{1^2 + \left( \frac{4}{7} \right)^2 + 49} \)
\( PQ = \sqrt{1 + \frac{16}{49} + 49} = \sqrt{\frac{65}{49} + 49} \)
\( PQ = \sqrt{\frac{2466}{49}} = \sqrt{66} \)
\( \Rightarrow PQ^2 = 66 \)