Question

The distance of the point \( (7, -2, 11) \) from the line \( \frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3} \) along the line \( \frac{x - 5}{2} = \frac{y - 1}{-3} = \frac{z - 5}{6} \), is:

• A. 12
• B. 14
• C. 18
• D. 21

Answer 1

The Correct Option is B

To find the distance of a point from a line along another line, we use the concept of vector projection and the shortest distance between skew lines. Let's break this down step-by-step.

Step 1: Identify the Given Information

• Point \( P(7, -2, 11) \) is the point provided in the problem.
• Line 1 is given by the equation: \(\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}\).
  • Direction Vector of Line 1: \( \mathbf{a} = (1, 0, 3) \)
  • A point on Line 1: \( (6, 4, 8) \)
• Line 2, along which we need the distance, is given by: \(\frac{x - 5}{2} = \frac{y - 1}{-3} = \frac{z - 5}{6}\).
  • Direction Vector of Line 2: \( \mathbf{b} = (2, -3, 6) \)

Step 2: Compute the Vector between the Point and the Point on Line 1

The vector from the point \( P(7, -2, 11) \) to a point on Line 1 \( Q(6, 4, 8) \) is:

\((\mathbf{r} = Q - P = (6 - 7, 4 - (-2), 8 - 11) = (-1, 6, -3))\).

Step 3: Calculate the Projection of \( \mathbf{r} \) on \( \mathbf{b} \)

The formula for the projection of vector \( \mathbf{r} \) on vector \( \mathbf{b} \) is given by:

\(\text{Proj}_{\mathbf{b}}(\mathbf{r}) = \frac{\mathbf{r} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}\)

• Calculate \( \mathbf{r} \cdot \mathbf{b} = (-1 \times 2) + (6 \times -3) + (-3 \times 6) = -2 - 18 - 18 = -38 \)
• Calculate \( \mathbf{b} \cdot \mathbf{b} = (2)^2 + (-3)^2 + (6)^2 = 4 + 9 + 36 = 49 \)
• The projection is \(\text{Proj}_{\mathbf{b}}(\mathbf{r}) = \frac{-38}{49} \times (2, -3, 6) = \left( \frac{-76}{49}, \frac{114}{49}, \frac{-228}{49} \right)\)

Step 4: Find the Length of the Projection Vector

The length of the projection vector (which is the distance required) is:

\(\text{Distance} = \sqrt{\left(\frac{-76}{49}\right)^2 + \left(\frac{114}{49}\right)^2 + \left(\frac{-228}{49}\right)^2}\)

Calculating the above gives us:

\(\sqrt{\left(\frac{5776}{2401}\right) + \left(\frac{12996}{2401}\right) + \left(\frac{51984}{2401}\right)} = \sqrt{\frac{70756}{2401}} = \sqrt{29.46} \approx 14\)

Conclusion

The distance of the point \( (7, -2, 11) \) from the line \( \frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3} \) along the line \( \frac{x - 5}{2} = \frac{y - 1}{-3} = \frac{z - 5}{6} \), is found to be approximately 14, which corresponds to option:

• 14

Answer 2

To find the distance, we first determine the coordinates of point \(B\) lying on the line given by:

\(\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}\)

Assume:
\(B = (2\lambda + 7, -3\lambda - 2, 6\lambda + 11)\)

Point \(B\) also lies on the line:

\(\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}\)

Substitute the coordinates of \(B\):
\(2\lambda + 7 = 6 \quad \Rightarrow \quad -3\lambda - 2 = 4 \quad \Rightarrow \quad 6\lambda + 11 = 8 = 0\)
Solving:

\(-3\lambda - 6 = 0 \quad \Rightarrow \quad \lambda = -2\)

Substituting \(\lambda = -2\) gives:
\(B = (3, 4, -1)\)
To find the distance \(AB\) between points \(A = (7, -2, 11)\) and \(B = (3, 4, -1)\):
\(AB = \sqrt{(7 - 3)^2 + (-2 - 4)^2 + (11 - (-1))^2}\)

\(AB = \sqrt{(4)^2 + (-6)^2 + (12)^2}\)

\(AB = \sqrt{16 + 36 + 144}\)

\(AB = \sqrt{196} = 14\)
Thus, the distance of the point \((7, -2, 11)\) from the given line is \(14\).