Question

The distance of the point (7, 10, 11) from the line \( \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} \) along the line \( \frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} \) is

• A. 18
• B. 14
• C. 12
• D. 16

Hint:

To find the distance of a point from a line along another line, assume a general point on the first line. The line joining the given point and this general point must be parallel to the second given line. Use the proportionality of direction ratios for parallel lines to find the coordinates of the point on the first line. Finally, calculate the distance between the given point and this point on the first line.

Answer 1

The Correct Option is B

The objective is to determine the distance from point P(7, 10, 11) to line \( L_1: \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} \), with the distance measured along line \( L_2: \frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} \). This is equivalent to finding the distance between point P and the intersection of lines \(L_1\) and \(L_2\).

Concept Used:

The following concepts are utilized to solve this problem:

1. Parametric form of a line: A line through \((x_0, y_0, z_0)\) with direction ratios \((a, b, c)\) is defined by: \[ x = x_0 + a\lambda, \quad y = y_0 + b\lambda, \quad z = z_0 + c\lambda \] where \(\lambda\) is a parameter.
2. Intersection of two lines in 3D: To find the intersection point of two lines, represent general points on each line using distinct parameters (e.g., \(\lambda\) and \(\mu\)). Equate corresponding coordinates to form a system of linear equations. A consistent solution for \(\lambda\) and \(\mu\) indicates intersection.
3. Distance formula in 3D: The distance between \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Step-by-Step Solution:

Step 1: Derive parametric equations for a general point on \(L_1\).

For line \(L_1\), \( \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} \). Setting this to \(\lambda\):

\[\frac{x - 4}{1} = \lambda \implies x = 4 + \lambda\]\[\frac{y - 4}{0} = \lambda \implies y = 4\]\[\frac{z - 2}{3} = \lambda \implies z = 2 + 3\lambda\]

A general point on \(L_1\) is \(Q_1 = (4 + \lambda, 4, 2 + 3\lambda)\).

Step 2: Derive parametric equations for a general point on \(L_2\).

For line \(L_2\), \( \frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} \). Setting this to \(\mu\):

\[\frac{x - 7}{2} = \mu \implies x = 7 + 2\mu\]\[\frac{y - 10}{3} = \mu \implies y = 10 + 3\mu\]\[\frac{z - 11}{6} = \mu \implies z = 11 + 6\mu\]

A general point on \(L_2\) is \(Q_2 = (7 + 2\mu, 10 + 3\mu, 11 + 6\mu)\).

Step 3: Determine \(\lambda\) and \(\mu\) at the intersection point.

Equating coordinates of \(Q_1\) and \(Q_2\) for intersection:

1. \(4 + \lambda = 7 + 2\mu \implies \lambda - 2\mu = 3 \quad \text{(i)}\)
2. \(4 = 10 + 3\mu \implies 3\mu = -6 \implies \mu = -2 \quad \text{(ii)}\)
3. \(2 + 3\lambda = 11 + 6\mu \implies 3\lambda - 6\mu = 9 \implies \lambda - 2\mu = 3 \quad \text{(iii)}\)

From equation (ii), \(\mu = -2\).

Step 4: Find \(\lambda\) using \(\mu\) and confirm intersection.

Substitute \(\mu = -2\) into equation (i):

\[\lambda - 2(-2) = 3 \implies \lambda + 4 = 3 \implies \lambda = -1\]

Equations (i) and (iii) are identical, confirming a consistent intersection at \(\lambda = -1\) and \(\mu = -2\).

Step 5: Find the coordinates of the intersection point.

Using \(\lambda = -1\) for \(L_1\):

\[x = 4 + (-1) = 3\]\[y = 4\]\[z = 2 + 3(-1) = -1\]

The intersection point is Q(3, 4, -1).

Step 6: Compute the required distance.

The distance between P(7, 10, 11) and Q(3, 4, -1) is calculated using the 3D distance formula:

\[d = \sqrt{(7 - 3)^2 + (10 - 4)^2 + (11 - (-1))^2}\]\[d = \sqrt{(4)^2 + (6)^2 + (12)^2}\]\[d = \sqrt{16 + 36 + 144}\]\[d = \sqrt{196}\]\[d = 14\]

The required distance is 14 units.