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 problem asks to find the distance of the point P(7, 10, 11) from the line \( L_1: \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} \), where the distance is measured along a second 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 point of intersection of lines \(L_1\) and \(L_2\).

Concept Used:

To solve this problem, we will use the following concepts:

1. Parametric form of a line: A line passing through the point \((x_0, y_0, z_0)\) with direction ratios \((a, b, c)\) can be represented by the parametric equations: \[ x = x_0 + a\lambda, \quad y = y_0 + b\lambda, \quad z = z_0 + c\lambda \] where \(\lambda\) is a parameter.
2. Intersection of two lines in 3D: To find the point of intersection of two lines, we write the coordinates of a general point on each line in terms of two different parameters (e.g., \(\lambda\) and \(\mu\)). We then equate the corresponding x, y, and z coordinates to get a system of linear equations. If this system has a consistent solution for \(\lambda\) and \(\mu\), the lines intersect.
3. Distance formula in 3D: The distance between two points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Step-by-Step Solution:

Step 1: Write the parametric equations for a general point on line \(L_1\).

The equation of line \(L_1\) is \( \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} \). Let's set this equal to a parameter \(\lambda\).

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

So, any point on line \(L_1\) can be represented as \(Q_1 = (4 + \lambda, 4, 2 + 3\lambda)\).

Step 2: Write the parametric equations for a general point on line \(L_2\).

The equation of line \(L_2\) is \( \frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} \). Let's set this equal to a parameter \(\mu\).

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

So, any point on line \(L_2\) can be represented as \(Q_2 = (7 + 2\mu, 10 + 3\mu, 11 + 6\mu)\).

Step 3: Find the values of \(\lambda\) and \(\mu\) at the point of intersection.

At the point of intersection, \(Q_1 = Q_2\). We can equate the corresponding coordinates:

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), we directly find \(\mu = -2\).

Step 4: Substitute the value of \(\mu\) to find \(\lambda\) and verify the intersection.

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

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

Equations (i) and (iii) are identical, so the system is consistent. The lines intersect at the point corresponding to \(\lambda = -1\) and \(\mu = -2\).

Step 5: Determine the coordinates of the point of intersection.

We can use either \(\lambda = -1\) in the coordinates for \(L_1\) or \(\mu = -2\) in the coordinates for \(L_2\). Let's use \(\lambda = -1\):

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

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

Step 6: Calculate the required distance.

The required distance is the distance between the given point P(7, 10, 11) and the point of intersection Q(3, 4, -1). 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.

Answer 2

\begin{center} \includegraphics{19S.png} \end{center} Let the given point be \( P(7, 10, 11) \). 
The equation of the line from which the distance is to be found is \( L_1: \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} = \lambda \). 
Any point Q on this line can be written as \( Q(\lambda + 4, 0\lambda + 4, 3\lambda + 2) = (\lambda + 4, 4, 3\lambda + 2) \). 
The distance is to be found along the line \( L_2 \) passing through P and Q, whose equation is given in the solution as parallel to \( \frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} \). 
The direction ratios of the line PQ are \( (\lambda + 4 - 7, 4 - 10, 3\lambda + 2 - 11) = (\lambda - 3, -6, 3\lambda - 9) \). Since the line PQ is parallel to the line with direction ratios \( (2, 3, 6) \), the direction ratios of PQ must be proportional to \( (2, 3, 6) \). \[ \frac{\lambda - 3}{2} = \frac{-6}{3} = \frac{3\lambda - 9}{6} \] From \( \frac{-6}{3} = -2 \), we have: \[ \frac{\lambda - 3}{2} = -2 \Rightarrow \lambda - 3 = -4 \Rightarrow \lambda = -1 \] \[ \frac{3\lambda - 9}{6} = -2 \Rightarrow 3\lambda - 9 = -12 \Rightarrow 3\lambda = -3 \Rightarrow \lambda = -1 \] So, the value of \( \lambda \) is \( -1 \). The coordinates of the point Q on the line \( L_1 \) are \( Q(-1 + 4, 4, 3(-1) + 2) = Q(3, 4, -1) \). The distance PQ is the distance between the points \( P(7, 10, 11) \) and \( Q(3, 4, -1) \). \[ PQ = \sqrt{(7 - 3)^2 + (10 - 4)^2 + (11 - (-1))^2} \] \[ PQ = \sqrt{(4)^2 + (6)^2 + (12)^2} \] \[ PQ = \sqrt{16 + 36 + 144} \] \[ PQ = \sqrt{196} = 14 \] The distance of the point (7, 10, 11) from the line along the given line is 14.