Question

Let \( P \) be the point of intersection of the lines \[ \frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} \quad \text{and} \quad \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}. \] Then, the shortest distance of \( P \) from the line \( 4x = 2y = z \) is:

• A. \( \frac{5\sqrt{14}}{7} \)
• B. \( \frac{\sqrt{14}}{7} \)
• C. \( \frac{3\sqrt{14}}{7} \)
• D. \( \frac{6\sqrt{14}}{7} \)

Answer 1

The Correct Option is C

Find the point of intersection \( P \):

Line \( L_1 \):

\[ \frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} = \lambda. \]

Parametric coordinates of \( P \) on \( L_1 \):

\[ x = \lambda + 2, \quad y = 5\lambda + 4, \quad z = \lambda + 2. \]

Line \( L_2 \):

\[ \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2} = \mu. \]

Parametric coordinates of \( P \) on \( L_2 \):

\[ x = 2\mu + 3, \quad y = 3\mu + 2, \quad z = 2\mu + 3. \]

Equate \( x, y, z \) for both lines:

\[ \lambda + 2 = 2\mu + 3 \implies \lambda = 2\mu + 1, \] \[ 5\lambda + 4 = 3\mu + 2 \implies 5(2\mu + 1) + 4 = 3\mu + 2, \] \[ 10\mu + 5 + 4 = 3\mu + 2 \implies 7\mu = -7 \implies \mu = -1. \]

Substituting \( \mu = -1 \):

\[ \lambda = 2(-1) + 1 = -1. \]

Coordinates of \( P \):

\[ P(1, -1, 1). \]

Find the shortest distance from \( P \) to the line \( 4x = 2y = z \):

The equation of the line \( L_3 \) in symmetric form is:

\[ \frac{x}{1/4} = \frac{y}{1/2} = \frac{z}{1}. \]

Let \( Q(k, 2k, 4k) \) be a point on \( L_3 \). Direction ratios of \( PQ \):

\[ \text{DRs of } PQ = (k - 1, 2k + 1, 4k - 1). \]

\( PQ \perp L_3 \): Solve using:

\[ (k - 1)\cdot\frac{1}{4} + (2k + 1)\cdot\frac{1}{2} + (4k - 1)\cdot 1 = 0. \]

Simplify:

\[ \frac{k - 1}{4} + \frac{4k + 2}{4} + 4k - 1 = 0, \] \[ \frac{k - 1}{4} + \frac{4k + 2}{4} + \frac{16k - 4}{4} = 0, \] \[ 21k - 3 = 0 \implies k = \frac{1}{7}. \]

Coordinates of \( Q \):

\[ Q\left(\frac{1}{7}, \frac{2}{7}, \frac{4}{7}\right). \]

Calculate \( PQ \):

\[ PQ = \sqrt{\left(\frac{1}{7} - 1\right)^2 + \left(\frac{2}{7} + 1\right)^2 + \left(\frac{4}{7} - 1\right)^2}. \]

Simplify:

\[ PQ = \sqrt{\left(-\frac{6}{7}\right)^2 + \left(\frac{9}{7}\right)^2 + \left(-\frac{3}{7}\right)^2}. \] \[ PQ = \sqrt{\frac{36}{49} + \frac{81}{49} + \frac{9}{49}} = \sqrt{\frac{126}{49}} = \sqrt{\frac{36}{7}} = \frac{3\sqrt{14}}{7}. \]

Option (3) is correct.

Answer 2

Step 1: Find the coordinates of point \( P \)
Given lines:
Line 1: \( \frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} \).
Line 2: \( \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2} \).
Let \( t \) be the parameter for line 1, then:
\[ x = 2 + t, y = 4 + 5t, z = 2 + t \] Similarly, for line 2, let \( s \) be the parameter:
\[ x = 3 + 2s, y = 2 + 3s, z = 3 + 2s \] Find intersection by solving:
\[ 2 + t = 3 + 2s \quad ...(i) \\ 4 + 5t = 2 + 3s \quad ...(ii) \\ 2 + t = 3 + 2s \quad ...(iii) \] From (i) and (iii), both are same: \( 2 + t = 3 + 2s \).
From (ii): \( 4 + 5t = 2 + 3s \)
Express \( t \) from (i): \( t = 3 + 2s - 2 = 1 + 2s \).
Substitute into (ii):
\[ 4 + 5(1 + 2s) = 2 + 3s \\ 4 + 5 + 10s = 2 + 3s \\ 9 + 10s = 2 + 3s \\ 7 = -7s \Rightarrow s = -1 \\ t = 1 + 2(-1) = -1 \] Now, point \( P \):
\[ x = 2 + t = 2 - 1 = 1 \\ y = 4 + 5t = 4 + 5(-1) = -1 \\ z = 2 + t = 2 - 1 = 1 \] So, \( P(1, -1, 1) \).
Step 2: Find the shortest distance from \( P \) to line \( 4x=2y=z \).
Express line as parametric equations:
Let \( u \) be parameter:
\[ x = u, \quad y = 2u, \quad z = 4u \] Point \( P(1, -1, 1) \). Distance \( D \) from point to line:
\[ D = \frac{\| (\vec{P} - \vec{A}) \times \vec{d} \|}{\| \vec{d} \|}, \text{ where } \vec{A} \text{ is any point on line, } \vec{d} \text{ is direction vector} \] Choose \( \vec{A} = (0, 0, 0) \), \( \vec{d} = (1, 2, 4) \).
Vector \( \vec{P} - \vec{A} = (1, -1, 1) \).
Cross product:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & 4 \end{vmatrix} = \hat{i}((-1)(4) - 1(2)) - \hat{j}(1(4) - 1(1)) + \hat{k}(1(2) - (-1)(1)) \\ = \hat{i}(-4 - 2) - \hat{j}(4 - 1) + \hat{k}(2 + 1) \\ = -6 \hat{i} - 3 \hat{j} + 3 \hat{k} \] Magnitude:
\[ \sqrt{(-6)^2 + (-3)^2 + 3^2} = \sqrt{36 + 9 + 9} = \sqrt{54} = 3 \sqrt{6} \] Distance:
\[ D = \frac{3 \sqrt{6}}{\sqrt{1^2 + 2^2 + 4^2}} = \frac{3 \sqrt{6}}{\sqrt{1 + 4 + 16}} = \frac{3 \sqrt{6}}{\sqrt{21}} = \frac{3 \sqrt{6}}{\sqrt{21}} = \frac{3 \sqrt{6}}{\sqrt{21}} \times \frac{\sqrt{21}}{\sqrt{21}} = \frac{3 \sqrt{6 \times 21}}{21} = \frac{3 \sqrt{126}}{21} = \frac{3 \times \sqrt{9 \times 14}}{21} = \frac{3 \times 3 \sqrt{14}}{21} = \frac{9 \sqrt{14}}{21} = \frac{3 \sqrt{14}}{7} \] Final answer:
\( \frac{3 \sqrt{14}}{7} \)