Question

Let \( L_1 : \frac{x - 1}{3} = \frac{y}{4} = \frac{z}{5} \) and \( L_2 : \frac{x - p}{2} = \frac{y}{3} = \frac{z}{4} \). If the shortest distance between \( L_1 \) and \( L_2 \) is \( \frac{1}{\sqrt{6}} \), then the possible value of \( p \) is:

• A. 3
• B. 2
• C. 5
• D. 7

Hint:

The shortest distance between two skew lines can be calculated using the formula involving the cross product of their direction vectors and the vector connecting a point on each line.

Answer 1

The Correct Option is B

The given problem involves finding the shortest distance between two skew lines \( L_1 \) and \( L_2 \). The parametric equations of the lines are as follows: For \( L_1 \): \[ \frac{x - 1}{3} = \frac{y}{4} = \frac{z}{5} \implies x = 3t + 1, \, y = 4t, \, z = 5t \quad \text{(for some parameter } t) \] For \( L_2 \): \[ \frac{x - p}{2} = \frac{y}{3} = \frac{z}{4} \implies x = 2s + p, \, y = 3s, \, z = 4s \quad \text{(for some parameter } s) \] The shortest distance between two skew lines is given by the formula: \[ d = \frac{|(\vec{b_1} - \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|} \] where \( \vec{b_1} \) and \( \vec{b_2} \) are the position vectors of any two points on \( L_1 \) and \( L_2 \), and \( \vec{a_1} \) and \( \vec{a_2} \) are direction vectors of \( L_1 \) and \( L_2 \). From the parametric equations: - The direction vector of \( L_1 \) is \( \vec{a_1} = \langle 3, 4, 5 \rangle \) - The direction vector of \( L_2 \) is \( \vec{a_2} = \langle 2, 3, 4 \rangle \) The position vectors are: - For \( L_1 \), \( \vec{b_1} = \langle 1, 0, 0 \rangle \) - For \( L_2 \), \( \vec{b_2} = \langle p, 0, 0 \rangle \) Now, calculate the cross product \( \vec{a_1} \times \vec{a_2} \): \[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 4 & 5
2 & 3 & 4 \end{vmatrix} = \langle -1, 2, -1 \rangle \] Next, calculate the dot product \( (\vec{b_1} - \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2}) \): \[ (\vec{b_1} - \vec{b_2}) = \langle 1 - p, 0, 0 \rangle \] \[ (\vec{b_1} - \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2}) = \langle 1 - p, 0, 0 \rangle \cdot \langle -1, 2, -1 \rangle = -(1 - p) \] The magnitude of \( \vec{a_1} \times \vec{a_2} \) is: \[ |\vec{a_1} \times \vec{a_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6} \] Now, substitute the known values into the formula for the shortest distance: \[ d = \frac{|-(1 - p)|}{\sqrt{6}} = \frac{|p - 1|}{\sqrt{6}} \] We are given that the shortest distance is \( \frac{1}{\sqrt{6}} \): \[ \frac{|p - 1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} \implies |p - 1| = 1 \] Solving this: \[ p - 1 = 1 \quad \text{or} \quad p - 1 = -1 \] \[ p = 2 \quad \text{or} \quad p = 0 \] Thus, the possible value of \( p \) is \( 2 \). Therefore, the correct answer is (2) 2.