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 problem requires determining the shortest distance between two skew lines, \( L_1 \) and \( L_2 \). The lines are defined by their parametric equations. For \( L_1 \): \[ \frac{x - 1}{3} = \frac{y}{4} = \frac{z}{5} \implies x = 3t + 1, \, y = 4t, \, z = 5t \quad \text{(for parameter } t) \] For \( L_2 \): \[ \frac{x - p}{2} = \frac{y}{3} = \frac{z}{4} \implies x = 2s + p, \, y = 3s, \, z = 4s \quad \text{(for parameter } s) \] The formula for the shortest distance \( d \) between two skew lines is: \[ d = \frac{|(\vec{b_1} - \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|} \] Here, \( \vec{b_1} \) and \( \vec{b_2} \) are position vectors of points on \( L_1 \) and \( L_2 \) respectively, and \( \vec{a_1} \) and \( \vec{a_2} \) are their direction vectors. From the parametric equations, we identify: - Direction vector of \( L_1 \): \( \vec{a_1} = \langle 3, 4, 5 \rangle \) - Direction vector of \( L_2 \): \( \vec{a_2} = \langle 2, 3, 4 \rangle \) The position vectors of points on the lines are: - For \( L_1 \): \( \vec{b_1} = \langle 1, 0, 0 \rangle \) - For \( L_2 \): \( \vec{b_2} = \langle p, 0, 0 \rangle \) Calculate the cross product \( \vec{a_1} \times \vec{a_2} \): \[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 4 & 5
2 & 3 & 4 \end{vmatrix} = \langle -1, 2, -1 \rangle \] Calculate the dot product \( (\vec{b_1} - \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2}) \): \[ (\vec{b_1} - \vec{b_2}) = \langle 1 - p, 0, 0 \rangle \] \[ (\vec{b_1} - \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2}) = \langle 1 - p, 0, 0 \rangle \cdot \langle -1, 2, -1 \rangle = -(1 - p) \] Determine the magnitude of \( \vec{a_1} \times \vec{a_2} \): \[ |\vec{a_1} \times \vec{a_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6} \] Substitute these values into the shortest distance formula: \[ d = \frac{|-(1 - p)|}{\sqrt{6}} = \frac{|p - 1|}{\sqrt{6}} \] Given that the shortest distance is \( \frac{1}{\sqrt{6}} \): \[ \frac{|p - 1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} \implies |p - 1| = 1 \] Solve for \( p \): \[ p - 1 = 1 \quad \text{or} \quad p - 1 = -1 \] \[ p = 2 \quad \text{or} \quad p = 0 \] The possible value of \( p \) is \( 2 \). Therefore, the correct answer is (2) 2.