Question

The shortest distance between the lines \[ \vec r=\frac13\hat i+2\hat j+\frac83\hat k+\lambda(2\hat i-5\hat j+6\hat k) \] \[ \vec r=\left(-\frac23\hat i-\frac13\hat k\right)+\mu(\hat j-\hat k), \quad \lambda,\mu\in\mathbb R \] is

• A. \(2\sqrt3\)
• B. \(3\)
• C. \(\sqrt{15}\)
• D. \(\sqrt5\)

Hint:

Shortest distance between two skew lines uses the scalar triple product formula \(\frac{|(\vec a_1-\vec a_2)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}\).

Answer 1

The Correct Option is B

Concept:
The shortest distance between two skew lines \[ \vec r = \vec a_1 + \lambda \vec b_1, \quad \vec r = \vec a_2 + \mu \vec b_2 \] is \[ SD = \frac{|(\vec a_1 - \vec a_2) \cdot (\vec b_1 \times \vec b_2)|}{|\vec b_1 \times \vec b_2|} \] Step 1: Identify vectors
\[ \vec a_1 = \frac13 \hat i + 2 \hat j + \frac83 \hat k \] \[ \vec a_2 = -\frac23 \hat i - \frac13 \hat k \] \[ \vec b_1 = 2 \hat i - 5 \hat j + 6 \hat k \] \[ \vec b_2 = \hat j - \hat k \] Step 2: Find cross product
\[ \vec b_1 \times \vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix} \] \[ = (-1) \hat i + 2 \hat j + 2 \hat k \] Step 3: Compute \(\vec a_1 - \vec a_2\)
\[ \vec a_1 - \vec a_2 = \hat i + 2 \hat j + 3 \hat k \] Step 4: Dot product
\[ (\vec a_1 - \vec a_2) \cdot (\vec b_1 \times \vec b_2) \] \[ = (1)(-1) + (2)(2) + (3)(2) \] \[ = 9 \] Step 5: Magnitude
\[ |\vec b_1 \times \vec b_2| = \sqrt{(-1)^2 + 2^2 + 2^2} \] \[ = 3 \] Step 6: Shortest distance
\[ SD = \frac{|9|}{3} \] \[ = 3 \]