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

To find the shortest distance between two skew lines, we apply the formula for the shortest distance between two lines given in vector form:

• The lines are given by their equations in vector form:
  \(\vec{r}_1 = \frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k} + \lambda (2\hat{i} - 5\hat{j} + 6\hat{k})\) and
  \(\vec{r}_2 = -\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k} + \mu (\hat{j} - \hat{k})\).
• Let the directional vectors of the lines be:
  • \(\vec{d}_1 = 2\hat{i} - 5\hat{j} + 6\hat{k}\)
  • \(\vec{d}_2 = \hat{j} - \hat{k}\)
• The vector connecting the initial points on the lines is: \(\vec{c} = \left(\frac{1}{3} - \left(-\frac{2}{3}\right)\right)\hat{i} + \left(2 - 0\right)\hat{j} + \left(\frac{8}{3} - \left(-\frac{1}{3}\right)\right)\hat{k}\)
  \(= \hat{i} + 2\hat{j} + 3\hat{k}\).
• The shortest distance \(d\) between two skew lines is given by: 

\[d = \frac{|\vec{c} \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}\]

• Calculate \(\vec{d}_1 \times \vec{d}_2\):
  • \(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix}\)
  • \(= \hat{i}((-5)(-1) - 6(1)) - \hat{j}(2(-1) - 6(0)) + \hat{k}(2(1) - (-5)(0))\)
  • \(= \hat{i}(5 - 6) - \hat{j}(-2) + \hat{k}(2)\)
  • \(= -\hat{i} + 2\hat{j} + 2\hat{k}\)
• Calculate \(|\vec{d}_1 \times \vec{d}_2|\): \(= \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = 3\).
• Calculate \(\vec{c} \cdot (\vec{d}_1 \times \vec{d}_2)\):
  • \(\vec{c} \cdot (\vec{d}_1 \times \vec{d}_2) = (1\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{i} + 2\hat{j} + 2\hat{k})\)
  • \(= (1)(-1) + (2)(2) + (3)(2)\)
  • \(= -1 + 4 + 6 = 9\)
• The shortest distance is then calculated as: \(d = \frac{|9|}{3} = 3\).

Thus, the shortest distance between the given lines is 3.