Question

If the shortest distance between the lines
\(\vec{r} = (-\hat{i} + 3\hat{k}) + \lambda(\hat{i} - a\hat{j})\) and \(\vec{r} = (-\hat{j} + 2\hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})\)
is \(\sqrt{\frac23}\), then the integral value of a is equal to ____________.

Hint:

Use shortest distance = |(a2 - a1) . (d1 x d2)| / |d1 x d2| and solve the resulting quadratic in a, keeping the integer root.

Answer 1

Correct Answer: 2

\(\vec{a}_1 = (-1, 0, 3)\), \(\vec{d}_1 = (1, -a, 0)\); \(\vec{a}_2 = (0, -1, 2)\), \(\vec{d}_2 = (1, -1, 1)\).
\(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix}\hat i & \hat j & \hat k \\ 1 & -a & 0 \\ 1 & -1 & 1\end{vmatrix} = (-a, -1, a - 1)\).
\(\vec{a}_2 - \vec{a}_1 = (1, -1, -1)\). Dot product \(= -a + 1 - (a - 1) = 2 - 2a\).
\(|\vec{d}_1 \times \vec{d}_2|^2 = a^2 + 1 + (a - 1)^2 = 2a^2 - 2a + 2\).
\(\text{SD}^2 = \frac{(2 - 2a)^2}{2a^2 - 2a + 2} = \frac{2(a - 1)^2}{a^2 - a + 1} = \frac23\).
\(3(a - 1)^2 = a^2 - a + 1 \Rightarrow 2a^2 - 5a + 2 = 0 \Rightarrow (2a - 1)(a - 2) = 0\), so \(a = \frac12\) or \(a = 2\).
The integral value is \(a = 2\).
Answer: 2