1. Key Concepts and Formulas

• Vector Form of a Line: A line in 3D space can be represented by the vector equation $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$, where $\overrightarrow a$ is the position vector of a point on the line, and $\overrightarrow b$ is the direction vector of the line.
• Shortest Distance Between Skew Lines: Skew lines are lines that are neither parallel nor intersecting. For two skew lines, Line 1: $\overrightarrow r = \overrightarrow{a_1} + \lambda \overrightarrow{b_1}$ and Line 2: $\overrightarrow r = \overrightarrow{a_2} + \mu \overrightarrow{b_2}$, the shortest distance $d$ between them is given by the formula: $d = \frac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2})|}{| \overrightarrow{b_1} \times \overrightarrow{b_2} |}$

2. Step-by-Step Solution

Step 1: Identify the position and direction vectors for each line. We are given the equations of two lines: Line 1: $\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$ Comparing this with $\overrightarrow r = \overrightarrow{a_1} + \lambda \overrightarrow{b_1}$: $\overrightarrow{a_1} = - \widehat i + 0\widehat j + 3\widehat k$ (position vector of a point on Line 1) $\overrightarrow{b_1} = \widehat i - a\widehat j + 0\widehat k$ (direction vector of Line 1)

Line 2: $\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$ Comparing this with $\overrightarrow r = \overrightarrow{a_2} + \mu \overrightarrow{b_2}$: $\overrightarrow{a_2} = 0\widehat i - \widehat j + 2\widehat k$ (position vector of a point on Line 2) $\overrightarrow{b_2} = \widehat i - \widehat j + \widehat k$ (direction vector of Line 2)

Step 2: Calculate the vector $(\overrightarrow{a_2} - \overrightarrow{a_1})$. This vector connects a point on Line 1 to a point on Line 2, and is crucial for the numerator of the shortest distance formula. $\overrightarrow{a_2} - \overrightarrow{a_1} = (0\widehat i - \widehat j + 2\widehat k) - (-\widehat i + 0\widehat j + 3\widehat k)$ $= (0 - (-1))\widehat i + (-1 - 0)\widehat j + (2 - 3)\widehat k$ $= \widehat i - \widehat j - \widehat k$

Step 3: Calculate the cross product $(\overrightarrow{b_1} \times \overrightarrow{b_2})$. This vector is perpendicular to both direction vectors and indicates the direction of the shortest distance. Its magnitude forms the denominator of the formula. $\overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & -a & 0 \\ 1 & -1 & 1 \end{vmatrix}$ $= \widehat i ((-a)(1) - (0)(-1)) - \widehat j ((1)(1) - (0)(1)) + \widehat k ((1)(-1) - (-a)(1))$ $= \widehat i (-a - 0) - \widehat j (1 - 0) + \widehat k (-1 + a)$ $= -a\widehat i - \widehat j + (a-1)\widehat k$

Step 4: Calculate the scalar triple product $(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2})$. This value forms the numerator of the shortest distance formula (taking its absolute value). $(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2}) = (\widehat i - \widehat j - \widehat k) \cdot (-a\widehat i - \widehat j + (a-1)\widehat k)$ $= (1)(-a) + (-1)(-1) + (-1)(a-1)$ $= -a + 1 - a + 1$ $= 2 - 2a$

Step 5: Calculate the magnitude of the cross product $| \overrightarrow{b_1} \times \overrightarrow{b_2} |$. This magnitude forms the denominator of the shortest distance formula. $| \overrightarrow{b_1} \times \overrightarrow{b_2} | = | -a\widehat i - \widehat j + (a-1)\widehat k |$ $= \sqrt{(-a)^2 + (-1)^2 + (a-1)^2}$ $= \sqrt{a^2 + 1 + a^2 - 2a + 1}$ $= \sqrt{2a^2 - 2a + 2}$

Step 6: Substitute the calculated values into the shortest distance formula and solve for 'a'. We are given that the shortest distance $d = \sqrt {{2 \over 3}}$. $d = \frac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2})|}{| \overrightarrow{b_1} \times \overrightarrow{b_2} |}$ $\sqrt {{2 \over 3}} = \frac{|2 - 2a|}{\sqrt{2a^2 - 2a + 2}}$ To eliminate the square roots, square both sides of the equation: $\left(\sqrt {{2 \over 3}}\right)^2 = \left(\frac{|2(1 - a)|}{\sqrt{2(a^2 - a + 1)}}\right)^2$ ${2 \over 3} = \frac{4(1 - a)^2}{2(a^2 - a + 1)}$ ${2 \over 3} = \frac{2(1 - a)^2}{a^2 - a + 1}$ Now, cross-multiply: $2(a^2 - a + 1) = 3 \cdot 2(1 - a)^2$ $2(a^2 - a + 1) = 6(1 - a)^2$ Divide both sides by 2: $a^2 - a + 1 = 3(1 - a)^2$ Expand the right side: $a^2 - a + 1 = 3(1 - 2a + a^2)$ $a^2 - a + 1 = 3 - 6a + 3a^2$ Rearrange the terms to form a quadratic equation: $0 = 3a^2 - a^2 - 6a + a + 3 - 1$ $0 = 2a^2 - 5a + 2$ Solve this quadratic equation using factorization or the quadratic formula. By factorization: $2a^2 - 4a - a + 2 = 0$ $2a(a - 2) - 1(a - 2) = 0$ $(2a - 1)(a - 2) = 0$ This gives two possible values for 'a': $2a - 1 = 0 \Rightarrow a = \frac{1}{2}$ $a - 2 = 0 \Rightarrow a = 2$

Step 7: Identify the integral value of 'a'. The problem asks for the integral value of 'a'. From the two solutions, $a = \frac{1}{2}$ and $a = 2$, the integral value is $a = 2$.

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating vector differences, cross products, and dot products. A single sign error can lead to an incorrect quadratic equation.
• Absolute Value: Remember to use the absolute value in the numerator of the shortest distance formula, as distance is always non-negative. Squaring both sides of the equation correctly handles the absolute value, but be mindful of its initial presence.
• Lines Intersecting: If the numerator $(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2})$ evaluates to zero, it implies the lines intersect, and thus the shortest distance is zero. In this problem, if $a=1$, the numerator becomes $2-2(1)=0$, meaning the lines would intersect, and the shortest distance would be 0, which contradicts the given distance $\sqrt{2/3}$. This indicates that $a=1$ is not a valid solution for the given shortest distance.

4. Summary

This problem involved finding an unknown parameter 'a' in the direction vector of a line, given the shortest distance between two skew lines. We began by identifying the position and direction vectors for both lines. Then, we systematically calculated the required components of the shortest distance formula: the vector connecting points on the lines, the cross product of the direction vectors, the scalar triple product, and the magnitude of the cross product. Substituting these into the formula and squaring both sides led to a quadratic equation in 'a'. Solving this quadratic equation yielded two values for 'a', from which the integral value was selected. The integral value of 'a' found was 2.

5. Final Answer

The final answer is $\boxed{2}$.