Key Concepts and Formulas

• Shortest Distance Between Skew Lines: For two skew lines given by $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$ and $\overrightarrow r = \overrightarrow c + \mu \overrightarrow d$, the shortest distance $D$ is given by: $D = \frac{|(\overrightarrow a - \overrightarrow c ) \cdot (\overrightarrow b \times \overrightarrow d )|}{|\overrightarrow b \times \overrightarrow d |}$ Here, $\overrightarrow a$ and $\overrightarrow c$ are position vectors of points on the respective lines, and $\overrightarrow b$ and $\overrightarrow d$ are their direction vectors. The term $(\overrightarrow a - \overrightarrow c)$ is a vector connecting a point on the first line to a point on the second line.
• Vector Operations: Essential vector operations include subtraction, dot product, and cross product. The cross product $\overrightarrow b \times \overrightarrow d$ yields a vector perpendicular to both $\overrightarrow b$ and $\overrightarrow d$. The scalar triple product $(\overrightarrow a - \overrightarrow c ) \cdot (\overrightarrow b \times \overrightarrow d )$ represents the volume of the parallelepiped formed by the three vectors.

Step-by-Step Solution

Step 1: Identify the position and direction vectors from the given line equations. We are given two lines: Line 1: $\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)$ Line 2: $\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k)$

Comparing with the standard form $\overrightarrow r = \overrightarrow p + t \overrightarrow v$: For Line 1:

• Position vector of a point on the line: $\overrightarrow a = \alpha \widehat i + 2\widehat j + 2\widehat k$
• Direction vector of the line: $\overrightarrow b = \widehat i - 2\widehat j + 2\widehat k$

For Line 2:

• Position vector of a point on the line: $\overrightarrow c = - 4\widehat i + 0\widehat j - \widehat k$
• Direction vector of the line: $\overrightarrow d = 3\widehat i - 2\widehat j - 2\widehat k$

We are given that the shortest distance $D = 9$ and $\alpha > 0$.

Step 2: Calculate the vector connecting a point on Line 1 to a point on Line 2. This vector is given by $(\overrightarrow a - \overrightarrow c)$. $\overrightarrow a - \overrightarrow c = (\alpha \widehat i + 2\widehat j + 2\widehat k) - (- 4\widehat i + 0\widehat j - \widehat k)$ $\overrightarrow a - \overrightarrow c = (\alpha - (-4))\widehat i + (2 - 0)\widehat j + (2 - (-1))\widehat k$ $\overrightarrow a - \overrightarrow c = (\alpha + 4)\widehat i + 2\widehat j + 3\widehat k$

Step 3: Calculate the cross product of the direction vectors $\overrightarrow b$ and $\overrightarrow d$. The cross product $\overrightarrow b \times \overrightarrow d$ gives a vector perpendicular to both lines, which is in the direction of the shortest distance. $\overrightarrow b \times \overrightarrow d = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix}$ $= \widehat i ((-2)(-2) - (2)(-2)) - \widehat j ((1)(-2) - (2)(3)) + \widehat k ((1)(-2) - (-2)(3))$ $= \widehat i (4 + 4) - \widehat j (-2 - 6) + \widehat k (-2 + 6)$ $= 8\widehat i + 8\widehat j + 4\widehat k$

Step 4: Calculate the magnitude of the cross product $|\overrightarrow b \times \overrightarrow d|$. This magnitude will be the denominator in the shortest distance formula. $|\overrightarrow b \times \overrightarrow d | = \sqrt{8^2 + 8^2 + 4^2}$ $= \sqrt{64 + 64 + 16}$ $= \sqrt{144}$ $|\overrightarrow b \times \overrightarrow d | = 12$

Step 5: Calculate the scalar triple product $(\overrightarrow a - \overrightarrow c ) \cdot (\overrightarrow b \times \overrightarrow d )$. This will be the numerator (before taking the absolute value) in the shortest distance formula. $(\overrightarrow a - \overrightarrow c ) \cdot (\overrightarrow b \times \overrightarrow d ) = ((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k) \cdot (8\widehat i + 8\widehat j + 4\widehat k)$ $= (\alpha + 4)(8) + (2)(8) + (3)(4)$ $= 8\alpha + 32 + 16 + 12$ $= 8\alpha + 60$

Step 6: Apply the shortest distance formula and solve for $\alpha$. We are given that the shortest distance $D = 9$. $D = \frac{|(\overrightarrow a - \overrightarrow c ) \cdot (\overrightarrow b \times \overrightarrow d )|}{|\overrightarrow b \times \overrightarrow d |}$ $9 = \frac{|8\alpha + 60|}{12}$ Multiply both sides by 12: $9 \times 12 = |8\alpha + 60|$ $108 = |8\alpha + 60|$ Divide by 4 inside the absolute value for simplification: $108 = |4(2\alpha + 15)|$ $108 = 4|2\alpha + 15|$ Divide both sides by 4: $\frac{108}{4} = |2\alpha + 15|$ $27 = |2\alpha + 15|$ This equation gives two possibilities: Case 1: $2\alpha + 15 = 27$ $2\alpha = 27 - 15$ $2\alpha = 12$ $\alpha = 6$ Case 2: $2\alpha + 15 = -27$ $2\alpha = -27 - 15$ $2\alpha = -42$ $\alpha = -21$

Step 7: Apply the given condition $\alpha > 0$ to find the correct value of $\alpha$. The problem states that $\alpha > 0$. From Case 1, $\alpha = 6$, which satisfies $\alpha > 0$. From Case 2, $\alpha = -21$, which does not satisfy $\alpha > 0$. Therefore, the only valid value for $\alpha$ is 6.

Common Mistakes & Tips

• Sign Errors in Cross Product: Double-check the determinant calculation for the cross product, as sign errors are common and can lead to incorrect results.
• Absolute Value: Always remember to take the absolute value of the scalar triple product in the numerator of the shortest distance formula, as distance is a non-negative quantity.
• Condition Interpretation: Ensure you use all given conditions (like $\alpha > 0$) to select the correct answer from the possible solutions derived from the absolute value equation.

Summary

To find the value of $\alpha$, we identified the position and direction vectors of the two given skew lines. We then applied the formula for the shortest distance between skew lines, which involves calculating the vector connecting points on the lines, the cross product of their direction vectors, and the magnitude of this cross product. By setting the given shortest distance (9) equal to the formula and solving the resulting absolute value equation, we obtained two potential values for $\alpha$. Applying the condition $\alpha > 0$, we determined the unique correct value of $\alpha$.

The final answer is $\boxed{6}$.