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Step-by-Step Solution
Step 1: Identify the Given Lines
We are given two lines in vector form:
Line 1:
\overrightarrow{r_1}
= (\alpha \hat{i} + 2\hat{j} + 2\hat{k})
+ \lambda \,(\hat{i} - 2\hat{j} + 2\hat{k})
Here,
\overrightarrow{a} = \alpha \hat{i} + 2\hat{j} + 2\hat{k},
\quad
\overrightarrow{b} = \hat{i} - 2\hat{j} + 2\hat{k}.
Line 2:
\overrightarrow{r_2}
= (-4\hat{i} - \hat{k})
+ \mu \,(3\hat{i} - 2\hat{j} - 2\hat{k})
Here,
\overrightarrow{c} = -4\hat{i} - \hat{k},
\quad
\overrightarrow{d} = 3\hat{i} - 2\hat{j} - 2\hat{k}.
Step 2: Write the Formula for Shortest Distance
The shortest distance L between two skew lines
\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}
and
\overrightarrow{r} = \overrightarrow{c} + \mu \overrightarrow{d}
is given by:
L
= \frac{\left| \bigl(\overrightarrow{a} - \overrightarrow{c}\bigr)\,\cdot
\bigl(\overrightarrow{b} \times \overrightarrow{d}\bigr) \right|}
{\bigl|\overrightarrow{b} \times \overrightarrow{d}\bigr|}.
Step 3: Compute the Vectors Needed
Compute
\overrightarrow{a} - \overrightarrow{c}:
\overrightarrow{a} = (\alpha, 2, 2),
\quad
\overrightarrow{c} = (-4, 0, -1).
Hence,
\overrightarrow{a} - \overrightarrow{c}
= (\alpha + 4, 2 - 0, 2 - (-1))
= (\alpha + 4, 2, 3).
Compute
\overrightarrow{b} \times \overrightarrow{d}:
\overrightarrow{b} = (1, -2, 2),
\quad
\overrightarrow{d} = (3, -2, -2).
The cross product
\overrightarrow{b} \times \overrightarrow{d}
can be determined (though we do not need each step shown explicitly here) and
a simplified form for
\frac{\overrightarrow{b} \times \overrightarrow{d}}
{\bigl|\overrightarrow{b} \times \overrightarrow{d}\bigr|}
is:
\frac{(2\hat{i} + 2\hat{j} + \hat{k})}{3}.
Step 4: Apply the Shortest Distance Formula
Since the shortest distance is given to be 9, we set up the equation:
\left(\overrightarrow{a} - \overrightarrow{c}\right)
\,\cdot
\frac{\overrightarrow{b} \times \overrightarrow{d}}
{\bigl|\overrightarrow{b} \times \overrightarrow{d}\bigr|}
= 9.
Substitute
\overrightarrow{a} - \overrightarrow{c}
= (\alpha + 4, 2, 3)
and
\frac{\overrightarrow{b} \times \overrightarrow{d}}
{|\overrightarrow{b} \times \overrightarrow{d}|}
= \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right).
Thus the dot product is:
(\alpha + 4,\, 2,\, 3)
\,\cdot
\left(\frac{2}{3}, \,\frac{2}{3}, \,\frac{1}{3}\right)
= \frac{1}{3}\bigl[2(\alpha + 4) + 2\times 2 + 3\bigr].
Simplify inside the bracket:
2(\alpha + 4) = 2\alpha + 8,
\quad
2 \times 2 = 4.
So the dot product becomes:
\frac{1}{3} \bigl(2\alpha + 8 + 4 + 3\bigr)
= \frac{1}{3} \bigl(2\alpha + 15\bigr).
We set this equal to 9:
\frac{2\alpha + 15}{3} = 9
\quad\Longrightarrow\quad
2\alpha + 15 = 27
\quad\Longrightarrow\quad
2\alpha = 12
\quad\Longrightarrow\quad
\alpha = 6.
Step 5: State the Final Answer
Hence, the value of
\alpha
satisfying the condition that the shortest distance between the given lines is 9 is
\boxed{6}.
