Question

If the shortest distance between the two skew lines \(\overline{r}=\overline{i}+\overline{j}+\overline{k}+t(3\overline{i}+2\overline{j}+\overline{k})\) and \(\overline{r}=\overline{i}-\overline{j}+x\overline{k}+s(\overline{i}+2\overline{j}+3\overline{k})\) is at most \(2\sqrt{6}\), then all values of \(x\) lie in the interval:

• A. \([-9, 15]\)
• B. \([-15, 9]\)
• C. \((-\infty, -15)\)
• D. \(R-[-11, 9]\)

Hint:

Always compute cross product first in skew line problems to simplify distance formula.

Answer 1

The Correct Option is A

Concept: Shortest distance between skew lines: \[ d=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|} \]

Step 1: Identify direction vectors and points.
\[ \overrightarrow{b_1}=(3,2,1), \quad \overrightarrow{b_2}=(1,2,3) \] \[ \overrightarrow{a_2}-\overrightarrow{a_1}=(0,-2,x-1) \]

Step 2: Compute cross product.
\[ \overrightarrow{b_1}\times\overrightarrow{b_2}=(4,-8,4) \] \[ |\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{16+64+16}=\sqrt{96}=4\sqrt{6} \]

Step 3: Compute scalar triple product.
\[ (0,-2,x-1)\cdot(4,-8,4)=16+4x-4=4x+12 \]

Step 4: Apply distance condition.
\[ d=\frac{|4x+12|}{4\sqrt{6}}=\frac{|x+3|}{\sqrt{6}} \] \[ \frac{|x+3|}{\sqrt{6}}\le 2\sqrt{6} \Rightarrow |x+3|\le 12 \] \[ -12\le x+3\le 12 \Rightarrow -15\le x\le 9 \]