Question

The shortest distance between the straight lines through the points \(A_1=(6,2,2)\) and \(A_2=(-4,0,-1)\), in the directions of \((1,-2,2)\) and \((3,-2,-2)\) is

• A. 6
• B. 8
• C. 12
• D. 9

Hint:

Shortest distance between skew lines uses triple product divided by magnitude of cross product: \(d=\frac{|(\Delta a)\cdot(b_1\times b_2)|}{|b_1\times b_2|}\).

Answer 1

The Correct Option is D

Step 1: Use formula for shortest distance between skew lines.
For lines:
\[ \vec{r}=\vec{a_1}+\lambda\vec{b_1},\quad \vec{r}=\vec{a_2}+\mu\vec{b_2} \]
Shortest distance:
\[ d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} \]
Step 2: Identify vectors.
\[ \vec{a_1}=(6,2,2),\quad \vec{a_2}=(-4,0,-1) \]
\[ \vec{b_1}=(1,-2,2),\quad \vec{b_2}=(3,-2,-2) \]
Step 3: Compute \(\vec{a_2}-\vec{a_1}\).
\[ \vec{a_2}-\vec{a_1}=(-10,-2,-3) \]
Step 4: Compute cross product \(\vec{b_1}\times\vec{b_2}\).
\[ \vec{b_1}\times\vec{b_2}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & 2\\ 3 & -2 & -2 \end{vmatrix} \]
\[ = \hat{i}[(-2)(-2)-2(-2)] - \hat{j}[1(-2)-2(3)] + \hat{k}[1(-2)-(-2)(3)] \]
\[ = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) \]
\[ = (8,8,4) \]
Step 5: Compute numerator.
\[ |(-10,-2,-3)\cdot(8,8,4)| = |-80-16-12| = | -108|=108 \]
Step 6: Compute denominator.
\[ |\vec{b_1}\times\vec{b_2}|=\sqrt{8^2+8^2+4^2} =\sqrt{64+64+16} =\sqrt{144}=12 \]
Step 7: Distance.
\[ d=\frac{108}{12}=9 \]
Final Answer:
\[ \boxed{9} \]