Question

Find the shortest distance between the skew lines \[ \vec r=(1,0,0)+\lambda(1,1,0) \] and \[ \vec r=(0,1,1)+\mu(0,1,1) \]

• A. \(1\)
• B. \(\sqrt2\)
• C. \(\frac1{\sqrt2}\)
• D. \(\sqrt3\)

Hint:

For shortest-distance problems, always verify the answer numerically because answer keys occasionally contain mistakes.

Answer 1

The Correct Option is C

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

Step 1: Identify vectors \[ \vec a_1=(1,0,0) \] \[ \vec a_2=(0,1,1) \] Direction vectors: \[ \vec b_1=(1,1,0) \] \[ \vec b_2=(0,1,1) \]

Step 2: Find cross product \[ \vec b_1\times\vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k 1& 1& 0 0& 1& 1 \end{vmatrix} \] \[ =\hat i-\hat j+\hat k \] Magnitude: \[ |\vec b_1\times\vec b_2| = \sqrt{1^2+(-1)^2+1^2} = \sqrt3 \]

Step 3: Find joining vector \[ \vec a_2-\vec a_1 = (-1,1,1) \]

Step 4: Apply formula \[ d= \frac{|(-1,1,1)\cdot(1,-1,1)|} {\sqrt3} \] \[ = \frac{|-1-1+1|} {\sqrt3} \] \[ = \frac1{\sqrt3} \] Thus \[ \boxed{\frac1{\sqrt3}} \] Hence the answer key provided in the question appears incorrect.