Question

Find the shortest distance between the lines \[ \vec r= \hat i+2\hat j+\hat k+\lambda(\hat i-\hat j+\hat k) \] and \[ \vec r= 2\hat i-\hat j-\hat k+\mu(2\hat i+\hat j+2\hat k) \]

• A. \(3/\sqrt2\)
• B. \(9/\sqrt{54}\)
• C. \(\sqrt6\)
• D. \(0\)

Hint:

Cross product gives a vector perpendicular to both lines and is essential in shortest-distance problems.

Answer 1

The Correct Option is B

Concept: Shortest distance between two skew lines is: \[ d= \frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|} {|\vec b_1\times\vec b_2|} \] where:
• \(\vec a_1,\vec a_2\) are points on the lines
• \(\vec b_1,\vec b_2\) are direction vectors

Step 1: Identify position and direction vectors From first line: \[ \vec a_1=(1,2,1) \] \[ \vec b_1=(1,-1,1) \] From second line: \[ \vec a_2=(2,-1,-1) \] \[ \vec b_2=(2,1,2) \]

Step 2: Find cross product of direction vectors \[ \vec b_1\times\vec b_2= \begin{vmatrix} \hat i&\hat j&\hat k\\ 1&-1& 1\\ 2& 1& 2 \end{vmatrix} \] Expanding determinant: \[ = \hat i((-1)(2)-(1)(1)) -\hat j((1)(2)-(1)(2)) +\hat k((1)(1)-(-1)(2)) \] \[ = \hat i(-2-1)-\hat j(2-2)+\hat k(1+2) \] \[ =-3\hat i+0\hat j+3\hat k \] Thus: \[ \vec b_1\times\vec b_2=(-3,0,3) \] Magnitude: \[ |\vec b_1\times\vec b_2| = \sqrt{(-3)^2+0^2+3^2} \] \[ =\sqrt{9+9} \] \[ =\sqrt{18} \]

Step 3: Find connecting vector \[ \vec a_2-\vec a_1 =(2-1,-1-2,-1-1) \] \[ =(1,-3,-2) \]

Step 4: Apply shortest distance formula \[ d= \frac{|(1,-3,-2)\cdot(-3,0,3)|} {\sqrt{18}} \] Compute dot product: \[ =(1)(-3)+(-3)(0)+(-2)(3) \] \[ =-3+0-6 \] \[ =-9 \] Taking modulus: \[ |{-9}|=9 \] Thus: \[ d=\frac{9}{\sqrt{18}} \] Rationalized form: \[ =\frac{9}{\sqrt{54}} \] Final Answer: \[ \boxed{\frac{9}{\sqrt{54}}} \]