Question

The shortest distance between the lines \[ \vec r=\vec a+t\vec b \] and \[ \vec r=\vec c+s\vec d \] where \[ \vec a=\hat i-2\hat j+2\hat k, \quad \vec b=3\hat i-2\hat j-2\hat k, \] \[ \vec c=6\hat i+2\hat j+2\hat k, \quad \vec d=-4\hat i-\hat k, \] is

• A. \(9\)
• B. \(\dfrac{6\sqrt3}{\sqrt7}\)
• C. \(\dfrac{\sqrt7}{2\sqrt3}\)
• D. \(\dfrac{5}{\sqrt3}\)

Hint:

For skew lines, always remember \[ D= \frac{|(\vec c-\vec a)\cdot(\vec b\times\vec d)|} {|\vec b\times\vec d|}. \] It is one of the most frequently tested formulas in three-dimensional geometry.

Answer 1

The Correct Option is B

Concept: The shortest distance between two skew lines \[ \vec r=\vec a+t\vec b \] and \[ \vec r=\vec c+s\vec d \] is given by \[ D= \frac{|(\vec c-\vec a)\cdot(\vec b\times\vec d)|} {|\vec b\times\vec d|}. \] This formula is obtained from the projection of the vector joining any two points on the lines onto the common perpendicular.

Step 1: Compute \(\vec c-\vec a\) \[ \vec c-\vec a = (6-1)\hat i + (2+2)\hat j + (2-2)\hat k. \] \[ \vec c-\vec a = 5\hat i+4\hat j. \]

Step 2: Compute \(\vec b\times\vec d\) \[ \vec b = (3,-2,-2), \qquad \vec d = (-4,0,-1). \] \[ \vec b\times\vec d = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -2 & -2\\ -4 & 0 & -1 \end{vmatrix}. \] Evaluating, \[ \vec b\times\vec d = 2\hat i+11\hat j-8\hat k. \]

Step 3: Compute the numerator \[ (\vec c-\vec a)\cdot(\vec b\times\vec d) = (5,4,0)\cdot(2,11,-8). \] \[ = 10+44. \] \[ =54. \]

Step 4: Compute the denominator \[ |\vec b\times\vec d| = \sqrt{2^2+11^2+(-8)^2}. \] \[ = \sqrt{4+121+64}. \] \[ = \sqrt{189}. \] \[ = 3\sqrt{21}. \]

Step 5: Determine the distance \[ D = \frac{54}{3\sqrt{21}} = \frac{18}{\sqrt{21}} = \frac{6\sqrt3}{\sqrt7}. \] Hence, \[ \boxed{\frac{6\sqrt3}{\sqrt7}}. \]