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

Step 1: Recall the formula.
For two lines $\vec r=\vec a+t\vec b$ and $\vec r=\vec c+s\vec d$, the shortest distance is \[ D=\frac{|(\vec c-\vec a)\cdot(\vec b\times\vec d)|}{|\vec b\times\vec d|}. \]

Step 2: Find the joining vector.
$\vec c-\vec a=(6-1)\hat i+(2+2)\hat j+(2-2)\hat k=5\hat i+4\hat j$.

Step 3: Compute the cross product $\vec b\times\vec d$.
With $\vec b=(3,-2,-2)$ and $\vec d=(-4,0,-1)$, \[ \vec b\times\vec d=((-2)(-1)-(-2)(0))\hat i-((3)(-1)-(-2)(-4))\hat j+((3)(0)-(-2)(-4))\hat k. \] This gives $\vec b\times\vec d=2\hat i+11\hat j-8\hat k$.

Step 4: Length of the cross product.
$|\vec b\times\vec d|=\sqrt{2^2+11^2+(-8)^2}=\sqrt{4+121+64}=\sqrt{189}=3\sqrt{21}$.

Step 5: Dot with the joining vector.
$(\vec c-\vec a)\cdot(\vec b\times\vec d)=(5)(2)+(4)(11)+(0)(-8)=10+44=54$.

Step 6: Put it all together.
\[ D=\frac{54}{3\sqrt{21}}=\frac{18}{\sqrt{21}}. \] Writing $\sqrt{21}=\sqrt3\,\sqrt7$ and using $\dfrac{18}{\sqrt3}=6\sqrt3$, this becomes $\dfrac{6\sqrt3}{\sqrt7}$.

Step 7: State the answer.
The shortest distance is \[ \frac{6\sqrt3}{\sqrt7}. \] \[ \boxed{\dfrac{6\sqrt3}{\sqrt7}} \]