Question

The shortest distance between the two lines \[ \vec r=(\hat i-\hat j)+s(\hat j+2\hat k) \] and \[ \vec r=(2\hat i+\hat k)+t(\hat i-\hat j+\hat k) \] is

• A. $4$
• B. $5$
• C. \[ \sqrt{\frac{6}{5}} \]
• D. \[ \sqrt{\frac{8}{7}} \]

Hint:

For skew lines, use $\displaystyle D=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}$.

Answer 1

The Correct Option is D

Step 1: Read off the line data.
Line 1 passes through $\vec a=(1,-1,0)$ with direction $\vec d_1=(0,1,2).$ Line 2 passes through $\vec b=(2,0,1)$ with direction $\vec d_2=(1,-1,1).$
Step 2: Recall the shortest distance formula.
For two skew lines the shortest distance is $D=\dfrac{|(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}.$
Step 3: Compute the cross product.
$\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&2\\1&-1&1\end{vmatrix}=(1\cdot1-2\cdot(-1),\,-(0\cdot1-2\cdot1),\,0\cdot(-1)-1\cdot1)=(3,2,-1).$
Step 4: Its length.
$|\vec d_1\times\vec d_2|=\sqrt{3^2+2^2+(-1)^2}=\sqrt{14}.$
Step 5: The joining vector and dot product.
$\vec b-\vec a=(1,1,1).$ Then $(\vec b-\vec a)\cdot(3,2,-1)=3+2-1=4.$
Step 6: Put it together.
$D=\dfrac{|4|}{\sqrt{14}}=\dfrac{4}{\sqrt{14}}=\sqrt{\dfrac{16}{14}}=\sqrt{\dfrac{8}{7}}.$ \[ \boxed{\sqrt{\dfrac{8}{7}}} \]