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Mathematics Question on Three Dimensional Geometry

Find the shortest distance between the lines whose vector equations are

r→=(1−t)i^+(t−2)j^+(3−2t)k^\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k and

r→=(s+1)i^+(2s−1)j^−(2s+1)k^\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k

Answer

The given lines are

r→=(1−t)i^+(t−2)j^+(3−2t)k^\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k
⇒r→=(i^−2j^+3k^)+t(−i^+j^−2k^)\Rightarrow\overrightarrow r=(\hat i-2\hat j+3\hat k)+t(-\hat i+\hat j-2\hat k)...(1)

r→=(s+1)i^+(2s−1)j^−(2s+1)k^\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k
⇒r→=(i^−j^+k^)+s(i^+2j^−2k^)\Rightarrow\overrightarrow r=(\hat i-\hat j+\hat k)+s(\hat i+2\hat j-2\hat k)...(2)

It is known that the shortest distance between the lines,
r→=a→1+λb→1\overrightarrow r=\overrightarrow a_1+\lambda \overrightarrow b_1 and r→=a→2+μb→2\overrightarrow r=\overrightarrow a_2+\mu \overrightarrow b_2, is given by,
d=|(b1→×b2→).(a2→-a2→)/|b1→×b2→||...(3)

For the given equations,
a→1=i^−2j^+3k^\overrightarrow a_1= \hat i-2\hat j+3\hat k
b→1=−i^+j^−2k^\overrightarrow b_1= -\hat i+\hat j-2\hat k
a→2=i^−j^−k^\overrightarrow a_2= \hat i-\hat j-\hat k
b→2=i^+2j^−2k^\overrightarrow b_2= \hat i+2\hat j-2\hat k

a→2−a→1\overrightarrow a_2-\overrightarrow a_1
=(i^−j^−k^)(\hat i-\hat j-\hat k)-(i^−2j^+3k^)(\hat i-2\hat j+3\hat k)=j^−4k^\hat j-4\hat k

b→1.b→2\overrightarrow b_1.\overrightarrow b_2 = ∣i^j^k^−11−2\12−2∣\begin{vmatrix}\hat i&\hat j&\hat k\\\\-1&1&-2\\\1&2&-2\end{vmatrix}

=(−2+4)i^−(2+2)j^+(−2−1)k^(-2+4)\hat i-(2+2)\hat j+(-2-1)\hat k
=2i^−4j^−3k^2\hat i-4\hat j-3\hat k

⇒∣b→1∗b→2∣\Rightarrow \mid \overrightarrow b_1*\overrightarrow b_2\mid
=(2)2+(−4)2+(−3)2\sqrt{(2)^2+(-4)^2+(-3)^2}
=4+16+9\sqrt{4+16+9}
=29\sqrt{29}

∴(b→1∗b→2 \overrightarrow b_1*\overrightarrow b_2).(a→2−a→1\overrightarrow a_2-\overrightarrow a_1)
=(2i^−4j^−3k^2\hat i-4\hat j-3\hat k).(j^−4k^\hat j-4\hat k^)
=-4+12
=8

Substituting all the values in equation (3), we obtain
d=|829\frac{8}{\sqrt{29}}|
=829\frac{8}{\sqrt{29}}

Therefore, the shortest distance between the lines is 829\frac{8}{\sqrt{29}} units.