\displaystyle \textbf{Question 1: }~\text{Find the shortest distance between the following} \\ \text{pairs of lines whose vector equations are:}
\displaystyle (i)\ \overrightarrow{r}=3\widehat{i}+8\widehat{j}+3\widehat{k}+\lambda(3\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=-3\widehat{i}-7\widehat{j}+6\widehat{k}+\mu(-3\widehat{i}+2\widehat{j}+4\widehat{k})

\displaystyle (ii)\ \overrightarrow{r}=(3\widehat{i}+5\widehat{j}+7\widehat{k})+\lambda(\widehat{i}-2\widehat{j}+7\widehat{k})\text{ and }\overrightarrow{r}=-\widehat{i}-\widehat{j}-\widehat{k}+\mu(7\widehat{i}-6\widehat{j}+\widehat{k})

\displaystyle (iii)\ \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}+4\widehat{j}+5\widehat{k})+\mu(3\widehat{i}+4\widehat{j}+5\widehat{k})

\displaystyle (iv)\ \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-t)\widehat{k}\text{ and }\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}

\displaystyle (v)\ \overrightarrow{r}=(\lambda-1)\widehat{i}+(\lambda+1)\widehat{j}-(1+\lambda)\widehat{k}\text{ and }\overrightarrow{r}=(1-\mu)\widehat{i}+(2\mu-1)\widehat{j}+(\mu+2)\widehat{k}

\displaystyle (vi)\ \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\lambda(2\widehat{i}-5\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\mu(\widehat{i}-\widehat{j}+\widehat{k})\text{ [CBSE 2008]}

\displaystyle (vii)\ \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(2\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{i}+\widehat{j}-\widehat{k}+\mu(3\widehat{i}-5\widehat{j}+2\widehat{k})\text{ [CBSE 2014]}

\displaystyle (viii)\ \overrightarrow{r}=(8+3\lambda)\widehat{i}-(9+16\lambda)\widehat{j}+(10+7\lambda)\widehat{k}\text{ and }\overrightarrow{r}=15\widehat{i}+29\widehat{j}+5\widehat{k}+\mu(3\widehat{i}+8\widehat{j}-5\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \overrightarrow{r}=3\widehat{i}+8\widehat{j}+3\widehat{k}+\lambda(3\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=-3\widehat{i}-7\widehat{j}+6\widehat{k}+\mu(-3\widehat{i}+2\widehat{j}+4\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \overrightarrow{a_1}=3\widehat{i}+8\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{a_2}=-3\widehat{i}-7\widehat{j}+6\widehat{k}
\displaystyle \overrightarrow{b_1}=3\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b_2}=-3\widehat{i}+2\widehat{j}+4\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=-6\widehat{i}-15\widehat{j}+3\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-1&1\\-3&2&4\end{vmatrix}
\displaystyle =-6\widehat{i}-15\widehat{j}+3\widehat{k}
\displaystyle =\sqrt{36+225+9}
\displaystyle =\sqrt{270}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-6\widehat{i}-15\widehat{j}+3\widehat{k})\cdot(-6\widehat{i}-15\widehat{j}+3\widehat{k})
\displaystyle =36+225+9
\displaystyle =270
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\frac{270}{\sqrt{270}}
\displaystyle =\sqrt{270}
\displaystyle \textbf{(ii)  }
\displaystyle \overrightarrow{r}=(3\widehat{i}+5\widehat{j}+7\widehat{k})+\lambda(\widehat{i}-2\widehat{j}+7\widehat{k})\text{ and }\overrightarrow{r}=-\widehat{i}-\widehat{j}-\widehat{k}+\mu(7\widehat{i}-6\widehat{j}+\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2},
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=3\widehat{i}+5\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{a_2}=-\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_1}=\widehat{i}-2\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{b_2}=7\widehat{i}-6\widehat{j}+\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=-4\widehat{i}-6\widehat{j}-8\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&7\\7&-6&1\end{vmatrix}
\displaystyle =40\widehat{i}+48\widehat{j}+8\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{40^2+48^2+8^2}
\displaystyle =\sqrt{1600+2304+64}
\displaystyle =\sqrt{3968}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-4\widehat{i}-6\widehat{j}-8\widehat{k})\cdot(40\widehat{i}+48\widehat{j}+8\widehat{k})
\displaystyle =-160-288-64
\displaystyle =-512
\displaystyle \text{The shortest distance between the lines,}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\frac{\left|-512\right|}{\sqrt{3968}}
\displaystyle =\frac{512}{\sqrt{3968}}
\displaystyle \textbf{(iii)  }
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}+4\widehat{j}+5\widehat{k})+\mu(3\widehat{i}+4\widehat{j}+5\widehat{k})
\displaystyle \text{Comparing the given equations with the equations,}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{a_2}=2\widehat{i}+4\widehat{j}+5\widehat{k}
\displaystyle \overrightarrow{b_1}=2\widehat{i}+3\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{b_2}=3\widehat{i}+4\widehat{j}+5\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&4\\3&4&5\end{vmatrix}
\displaystyle =-\widehat{i}+2\widehat{j}-\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-1)^2+2^2+(-1)^2}
\displaystyle =\sqrt{1+4+1}
\displaystyle =\sqrt{6}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{i}+2\widehat{j}+2\widehat{k})\cdot(-\widehat{i}+2\widehat{j}-\widehat{k})
\displaystyle =-1+4-2
\displaystyle =1
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{1}{\sqrt{6}}\right|
\displaystyle =\frac{1}{\sqrt{6}}
\displaystyle \textbf{(iv)  }
\displaystyle \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-t)\widehat{k}\text{ and }\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}
\displaystyle \text{The vector equations of the given lines can be re-written as}
\displaystyle \overrightarrow{r}=\widehat{i}-2\widehat{j}+3\widehat{k}+t(-\widehat{i}+\widehat{j}-\widehat{k})\text{ and }\overrightarrow{r}=\widehat{i}-\widehat{j}-\widehat{k}+s(\widehat{i}+2\widehat{j}-2\widehat{k})
\displaystyle \text{Comparing the given equations with the equations }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}-2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{a_2}=\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_1}=-\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_2}=\widehat{i}+2\widehat{j}-2\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{j}-4\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&1&-1\\1&2&-2\end{vmatrix}
\displaystyle =-3\widehat{j}-3\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-3)^2+(-3)^2}
\displaystyle =\sqrt{9+9}
\displaystyle =3\sqrt{2}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{j}-4\widehat{k})\cdot(-3\widehat{j}-3\widehat{k})
\displaystyle =-3+12
\displaystyle =9
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{9}{3\sqrt{2}}\right|
\displaystyle =\frac{3}{\sqrt{2}}
\displaystyle \textbf{(v)  }
\displaystyle \overrightarrow{r}=(\lambda-1)\widehat{i}+(\lambda+1)\widehat{j}-(1+\lambda)\widehat{k}\text{ and }\overrightarrow{r}=(1-\mu)\widehat{i}+(2\mu-1)\widehat{j}+(\mu+2)\widehat{k}
\displaystyle \text{The vector equations of the given lines can be re-written as}
\displaystyle \overrightarrow{r}=-\widehat{i}+\widehat{j}-\widehat{k}+\lambda(\widehat{i}+\widehat{j}-\widehat{k})\text{ and }\overrightarrow{r}=\widehat{i}-\widehat{j}+2\widehat{k}+\mu(-\widehat{i}+2\widehat{j}+\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=-\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{a_2}=\widehat{i}-\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_1}=\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_2}=-\widehat{i}+2\widehat{j}+\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=2\widehat{i}-2\widehat{j}+3\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&1&-1\\-1&2&1\end{vmatrix}
\displaystyle =3\widehat{i}+3\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{3^2+3^2}
\displaystyle =3\sqrt{2}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(2\widehat{i}-2\widehat{j}+3\widehat{k})\cdot(3\widehat{i}+3\widehat{k})
\displaystyle =6+9
\displaystyle =15
\displaystyle \text{The shortest distance between the lines,}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{15}{3\sqrt{2}}\right|
\displaystyle =\frac{5}{\sqrt{2}}
\displaystyle \textbf{(vi)  }
\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\lambda(2\widehat{i}-5\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\mu(\widehat{i}-\widehat{j}+\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=2\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{a_2}=\widehat{i}+2\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b_1}=2\widehat{i}-5\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_2}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=-\widehat{i}+3\widehat{j}+2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&-5&2\\1&-1&1\end{vmatrix}
\displaystyle =-3\widehat{i}+3\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-3)^2+3^2}
\displaystyle =\sqrt{9+9}
\displaystyle =3\sqrt{2}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-\widehat{i}+3\widehat{j}+2\widehat{k})\cdot(-3\widehat{i}+3\widehat{k})
\displaystyle =3+6
\displaystyle =9
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{9}{3\sqrt{2}}\right|
\displaystyle =\frac{3}{\sqrt{2}}
\displaystyle \textbf{(vii)  }
\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(2\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{i}+\widehat{j}-\widehat{k}+\mu(3\widehat{i}-5\widehat{j}+2\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{We get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}+\widehat{j}
\displaystyle \overrightarrow{a_2}=2\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_1}=2\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b_2}=3\widehat{i}-5\widehat{j}+2\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}-\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&-1&1\\3&-5&2\end{vmatrix}
\displaystyle =3\widehat{i}-\widehat{j}-7\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{3^2+(-1)^2+(-7)^2}
\displaystyle =\sqrt{9+1+49}
\displaystyle =\sqrt{59}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{i}-\widehat{k})\cdot(3\widehat{i}-\widehat{j}-7\widehat{k})
\displaystyle =3+7
\displaystyle =10
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{10}{\sqrt{59}}\right|
\displaystyle =\frac{10}{\sqrt{59}}
\displaystyle \textbf{(viii)  }
\displaystyle \text{The vector equations of the given lines can be re-written as}
\displaystyle \overrightarrow{r}=8\widehat{i}-9\widehat{j}+10\widehat{k}+\lambda(3\widehat{i}-16\widehat{j}+7\widehat{k})\text{ and }\overrightarrow{r}=15\widehat{i}+29\widehat{j}+5\widehat{k}+\mu(3\widehat{i}+8\widehat{j}-5\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{We get,}
\displaystyle \overrightarrow{a_1}=8\widehat{i}-9\widehat{j}+10\widehat{k}
\displaystyle \overrightarrow{b_1}=3\widehat{i}-16\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{a_2}=15\widehat{i}+29\widehat{j}+5\widehat{k}
\displaystyle \overrightarrow{b_2}=3\widehat{i}+8\widehat{j}-5\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=(15\widehat{i}+29\widehat{j}+5\widehat{k})-(8\widehat{i}-9\widehat{j}+10\widehat{k})=7\widehat{i}+38\widehat{j}-5\widehat{k}
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-16&7\\3&8&-5\end{vmatrix}=24\widehat{i}+36\widehat{j}+72\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{24^2+36^2+72^2}=\sqrt{576+1296+5184}=\sqrt{7056}=84
\displaystyle \text{Also,}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(7\widehat{i}+38\widehat{j}-5\widehat{k})\cdot(24\widehat{i}+36\widehat{j}+72\widehat{k})
\displaystyle =7\times24+38\times36+(-5)\times72
\displaystyle =168+1368-360
\displaystyle =1176
\displaystyle \text{We know that the shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle \text{Required shortest distance between the given pairs of lines,}
\displaystyle d=\left|\frac{1176}{84}\right|
\displaystyle =14

\displaystyle \textbf{Question 2: }~\text{Find the shortest distance between the following pairs of} \\ \text{lines whose cartesian equations are:}
\displaystyle (i)\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\text{ and }\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{5}\text{ [CBSE 2005]}
\displaystyle (ii)\ \frac{x-1}{2}=\frac{y+1}{3}=z\text{ and }\frac{x+1}{3}=\frac{y-2}{1},\ z=2
\displaystyle (iii)\ \frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}\text{ and }\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}
\displaystyle (iv)\ \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\text{ and }\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\text{ [CBSE 2008, 2014]}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{The equations of the given lines are}
\displaystyle \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\qquad\text{...(1)}
\displaystyle \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{5}\qquad\text{...(2)}
\displaystyle \text{Since line (1) passes through the point }(1,2,3)\text{ and has direction ratios proportional to } \\ (2,3,4),\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b_1}=2\widehat{i}+3\widehat{j}+4\widehat{k}
\displaystyle \text{Also, line (2) passes through the point }(2,3,5)\text{ and has direction ratios proportional to }(3,4,5).
\displaystyle \text{Its vector equation is }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=2\widehat{i}+3\widehat{j}+5\widehat{k}
\displaystyle \overrightarrow{b_2}=3\widehat{i}+4\widehat{j}+5\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}+\widehat{j}+2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&4\\3&4&5\end{vmatrix}
\displaystyle =-\widehat{i}+2\widehat{j}-\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-1)^2+2^2+(-1)^2}
\displaystyle =\sqrt{1+4+1}
\displaystyle =\sqrt{6}
\displaystyle \text{and }(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{i}+\widehat{j}+2\widehat{k})\cdot(-\widehat{i}+2\widehat{j}-\widehat{k})
\displaystyle =-1+2-2
\displaystyle =-1
\displaystyle \text{The shortest distance between the lines }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{-1}{\sqrt{6}}\right|
\displaystyle =\frac{1}{\sqrt{6}}
\displaystyle \textbf{(ii)  }
\displaystyle \text{The equations of the given lines are}
\displaystyle \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1}\qquad\text{...(1)}
\displaystyle \frac{x+1}{3}=\frac{y-2}{1}=\frac{z-2}{0}\qquad\text{...(2)}
\displaystyle \text{Since line (1) passes through the point }(1,-1,0)\text{ and has direction ratios} \\ \text{proportional to }(2,3,1),\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=\widehat{i}-\widehat{j}+0\widehat{k}
\displaystyle \overrightarrow{b_1}=2\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle \text{Also, line (2) passes through the point }(-1,2,2)\text{ and has direction ratios} \\ \text{proportional to }(3,1,0).\text{ Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=-\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_2}=3\widehat{i}+\widehat{j}+0\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-2\widehat{i}+3\widehat{j}+2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&1\\3&1&0\end{vmatrix}
\displaystyle =-\widehat{i}+3\widehat{j}-7\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-1)^2+3^2+(-7)^2}
\displaystyle =\sqrt{1+9+49}
\displaystyle =\sqrt{59}
\displaystyle \text{and }(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-2\widehat{i}+3\widehat{j}+2\widehat{k})\cdot(-\widehat{i}+3\widehat{j}-7\widehat{k})
\displaystyle =2+9-14
\displaystyle =-3
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{-3}{\sqrt{59}}\right|
\displaystyle =\frac{3}{\sqrt{59}}
\displaystyle \textbf{(iii)  }
\displaystyle \frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}\qquad\text{...(1)}
\displaystyle \frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}\qquad\text{...(2)}
\displaystyle \text{Since line (1) passes through the point }(1,-2,3)\text{ and has direction ratios} \\ \text{proportional to }(-1,1,-2),\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=\widehat{i}-2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b_1}=-\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \text{Also, line (2) passes through the point }(1,-1,-1)\text{ and has direction ratios} \\ \text{proportional to }(1,2,-2).\text{ Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_2}=\widehat{i}+2\widehat{j}-2\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{j}-4\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&1&-2\\1&2&-2\end{vmatrix}
\displaystyle =2\widehat{i}-4\widehat{j}-3\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{2^2+(-4)^2+(-3)^2}
\displaystyle =\sqrt{4+16+9}
\displaystyle =\sqrt{29}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{j}-4\widehat{k})\cdot(2\widehat{i}-4\widehat{j}-3\widehat{k})
\displaystyle =-4+12
\displaystyle =8
\displaystyle \text{The shortest distance between the lines }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{8}{\sqrt{29}}\right|
\displaystyle =\frac{8}{\sqrt{29}}
\displaystyle \textbf{(iv)  }
\displaystyle \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\qquad\text{...(1)}
\displaystyle \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\qquad\text{...(2)}
\displaystyle \text{Since line (1) passes through the point }(3,5,7)\text{ and has direction ratios} \\ \text{proportional to }(1,-2,1),\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=3\widehat{i}+5\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{b_1}=\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \text{Also, line (2) passes through the point }(-1,-1,-1)\text{ and has direction ratios} \\ \text{proportional to }(7,-6,1).\text{ Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=-\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_2}=7\widehat{i}-6\widehat{j}+\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-4\widehat{i}-6\widehat{j}-8\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&1\\7&-6&1\end{vmatrix}
\displaystyle =4\widehat{i}+6\widehat{j}+8\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{4^2+6^2+8^2}
\displaystyle =\sqrt{16+36+64}
\displaystyle =\sqrt{116}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-4\widehat{i}-6\widehat{j}-8\widehat{k})\cdot(4\widehat{i}+6\widehat{j}+8\widehat{k})
\displaystyle =-16-36-64
\displaystyle =-116
\displaystyle \text{The shortest distance between the lines}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle =\left|\frac{-116}{\sqrt{116}}\right|
\displaystyle =\sqrt{116}
\displaystyle =2\sqrt{29}

\displaystyle \textbf{Question 3: }~\text{By computing the shortest distance determine whether the} \\ \text{following pairs of lines intersect or not:}
\displaystyle (i)\ \overrightarrow{r}=(\widehat{i}-\widehat{j})+\lambda(2\widehat{i}+\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}-\widehat{j})+\mu(\widehat{i}+\widehat{j}-\widehat{k})
\displaystyle (ii)\ \overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})\text{ and }\overrightarrow{r}=(4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})
\displaystyle (iii)\ \frac{x-1}{2}=\frac{y+1}{3}=z\text{ and }\frac{x+1}{5}=\frac{y-2}{1},\ z=2
\displaystyle (iv)\ \frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}\text{ and }\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j})+\lambda(2\widehat{i}+\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}-\widehat{j})+\mu(\widehat{i}+\widehat{j}-\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{We get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}-\widehat{j}
\displaystyle \overrightarrow{a_2}=2\widehat{i}-\widehat{j}
\displaystyle \overrightarrow{b_1}=2\widehat{i}+\widehat{k}
\displaystyle \overrightarrow{b_2}=\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&1\\1&1&-1\end{vmatrix}
\displaystyle =-\widehat{i}+3\widehat{j}+2\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{i})\cdot(-\widehat{i}+3\widehat{j}+2\widehat{k})
\displaystyle =-1
\displaystyle \text{We observe}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})\neq0
\displaystyle \text{Thus, the given lines do not intersect.}
\displaystyle \textbf{(ii)  }
\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})\text{ and }\overrightarrow{r}=(4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{a_2}=4\widehat{i}-\widehat{k}
\displaystyle \overrightarrow{b_1}=3\widehat{i}-\widehat{j}
\displaystyle \overrightarrow{b_2}=2\widehat{i}+3\widehat{k}
\displaystyle \therefore\ \overrightarrow{a_2}-\overrightarrow{a_1}=3\widehat{i}-\widehat{j}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-1&0\\2&0&3\end{vmatrix}
\displaystyle =-3\widehat{i}-9\widehat{j}+2\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(3\widehat{i}-\widehat{j})\cdot(-3\widehat{i}-9\widehat{j}+2\widehat{k})
\displaystyle =-9+9
\displaystyle =0
\displaystyle \text{We observe}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=0
\displaystyle \text{Thus, the given lines intersect.}
\displaystyle \textbf{(iii)  }
\displaystyle \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1}\text{ and }\frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}
\displaystyle \text{Since the first line passes through the point }(1,-1,0)\text{ and has direction ratios} \\ \text{proportional to }(2,3,1),\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{...(1)}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=\widehat{i}-\widehat{j}+0\widehat{k}
\displaystyle \overrightarrow{b_1}=2\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle \text{Also, the second line passes through the point }(-1,2,2)\text{ and has direction ratios} \\ \text{proportional to }(5,1,0).\text{ Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{...(2)}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=-\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_2}=5\widehat{i}+\widehat{j}+0\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-2\widehat{i}+3\widehat{j}+2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&1\\5&1&0\end{vmatrix}
\displaystyle =-\widehat{i}+5\widehat{j}-13\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-2\widehat{i}+3\widehat{j}+2\widehat{k})\cdot(-\widehat{i}+5\widehat{j}-13\widehat{k})
\displaystyle =2+15-26
\displaystyle =-9
\displaystyle \text{We observe}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})\neq0
\displaystyle \text{Thus, the given lines do not intersect.}
\displaystyle \textbf{(iv)  }
\displaystyle \frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}\text{ and }\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}
\displaystyle \text{Since the first line passes through the point }(5,7,-3)\text{ and has direction ratios} \\ \text{proportional to }(4,-5,-5),\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{...(1)}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=5\widehat{i}+7\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{b_1}=4\widehat{i}-5\widehat{j}-5\widehat{k}
\displaystyle \text{Also, the second line passes through the point }(8,7,5)\text{ and has direction ratios} \\ \text{proportional to }(7,1,3).
\displaystyle \text{Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{...(2)}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=8\widehat{i}+7\widehat{j}+5\widehat{k}
\displaystyle \overrightarrow{b_2}=7\widehat{i}+\widehat{j}+3\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=3\widehat{i}+8\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&-5&-5\\7&1&3\end{vmatrix}
\displaystyle =-10\widehat{i}-47\widehat{j}+39\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(3\widehat{i}+8\widehat{k})\cdot(-10\widehat{i}-47\widehat{j}+39\widehat{k})
\displaystyle =-30+312
\displaystyle =282
\displaystyle \text{We observe}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})\neq0
\displaystyle \text{Thus, the given lines do not intersect.}

\displaystyle \textbf{Question 4: }~\text{Find the shortest distance between the following pairs of} \\ \text{parallel lines whose equations are:}
\displaystyle (i)\ \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu(-\widehat{i}+\widehat{j}-\widehat{k})
\displaystyle (ii)\ \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(2\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}+\widehat{k})+\mu(4\widehat{i}-2\widehat{j}+2\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}-\widehat{j}+\widehat{k})\text{...(1)}
\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu(\widehat{i}+\widehat{j}-\widehat{k})
\displaystyle =(2\widehat{i}-\widehat{j}-\widehat{k})-\mu(\widehat{i}-\widehat{j}+\widehat{k})\text{...(2)}
\displaystyle \text{These two lines pass through the points having position vectors }\overrightarrow{a_1}=\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{a_2}=2\widehat{i}-\widehat{j}-\widehat{k}\text{ and are parallel to the vector }\overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}-3\widehat{j}-4\widehat{k}
\displaystyle \text{and}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}=(\widehat{i}-3\widehat{j}-4\widehat{k})\times(\widehat{i}-\widehat{j}+\widehat{k})
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&-4\\1&-1&1\end{vmatrix}
\displaystyle =-7\widehat{i}-5\widehat{j}+2\widehat{k}
\displaystyle \Rightarrow \left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}\right|=\sqrt{(-7)^2+(-5)^2+2^2}
\displaystyle =\sqrt{49+25+4}
\displaystyle =\sqrt{78}
\displaystyle \text{The shortest distance between the two lines is given by}
\displaystyle \frac{\left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}\right|}{|\overrightarrow{b}|}=\frac{\sqrt{78}}{\sqrt{3}}
\displaystyle =\sqrt{26}
\displaystyle \textbf{(ii)  }
\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(2\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}+\widehat{j}-\widehat{k})+\mu(4\widehat{i}-2\widehat{j}+2\widehat{k})
\displaystyle \text{or }\overrightarrow{r}=(2\widehat{i}+\widehat{j}-\widehat{k})+2\mu(2\widehat{i}-\widehat{j}+\widehat{k})
\displaystyle \text{These two lines pass through the points having position vectors }\overrightarrow{a_1}=\widehat{i}+\widehat{j}\text{ and }\overrightarrow{a_2}=2\widehat{i}+\widehat{j}-\widehat{k}\text{ and are parallel to the vector }\overrightarrow{b}=2\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}-\widehat{k}
\displaystyle \text{and}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}=(\widehat{i}-\widehat{k})\times(2\widehat{i}-\widehat{j}+\widehat{k})
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&0&-1\\2&-1&1\end{vmatrix}
\displaystyle =-\widehat{i}-3\widehat{j}-\widehat{k}
\displaystyle \Rightarrow \left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}\right|=\sqrt{(-1)^2+(-3)^2+(-1)^2}
\displaystyle =\sqrt{1+9+1}
\displaystyle =\sqrt{11}
\displaystyle \text{The shortest distance between the two lines is given by}
\displaystyle \frac{\left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}\right|}{|\overrightarrow{b}|}=\frac{\sqrt{11}}{\sqrt{6}}

\displaystyle \textbf{Question 5: }~\text{Find the equations of the lines joining the following pairs of} \\ \text{vertices and then find the shortest distance between the lines:}
\displaystyle (i)\ (0,0,0)\text{ and }(1,0,2)\quad (ii)\ (1,3,0)\text{ and }(0,3,0)
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{The equation of the line passing through the points }(0,0,0)\text{ and }(1,0,2)\text{ is}
\displaystyle \frac{x-0}{1-0}=\frac{y-0}{0-0}=\frac{z-0}{2-0}
\displaystyle \Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{2}
\displaystyle \textbf{(ii)  }
\displaystyle \text{The equation of the line passing through the points }(1,3,0)\text{ and }(0,3,0)\text{ is }
\displaystyle \frac{x-1}{0-1}=\frac{y-3}{3-3}=\frac{z-0}{0-0}
\displaystyle \Rightarrow \frac{x-1}{-1}=\frac{y-3}{0}=\frac{z}{0}
\displaystyle \text{Since the first line passes through }(0,0,0)\text{ and has direction ratios }1,0,2,
\displaystyle \vec r=\vec a_1+\lambda\vec b_1
\displaystyle \vec a_1=0\hat i+0\hat j+0\hat k, \qquad \vec b_1=\hat i+0\hat j+2\hat k
\displaystyle \text{Also, the second line passes through }(1,3,0)\text{ and has direction ratios }-1,0,0.
\displaystyle \vec r=\vec a_2+\mu\vec b_2
\displaystyle \vec a_2=\hat i+3\hat j+0\hat k, \qquad \vec b_2=-\hat i+0\hat j+0\hat k
\displaystyle \vec a_2-\vec a_1=\hat i+3\hat j+0\hat k
\displaystyle \vec b_1\times\vec b_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&0&2\\ -1&0&0 \end{vmatrix} =0\hat i-2\hat j+0\hat k
\displaystyle |\vec b_1\times\vec b_2| =\sqrt{0^2+(-2)^2+0^2}=2
\displaystyle (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) =(\hat i+3\hat j)(0\hat i-2\hat j) =-6
\displaystyle \text{The shortest distance between the lines is}
\displaystyle d=\left| \frac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)} {|\vec b_1\times\vec b_2|} \right|
\displaystyle =\left|\frac{-6}{2}\right|=3
\displaystyle \therefore d=3\text{ units.}

\displaystyle \textbf{Question 6: }~\text{Write the vector equations of the following lines and hence} \\ \text{determine the distance between them } \\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\text{ and }\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}. \text{ [CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{We have}
\displaystyle \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}
\displaystyle \frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}
\displaystyle \text{Since the first line passes through the point }(1,2,-4)\text{ and has direction ratios} \\ \text{proportional to }2,3,6,\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\quad ...(1)
\displaystyle \overrightarrow{r}=\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{Also, the second line passes through the point }(3,3,-5)\text{ and has direction ratios} \\ \text{proportional to }4,6,12.
\displaystyle \text{Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\quad ...(2)
\displaystyle \overrightarrow{r}=3\widehat{i}+3\widehat{j}-5\widehat{k}+\mu(4\widehat{i}+6\widehat{j}+12\widehat{k})
\displaystyle \overrightarrow{r}=3\widehat{i}+3\widehat{j}-5\widehat{k}+2\mu(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{These two lines pass through the points having position vectors }\overrightarrow{a_1}=\widehat{i}+2\widehat{j}-4\widehat{k}\text{ and }\overrightarrow{a_2}=3\widehat{i}+3\widehat{j}-5\widehat{k}\text{ and are parallel to the vector}
\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=2\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=(2\widehat{i}+\widehat{j}-\widehat{k})\times(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&-1\\2&3&6\end{vmatrix}
\displaystyle =9\widehat{i}-14\widehat{j}+4\widehat{k}
\displaystyle \Rightarrow \left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}\right|=\sqrt{9^2+(-14)^2+4^2}
\displaystyle =\sqrt{81+196+16}
\displaystyle =\sqrt{293}
\displaystyle \text{and }\left|\overrightarrow{b}\right|=\sqrt{2^2+3^2+6^2}
\displaystyle =\sqrt{4+9+36}
\displaystyle =7
\displaystyle \text{The shortest distance between the two lines is given by}
\displaystyle \frac{\left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}\right|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{293}}{7}\text{ units}

\displaystyle \textbf{Question 7: }~\text{Find the shortest distance between the lines:}
\displaystyle (i)\ \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda(\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{i}-\widehat{j}-\widehat{k}+\mu(2\widehat{i}+\widehat{j}+2\widehat{k}) 
\displaystyle (ii)\ \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\text{ and }\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1} 
\displaystyle (iii)\ \overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda(\widehat{i}-3\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=4\widehat{i}+5\widehat{j}+6\widehat{k}+\mu(2\widehat{i}+3\widehat{j}+\widehat{k})\text{ [CBSE 2014]}
\displaystyle (iv)\ \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda(\widehat{i}-2\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}) 
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda(\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{i}-\widehat{j}-\widehat{k}+\mu(2\widehat{i}+\widehat{j}+2\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}+2\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{a_2}=2\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_1}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b_2}=2\widehat{i}+\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}-3\widehat{j}-2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&1\\2&1&2\end{vmatrix}
\displaystyle =-3\widehat{i}+3\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-3)^2+3^2}
\displaystyle =\sqrt{9+9}
\displaystyle =3\sqrt{2}
\displaystyle \text{and }(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(\widehat{i}-3\widehat{j}-2\widehat{k})\cdot(-3\widehat{i}+3\widehat{k})
\displaystyle =-3-6
\displaystyle =-9
\displaystyle \text{The shortest distance between the lines }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle \Rightarrow d=\left|\frac{-9}{3\sqrt{2}}\right|
\displaystyle =\frac{3}{\sqrt{2}}
\displaystyle \textbf{(ii)  }
\displaystyle \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\text{ and }\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}
\displaystyle \text{Since the first line passes through the point }(-1,-1,-1)\text{ and has direction ratios} \\ \text{proportional to }7,-6,1,\text{ its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_1}=-\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b_1}=7\widehat{i}-6\widehat{j}+\widehat{k}
\displaystyle \text{Also, the second line passing through the point }(3,5,7)\text{ has direction ratios} \\ \text{proportional to }1,-2,1.\text{ Its vector equation is}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a_2}=3\widehat{i}+5\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{b_2}=\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=4\widehat{i}+6\widehat{j}+8\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\7&-6&1\\1&-2&1\end{vmatrix}
\displaystyle =-4\widehat{i}-6\widehat{j}-8\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-4)^2+(-6)^2+(-8)^2}
\displaystyle =\sqrt{16+36+64}
\displaystyle =\sqrt{116}
\displaystyle \text{and }(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(4\widehat{i}+6\widehat{j}+8\widehat{k})\cdot(-4\widehat{i}-6\widehat{j}-8\widehat{k})
\displaystyle =-16-36-64
\displaystyle =-116
\displaystyle \text{The shortest distance between the lines }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle \Rightarrow d=\left|\frac{-116}{\sqrt{116}}\right|
\displaystyle =\sqrt{116}
\displaystyle =2\sqrt{29}
\displaystyle \textbf{(iii)  }
\displaystyle \overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda(\widehat{i}-3\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=4\widehat{i}+5\widehat{j}+6\widehat{k}+\mu(2\widehat{i}+3\widehat{j}+\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{a_2}=4\widehat{i}+5\widehat{j}+6\widehat{k}
\displaystyle \overrightarrow{b_1}=\widehat{i}-3\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_2}=2\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=3\widehat{i}+3\widehat{j}+3\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&2\\2&3&1\end{vmatrix}
\displaystyle =-9\widehat{i}+3\widehat{j}+9\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{(-9)^2+3^2+9^2}
\displaystyle =\sqrt{81+9+81}
\displaystyle =\sqrt{171}
\displaystyle \text{and }(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(3\widehat{i}+3\widehat{j}+3\widehat{k})\cdot(-9\widehat{i}+3\widehat{j}+9\widehat{k})
\displaystyle =-27+9+27
\displaystyle =9
\displaystyle \text{The shortest distance between the lines }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle \Rightarrow d=\left|\frac{9}{\sqrt{171}}\right|
\displaystyle =\frac{3}{\sqrt{19}}
\displaystyle \textbf{(iv)  }
\displaystyle \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda(\widehat{i}-2\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})
\displaystyle \text{Comparing the given equations with the equations}
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}
\displaystyle \text{we get,}
\displaystyle \overrightarrow{a_1}=6\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{a_2}=-4\widehat{i}-\widehat{k}
\displaystyle \overrightarrow{b_1}=\widehat{i}-2\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_2}=3\widehat{i}-2\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-10\widehat{i}-2\widehat{j}-3\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&2\\3&-2&-2\end{vmatrix}
\displaystyle =8\widehat{i}+8\widehat{j}+4\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|=\sqrt{8^2+8^2+4^2}
\displaystyle =\sqrt{64+64+16}
\displaystyle =\sqrt{144}
\displaystyle =12
\displaystyle \text{and }(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(-10\widehat{i}-2\widehat{j}-3\widehat{k})\cdot(8\widehat{i}+8\widehat{j}+4\widehat{k})
\displaystyle =-80-16-12
\displaystyle =-108
\displaystyle \text{The shortest distance between the lines }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ is given by}
\displaystyle d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})}{\left|\overrightarrow{b_1}\times\overrightarrow{b_2}\right|}\right|
\displaystyle \Rightarrow d=\left|\frac{-108}{12}\right|
\displaystyle =9

\displaystyle \textbf{Question 8: }~\text{Find the distance between the lines }l_1\text{ and }l_2\text{ given by }\overrightarrow{r}=\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})\text{ and }\overrightarrow{r}=3\widehat{i}+3\widehat{j}-5\widehat{k}+\mu(2\widehat{i}+3\widehat{j}+6\widehat{k}). \text{ [CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{r}=\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \overrightarrow{r}=3\widehat{i}+3\widehat{j}-5\widehat{k}+\mu(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{These two lines pass through the points having position vectors}
\displaystyle \overrightarrow{a_1}=\widehat{i}+2\widehat{j}-4\widehat{k}\text{ and }\overrightarrow{a_2}=3\widehat{i}+3\widehat{j}-5\widehat{k}\text{ and are parallel to the vector}
\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=2\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \text{and}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}=(2\widehat{i}+\widehat{j}-\widehat{k})\times(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&-1\\2&3&6\end{vmatrix}
\displaystyle =9\widehat{i}-14\widehat{j}+4\widehat{k}
\displaystyle \Rightarrow \left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}\right|=\sqrt{9^2+(-14)^2+4^2}
\displaystyle =\sqrt{81+196+16}
\displaystyle =\sqrt{293}
\displaystyle \text{and }\left|\overrightarrow{b}\right|=\sqrt{2^2+3^2+6^2}
\displaystyle =\sqrt{4+9+36}
\displaystyle =7
\displaystyle \text{The shortest distance between the two lines is given by}
\displaystyle \frac{\left|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}\right|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{293}}{7}

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