\displaystyle \textbf{Question 1: }~\text{Show that the lines }\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}\text{ and }\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}\text{ intersect and find their point of intersection.}
\displaystyle \text{Answer:}
\displaystyle \text{The coordinates of any point on the first line are given by}
\displaystyle \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}=\lambda
\displaystyle x=\lambda
\displaystyle y=2\lambda+2
\displaystyle z=3\lambda-3
\displaystyle \text{The coordinates of a general point on the first line are}
\displaystyle (\lambda,2\lambda+2,3\lambda-3)
\displaystyle \text{Also, the coordinates of any point on the second line are given by}
\displaystyle \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}=\mu
\displaystyle x=2\mu+2
\displaystyle y=3\mu+6
\displaystyle z=4\mu+3
\displaystyle \text{The coordinates of a general point on the second line are}
\displaystyle (2\mu+2,3\mu+6,4\mu+3)
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle \lambda=2\mu+2,2\lambda+2=3\mu+6,3\lambda-3=4\mu+3
\displaystyle \lambda-2\mu=2\quad\text{(1)}
\displaystyle 2\lambda-3\mu=4\quad\text{(2)}
\displaystyle 3\lambda-4\mu=6\quad\text{(3)}
\displaystyle \text{Solving (1) and (2), we get}
\displaystyle \lambda=2\text{ and }\mu=0
\displaystyle \text{Substituting }\lambda=2\text{ and }\mu=0\text{ in (3), we get}
\displaystyle \text{LHS}=3\lambda-4\mu
\displaystyle =3(2)-4(0)
\displaystyle =6
\displaystyle =\text{RHS}
\displaystyle \text{Since }\lambda=2\text{ and }\mu=0\text{ satisfy the third equation, the given lines intersect at }(2,6,3)

\displaystyle \textbf{Question 2: }~\text{Show that the lines }\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}\text{ and }\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}\text{ do not intersect.}
\displaystyle \text{Answer:}
\displaystyle \text{The coordinates of any point on the first line are given by}
\displaystyle \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda
\displaystyle x=3\lambda+1
\displaystyle y=2\lambda-1
\displaystyle z=5\lambda+1
\displaystyle \text{The coordinates of a general point on the first line are}
\displaystyle (3\lambda+1,2\lambda-1,5\lambda+1)
\displaystyle \text{The coordinates of any point on the second line are given by}
\displaystyle \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu
\displaystyle x=4\mu-2
\displaystyle y=3\mu+1
\displaystyle z=-2\mu-1
\displaystyle \text{The coordinates of a general point on the second line are}
\displaystyle (4\mu-2,3\mu+1,-2\mu-1)
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle 3\lambda+1=4\mu-2,2\lambda-1=3\mu+1,5\lambda+1=-2\mu-1
\displaystyle 3\lambda-4\mu=-3\quad\text{(1)}
\displaystyle 2\lambda-3\mu=2\quad\text{(2)}
\displaystyle 5\lambda+2\mu=-2\quad\text{(3)}
\displaystyle \text{Solving (1) and (2), we get}
\displaystyle \lambda=-17
\displaystyle \mu=-12
\displaystyle \text{Substituting }\lambda=-17\text{ and }\mu=-12\text{ in (3), we get}
\displaystyle \text{LHS}=5\lambda+2\mu
\displaystyle =5(-17)+2(-12)
\displaystyle =-109
\displaystyle \neq-2
\displaystyle \text{LHS}\neq\text{RHS}
\displaystyle \text{Hence, the given lines do not intersect.}

\displaystyle \textbf{Question 3: }~\text{Show that the lines }\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\text{ and }\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}\text{ intersect. Find their point of intersection. \ [CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The coordinates of any point on the first line are given by}
\displaystyle \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda
\displaystyle x=3\lambda-1
\displaystyle y=5\lambda-3
\displaystyle z=7\lambda-5
\displaystyle \text{The coordinates of a general point on the first line are}
\displaystyle (3\lambda-1,5\lambda-3,7\lambda-5)
\displaystyle \text{The coordinates of any point on the second line are given by}
\displaystyle \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu
\displaystyle x=\mu+2
\displaystyle y=3\mu+4
\displaystyle z=5\mu+6
\displaystyle \text{The coordinates of a general point on the second line are}
\displaystyle (\mu+2,3\mu+4,5\mu+6)
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle 3\lambda-1=\mu+2,5\lambda-3=3\mu+4,7\lambda-5=5\mu+6
\displaystyle 3\lambda-\mu=3\quad\text{(1)}
\displaystyle 5\lambda-3\mu=7\quad\text{(2)}
\displaystyle 7\lambda-5\mu=11\quad\text{(3)}
\displaystyle \text{Solving (1) and (2), we get}
\displaystyle \lambda=\frac{1}{2}
\displaystyle \mu=-\frac{3}{2}
\displaystyle \text{Substituting }\lambda=\frac{1}{2}\text{ and }\mu=-\frac{3}{2}\text{ in (3), we get}
\displaystyle \text{LHS}=7\lambda-5\mu
\displaystyle =7\left(\frac{1}{2}\right)-5\left(-\frac{3}{2}\right)
\displaystyle =11
\displaystyle =\text{RHS}
\displaystyle \text{Since }\lambda=\frac{1}{2}\text{ and }\mu=-\frac{3}{2}\text{ satisfy (3), the given lines intersect.}
\displaystyle \text{Substituting the value of }\lambda\text{ in the general coordinates of the first line, we get}
\displaystyle x=\frac{1}{2}
\displaystyle y=-\frac{1}{2}
\displaystyle z=-\frac{3}{2}
\displaystyle \text{Hence, the given lines intersect at the point }\left(\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\right)

\displaystyle \textbf{Question 4: }~\text{Prove that the line through }A(0,-1,-1)\text{ and }B(4,5,1) \\ \text{ intersects the line through }C(3,9,4)\text{ and }D(-4,4,4). \\ \text{ Also, find their point of intersection. \ [CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{The coordinates of any point on the line }AB\text{ are given by}
\displaystyle \frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}=\lambda
\displaystyle x=4\lambda
\displaystyle y=6\lambda-1
\displaystyle z=2\lambda-1
\displaystyle \text{The coordinates of a general point on }AB\text{ are}
\displaystyle (4\lambda,6\lambda-1,2\lambda-1)
\displaystyle \text{The coordinates of any point on the line }CD\text{ are given by}
\displaystyle \frac{x-3}{3+4}=\frac{y-9}{9-4}=\frac{z-4}{4-4}=\mu
\displaystyle x=7\mu+3
\displaystyle y=5\mu+9
\displaystyle z=4
\displaystyle \text{The coordinates of a general point on }CD\text{ are}
\displaystyle (7\mu+3,5\mu+9,4)
\displaystyle \text{If the lines }AB\text{ and }CD\text{ intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle 4\lambda=7\mu+3,6\lambda-1=5\mu+9,2\lambda-1=4
\displaystyle 4\lambda-7\mu=3\quad\text{(1)}
\displaystyle 6\lambda-5\mu=10\quad\text{(2)}
\displaystyle \lambda=\frac{5}{2}\quad\text{(3)}
\displaystyle \text{Solving (2) and (3), we get}
\displaystyle \lambda=\frac{5}{2}
\displaystyle \mu=1
\displaystyle \text{Substituting }\lambda=\frac{5}{2}\text{ and }\mu=1\text{ in (1), we get}
\displaystyle \text{LHS}=4\lambda-7\mu
\displaystyle =4\left(\frac{5}{2}\right)-7(1)
\displaystyle =3
\displaystyle =\text{RHS}
\displaystyle \text{Since }\lambda=\frac{5}{2}\text{ and }\mu=1\text{ satisfy the equations, the given lines intersect.}
\displaystyle \text{Substituting the value of }\lambda\text{ in the coordinates of a general point on the line }AB\text{, we get}
\displaystyle x=10
\displaystyle y=14
\displaystyle z=4
\displaystyle \text{Hence, }AB\text{ and }CD\text{ intersect at the point }(10,14,4)

\displaystyle \textbf{Question 5: }~\text{Prove that the lines }\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})\text{ intersect and find their point of intersection.}
\displaystyle \text{Answer:}
\displaystyle \text{The position vectors of two arbitrary points on the given lines are}
\displaystyle (\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})=(1+3\lambda)\widehat{i}+(1-\lambda)\widehat{j}-\widehat{k}
\displaystyle (4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})=(4+2\mu)\widehat{i}+0\widehat{j}+(3\mu-1)\widehat{k}
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle (1+3\lambda)\widehat{i}+(1-\lambda)\widehat{j}-\widehat{k}=(4+2\mu)\widehat{i}+0\widehat{j}+(3\mu-1)\widehat{k}
\displaystyle \text{Equating the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{ we get}
\displaystyle 1+3\lambda=4+2\mu\quad\text{(1)}
\displaystyle 1-\lambda=0\quad\text{(2)}
\displaystyle 3\mu-1=-1\quad\text{(3)}
\displaystyle \text{Solving (2) and (3), we get}
\displaystyle \lambda=1
\displaystyle \mu=0
\displaystyle \text{Substituting the values }\lambda=1\text{ and }\mu=0
\displaystyle \text{we get}
\displaystyle \text{LHS}=1+3\lambda
\displaystyle =1+3(1)
\displaystyle =4
\displaystyle \text{RHS}=4+2\mu
\displaystyle =4+2(0)
\displaystyle =4
\displaystyle \text{LHS}=\text{RHS}
\displaystyle \text{Since }\lambda=1\text{ and }\mu=0\text{ satisfy (3), the given lines intersect.}
\displaystyle \text{Substituting }\mu=0\text{ in the second line, we get }\overrightarrow{r}=4\widehat{i}+0\widehat{j}-\widehat{k}\text{ as the position vector of the point of intersection.}
\displaystyle \text{Thus, the coordinates of the point of intersection are }(4,0,-1)

\displaystyle \textbf{Question 6: }~\text{Determine whether the following pair 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)\ \frac{x-1}{2}=\frac{y+1}{3}=z\text{ and }\frac{x+1}{5}=\frac{y-2}{1};\ z=2
\displaystyle (iii)\ \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\text{ and }\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}
\displaystyle (iv)\ \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\text{ and }\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}.\ \text{[CBSE\ 2002]}
\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{The position vectors of two arbitrary points on the given lines are}
\displaystyle (\widehat{i}-\widehat{j})+\lambda(2\widehat{i}+\widehat{k})=(1+2\lambda)\widehat{i}-\widehat{j}+\lambda\widehat{k}
\displaystyle (2\widehat{i}-\widehat{j})+\mu(\widehat{i}+\widehat{j}-\widehat{k})=(2+\mu)\widehat{i}+(-1+\mu)\widehat{j}-\mu\widehat{k}
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle (1+2\lambda)\widehat{i}-\widehat{j}+\lambda\widehat{k}=(2+\mu)\widehat{i}+(-1+\mu)\widehat{j}-\mu\widehat{k}
\displaystyle \text{Equating the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{ we get}
\displaystyle 1+2\lambda=2+\mu\quad\text{(1)}
\displaystyle -1=-1+\mu\quad\text{(2)}
\displaystyle \lambda=-\mu\quad\text{(3)}
\displaystyle \text{Solving (2) and (3), we get}
\displaystyle \lambda=0
\displaystyle \mu=0
\displaystyle \text{Substituting }\lambda=0\text{ and }\mu=0\text{ in (1), we get}
\displaystyle \text{LHS}=1+2\lambda
\displaystyle =1+2(0)
\displaystyle =1
\displaystyle \text{RHS}=2+\mu
\displaystyle =2+0
\displaystyle =2
\displaystyle \text{LHS}\neq\text{RHS}
\displaystyle \text{Since }\lambda=0\text{ and }\mu=0\text{ do not satisfy (1), the given lines do not intersect.}
\displaystyle \textbf{(ii)  }
\displaystyle \frac{x-1}{2}=\frac{y+1}{3}=z\text{ and }\frac{x+1}{5}=\frac{y-2}{1},z=2
\displaystyle \text{The coordinates of any point on the first line are given by}
\displaystyle \frac{x-1}{2}=\frac{y+1}{3}=z=\lambda
\displaystyle x=2\lambda+1
\displaystyle y=3\lambda-1
\displaystyle z=\lambda
\displaystyle \text{The coordinates of a general point on the first line are }(2\lambda+1,3\lambda-1,\lambda)
\displaystyle \text{Also, the coordinates of any point on the second line are given by}
\displaystyle \frac{x+1}{5}=\frac{y-2}{1}=\mu,z=2
\displaystyle x=5\mu-1
\displaystyle y=\mu+2
\displaystyle z=2
\displaystyle \text{The coordinates of a general point on the second line are }(5\mu-1,\mu+2,2)
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle 2\lambda+1=5\mu-1,3\lambda-1=\mu+2,\lambda=2
\displaystyle 2\lambda-5\mu=-2\quad\text{(1)}
\displaystyle 3\lambda-\mu=3\quad\text{(2)}
\displaystyle \lambda=2\quad\text{(3)}
\displaystyle \text{Solving (2) and (3), we get}
\displaystyle \lambda=2
\displaystyle \mu=3
\displaystyle \text{Substituting }\lambda=2\text{ and }\mu=3\text{ in (1), we get}
\displaystyle \text{LHS}=2\lambda-5\mu
\displaystyle =2(2)-5(3)
\displaystyle =4-15
\displaystyle =-11\neq-2
\displaystyle \text{LHS}\neq\text{RHS}
\displaystyle \text{Since }\lambda=2\text{ and }\mu=3\text{ do not satisfy (1), the given lines do not intersect.}
\displaystyle \textbf{(iii)  }
\displaystyle \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\text{ and }\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}
\displaystyle \text{The coordinates of any point on the first line are given by}
\displaystyle \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda
\displaystyle x=3\lambda+1
\displaystyle y=-\lambda+1
\displaystyle z=-1
\displaystyle \text{The coordinates of a general point on the first line are }(3\lambda+1,-\lambda+1,-1)
\displaystyle \text{Also, the coordinates of any point on the second line are given by}
\displaystyle \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}=\mu
\displaystyle x=2\mu+4
\displaystyle y=0
\displaystyle z=3\mu-1
\displaystyle \text{The coordinates of a general point on the second line are }(2\mu+4,0,3\mu-1)
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle 3\lambda+1=2\mu+4,-\lambda+1=0,-1=3\mu-1
\displaystyle 3\lambda-2\mu=3\quad\text{(1)}
\displaystyle \lambda=1\quad\text{(2)}
\displaystyle \mu=0\quad\text{(3)}
\displaystyle \text{From (2) and (3), we get}
\displaystyle \lambda=1
\displaystyle \mu=0
\displaystyle \text{Substituting }\lambda=1\text{ and }\mu=0\text{ in (1), we get}
\displaystyle \text{LHS}=3\lambda-2\mu
\displaystyle =3(1)-2(0)
\displaystyle =3
\displaystyle =\text{RHS}
\displaystyle \text{Since }\lambda=1\text{ and }\mu=0\text{ satisfy (1), the lines intersect.}
\displaystyle \text{Substituting }\lambda=1\text{ and }\mu=0\text{ in the coordinates of a general point on the first line, we get}
\displaystyle x=4
\displaystyle y=0
\displaystyle z=-1
\displaystyle \text{Hence, the given lines intersect at }(4,0,-1)
\displaystyle \textbf{(iv)  }
\displaystyle \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\text{ and }\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}
\displaystyle \text{The coordinates of any point on the first line are given by}
\displaystyle \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda
\displaystyle x=4\lambda+5
\displaystyle y=4\lambda+7
\displaystyle z=-5\lambda-3
\displaystyle \text{The coordinates of a general point on the first line are }(4\lambda+5,4\lambda+7,-5\lambda-3)
\displaystyle \text{The coordinates of any point on the second line are given by}
\displaystyle \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu
\displaystyle x=7\mu+8
\displaystyle y=\mu+4
\displaystyle z=3\mu+5
\displaystyle \text{The coordinates of a general point on the second line are }(7\mu+8,\mu+4,3\mu+5)
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle 4\lambda+5=7\mu+8,4\lambda+7=\mu+4,-5\lambda-3=3\mu+5
\displaystyle 4\lambda-7\mu=3\quad\text{(1)}
\displaystyle 4\lambda-\mu=-3\quad\text{(2)}
\displaystyle 5\lambda+3\mu=-8\quad\text{(3)}
\displaystyle \text{From (1) and (2), we get}
\displaystyle \lambda=-1
\displaystyle \mu=-1
\displaystyle \text{Substituting }\lambda=-1\text{ and }\mu=-1\text{ in (3), we get}
\displaystyle \text{LHS}=5\lambda+3\mu
\displaystyle =5(-1)+3(-1)
\displaystyle =-8
\displaystyle =\text{RHS}
\displaystyle \text{Since }\lambda=-1\text{ and }\mu=-1\text{ satisfy (3), the lines intersect.}
\displaystyle \text{Substituting }\lambda=-1\text{ and }\mu=-1\text{ in the coordinates of a general point on the first line, we get}
\displaystyle x=1
\displaystyle y=3
\displaystyle z=2
\displaystyle \text{Hence, the given lines intersect at }(1,3,2)
\displaystyle \\
\displaystyle \textbf{Question 7: }~\text{Show that the lines }\overrightarrow{r}=3\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=5\widehat{i}-2\widehat{j}+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k})\text{ are intersecting. Hence, find their point of intersection. \ [CBSE\ 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The position vectors of two arbitrary points on the given lines are}
\displaystyle 3\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k})=(3+\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(2\lambda-4)\widehat{k}
\displaystyle 5\widehat{i}-2\widehat{j}+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k})=(5+3\mu)\widehat{i}+(-2+2\mu)\widehat{j}+6\mu\widehat{k}
\displaystyle \text{If the lines intersect, then they have a common point, so for some values of }\lambda\text{ and }\mu\text{ we must have}
\displaystyle (3+\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(2\lambda-4)\widehat{k}=(5+3\mu)\widehat{i}+(-2+2\mu)\widehat{j}+6\mu\widehat{k}
\displaystyle \text{Equating the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{ we get}
\displaystyle 3+\lambda=5+3\mu\quad\text{(1)}
\displaystyle 2+2\lambda=-2+2\mu\quad\text{(2)}
\displaystyle 2\lambda-4=6\mu\quad\text{(3)}
\displaystyle \text{Solving (1) and (2), we get}
\displaystyle \lambda=-4
\displaystyle \mu=-2
\displaystyle \text{Substituting }\lambda=-4\text{ and }\mu=-2\text{ in (3), we get}
\displaystyle \text{LHS}=2\lambda-4
\displaystyle =2(-4)-4
\displaystyle =-12
\displaystyle \text{RHS}=6\mu
\displaystyle =6(-2)
\displaystyle =-12
\displaystyle \text{LHS}=\text{RHS}
\displaystyle \text{Since }\lambda=-4\text{ and }\mu=-2\text{ satisfy (3), the lines intersect.}
\displaystyle \text{Substituting }\mu=-2\text{ in the second line, we get }\overrightarrow{r}=5\widehat{i}-2\widehat{j}-6\widehat{i}-4\widehat{j}-12\widehat{k}=-\widehat{i}-6\widehat{j}-12\widehat{k}\text{ as the position vector of the point of intersection.}
\displaystyle \text{Thus, the coordinates of the point of intersection are }(-1,-6,-12)

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