\displaystyle \textbf{Question 1: }~\text{Find the vector and cartesian equations of the line through the point } \\ (5,2,-4)\text{ and which is parallel to the vector }3\hat{i}+2\hat{j}-8\hat{k}.   
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector }\\ \overrightarrow{a}\text{ and parallel to vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}.
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=5\widehat{i}+2\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{b}=3\widehat{i}+2\widehat{j}-8\widehat{k}
\displaystyle \text{Vector equation of the required line is given by}
\displaystyle \overrightarrow{r}=(5\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(3\widehat{i}+2\widehat{j}-8\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(5\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(3\widehat{i}+2\widehat{j}-8\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(5+3\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(-4-8\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=5+3\lambda,\;y=2+2\lambda,\;z=-4-8\lambda
\displaystyle \Rightarrow \frac{x-5}{3}=\lambda,\;\frac{y-2}{2}=\lambda,\;\frac{z+4}{-8}=\lambda
\displaystyle \Rightarrow \frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}

\displaystyle \textbf{Question 2: }~\text{Find the vector equation of the line passing through the points }(-1,0,2) \\ \text{and }(3,4,6). \   
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through the points with position vectors } \\ \overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})\text{ where }\lambda\text{ is a scalar.}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=-1\widehat{i}+0\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b}=3\widehat{i}+4\widehat{j}+6\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(-1\widehat{i}+0\widehat{j}+2\widehat{k})+\lambda\{(3\widehat{i}+4\widehat{j}+6\widehat{k})-(-1\widehat{i}+0\widehat{j}+2\widehat{k})\}
\displaystyle \Rightarrow \overrightarrow{r}=(-1\widehat{i}+0\widehat{j}+2\widehat{k})+\lambda(4\widehat{i}+4\widehat{j}+4\widehat{k})
\displaystyle \text{Here, }\lambda\text{ is a parameter.}

\displaystyle \textbf{Question 3: }~\text{Find the vector equation of a line which is parallel to the vector }2\hat{i}-\hat{j}+3\hat{k}\text{ and which passes through the point }(5,-2,4).\text{ Also, reduce it to cartesian form.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=5\widehat{i}-2\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}-\widehat{j}+3\widehat{k}
\displaystyle \text{So, the vector equation of the required line is}
\displaystyle \overrightarrow{r}=(5\widehat{i}-2\widehat{j}+4\widehat{k})+\lambda(2\widehat{i}-\widehat{j}+3\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(5\widehat{i}-2\widehat{j}+4\widehat{k})+\lambda(2\widehat{i}-\widehat{j}+3\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(5+2\lambda)\widehat{i}+(-2-\lambda)\widehat{j}+(4+3\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=5+2\lambda,\;y=-2-\lambda,\;z=4+3\lambda
\displaystyle \Rightarrow \frac{x-5}{2}=\lambda,\;\frac{y+2}{-1}=\lambda,\;\frac{z-4}{3}=\lambda
\displaystyle \Rightarrow \frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}

\displaystyle \textbf{Question 4: }~\text{A line passes through the point with position vector } \\ 2\hat{i}-3\hat{j}+4\hat{k}\text{and is in the direction of }3\hat{i}+4\hat{j}-5\hat{k}.\text{Find equations of the line in vector} \\ \text{and cartesian form.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{b}=3\widehat{i}+4\widehat{j}-5\widehat{k}
\displaystyle \text{So, the vector equation of the required line is}
\displaystyle \overrightarrow{r}=(2\widehat{i}-3\widehat{j}+4\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}-5\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(2\widehat{i}-3\widehat{j}+4\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}-5\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(2+3\lambda)\widehat{i}+(-3+4\lambda)\widehat{j}+(4-5\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=2+3\lambda,\;y=-3+4\lambda,\;z=4-5\lambda
\displaystyle \Rightarrow \frac{x-2}{3}=\lambda,\;\frac{y+3}{4}=\lambda,\;\frac{z-4}{-5}=\lambda
\displaystyle \Rightarrow \frac{x-2}{3}=\frac{y+3}{4}=\frac{z-4}{-5}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-2}{3}=\frac{y+3}{4}=\frac{z-4}{-5}

\displaystyle \textbf{Question 5: }~ABCD\text{ is a parallelogram. The position vectors of the points }A,B\text{ and }C \\ \text{are respectively }4\hat{i}+5\hat{j}-10\hat{k},\ 2\hat{i}-3\hat{j}+4\hat{k}\text{ and }-\hat{i}+2\hat{j}+\hat{k}.\text{Find the vector equation} \\ \text{of the line }BD.\text{ Also, reduce it to cartesian form.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the position vector of the mid-point of }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\overrightarrow{a}+\overrightarrow{b}}{2}. \\ \text{Let the position vector of point }D\text{ be }x\widehat{i}+y\widehat{j}+z\widehat{k}
\displaystyle \text{Position vector of mid-point of }A\text{ and }C=\text{Position vector of mid-point of }B\text{ and }D
\displaystyle \frac{(4\widehat{i}+5\widehat{j}-10\widehat{k})+(-\widehat{i}+2\widehat{j}+\widehat{k})}{2}=\frac{(2\widehat{i}-3\widehat{j}+4\widehat{k})+(x\widehat{i}+y\widehat{j}+z\widehat{k})}{2}
\displaystyle \Rightarrow \frac{3}{2}\widehat{i}+\frac{7}{2}\widehat{j}-\frac{9}{2}\widehat{k}=\frac{x+2}{2}\widehat{i}+\frac{-3+y}{2}\widehat{j}+\frac{4+z}{2}\widehat{k}
\displaystyle \text{Comparing the coefficient of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle \frac{x+2}{2}=\frac{3}{2}
\displaystyle \Rightarrow x=1
\displaystyle \frac{-3+y}{2}=\frac{7}{2}
\displaystyle \Rightarrow y=10
\displaystyle \frac{4+z}{2}=-\frac{9}{2}
\displaystyle \Rightarrow z=-13
\displaystyle \text{Position vector of point }D=\widehat{i}+10\widehat{j}-13\widehat{k}
\displaystyle \text{The vector equation of line }BD\text{ passing through the points with position vectors } \\ \overrightarrow{a}(B)\text{ and }\overrightarrow{b}(D)\text{ is}
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+10\widehat{j}-13\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(2\widehat{i}-3\widehat{j}+4\widehat{k})+\lambda\{(\widehat{i}+10\widehat{j}-13\widehat{k})-(2\widehat{i}-3\widehat{j}+4\widehat{k})\}
\displaystyle \Rightarrow \overrightarrow{r}=(2\widehat{i}-3\widehat{j}+4\widehat{k})+\lambda(-\widehat{i}+13\widehat{j}-17\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(2\widehat{i}-3\widehat{j}+4\widehat{k})+\lambda(-\widehat{i}+13\widehat{j}-17\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(2-\lambda)\widehat{i}+(-3+13\lambda)\widehat{j}+(4-17\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=2-\lambda,\;y=-3+13\lambda,\;z=4-17\lambda
\displaystyle \Rightarrow \frac{x-2}{-1}=\lambda,\;\frac{y+3}{13}=\lambda,\;\frac{z-4}{-17}=\lambda
\displaystyle \Rightarrow \frac{x-2}{-1}=\frac{y+3}{13}=\frac{z-4}{-17}=\lambda
\displaystyle \Rightarrow \frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}=-\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}

\displaystyle \textbf{Question 6: }~\text{Find in vector form as well as in cartesian form, the equation of the line} \\ \text{passing through the points }A(1,2,-1)\text{ and }B(2,1,1).
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through the points with position vectors } \\ \overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})\text{, where }\lambda\text{ is a scalar.}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-\widehat{k})+\lambda\{(2\widehat{i}+\widehat{j}+\widehat{k})-(\widehat{i}+2\widehat{j}-\widehat{k})\}
\displaystyle \Rightarrow \overrightarrow{r}=(\widehat{i}+2\widehat{j}-\widehat{k})+\lambda(\widehat{i}-\widehat{j}+2\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(\widehat{i}+2\widehat{j}-\widehat{k})+\lambda(\widehat{i}-\widehat{j}+2\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(1+\lambda)\widehat{i}+(2-\lambda)\widehat{j}+(-1+2\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=1+\lambda,\;y=2-\lambda,\;z=-1+2\lambda
\displaystyle \Rightarrow \frac{x-1}{1}=\lambda,\;\frac{y-2}{-1}=\lambda,\;\frac{z+1}{2}=\lambda
\displaystyle \Rightarrow \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}

\displaystyle \textbf{Question 7: }~\text{Find the vector equation for the line which passes through the point } \\ (1,2,3) \text{ and parallel to the vector }\hat{i}-2\hat{j}+3\hat{k}.\text{ Reduce the corresponding equation in} \\ \text{cartesian form.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}-2\widehat{j}+3\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}-2\widehat{j}+3\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}-2\widehat{j}+3\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(1+\lambda)\widehat{i}+(2-2\lambda)\widehat{j}+(3+3\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=1+\lambda,\;y=2-2\lambda,\;z=3+3\lambda
\displaystyle \Rightarrow \frac{x-1}{1}=\lambda,\;\frac{y-2}{-2}=\lambda,\;\frac{z-3}{3}=\lambda
\displaystyle \Rightarrow \frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{3}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{3}

\displaystyle \textbf{Question 8: }~\text{Find the vector equation of a line passing through }(2,-1,1)\text{ and parallel} \\ \text{to the linewhose equations are }\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}.
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}+7\widehat{j}-3\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}+\widehat{k})+\lambda(2\widehat{i}+7\widehat{j}-3\widehat{k})
\displaystyle \text{Here, }\lambda\text{ is a parameter.}

\displaystyle \textbf{Question 9: }~\text{The cartesian equations of a line are }\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}. \\ \text{Find a vector equation for the line.  }
\displaystyle \text{Answer:}
\displaystyle \text{The Cartesian equation of the given line is }\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}
\displaystyle \text{It can be re-written as}
\displaystyle \frac{x-5}{3}=\frac{y-(-4)}{7}=\frac{z-6}{2}
\displaystyle \text{Thus, the given line passes through the point having position vector }\overrightarrow{a}=5\widehat{i}-4\widehat{j}+6\widehat{k}\text{ and is parallel to the vector }
\displaystyle \overrightarrow{b}=3\widehat{i}+7\widehat{j}+2\widehat{k}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector }\overrightarrow{a} \\ \text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(5\widehat{i}-4\widehat{j}+6\widehat{k})+\lambda(3\widehat{i}+7\widehat{j}+2\widehat{k})
\displaystyle \text{Here, }\lambda\text{ is a parameter.}

\displaystyle \textbf{Question 10: }~\text{Find the cartesian equation of a line passing through }(1,-1,2)\text{ and parallel} \\ \text{to the line whose equations are }\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}.\text{ Also, reducethe equation obtained} \\ \text{in vector form.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the Cartesian equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{m}\text{ is } \frac{x-x_1}{a}=\frac{y-y_2}{b}=\frac{z-z_3}{c}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=x_1\widehat{i}+y_1\widehat{j}+z_1\widehat{k}
\displaystyle \overrightarrow{m}=a\widehat{i}+b\widehat{j}+c\widehat{k}
\displaystyle \text{Here, }\overrightarrow{a}=\widehat{i}-\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}
\displaystyle \text{Cartesian equation of the required line is}
\displaystyle \frac{x-1}{1}=\frac{y-(-1)}{2}=\frac{z-2}{-2}
\displaystyle \Rightarrow \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}
\displaystyle \text{We know that the Cartesian equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{m}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{m}
\displaystyle \text{Here, the line is passing through the point }(1,1,-2)\text{ and its direction ratios are proportional to } \\ 1,2,-2
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j}+2\widehat{k})+\lambda(\widehat{i}+2\widehat{j}-2\widehat{k})

\displaystyle \textbf{Question 11: }~\text{Find the direction cosines of the line }\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.\text{ Also, reduce} \\ \text{it to vector form.}
\displaystyle \text{Answer:}
\displaystyle \text{The Cartesian equation of the given line is }\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}
\displaystyle \text{It can be re-written as}
\displaystyle \frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3}
\displaystyle \text{This shows that the given line passes through the point }(4,0,1)\text{ and its direction ratios are proportional to }-2,6,-3
\displaystyle \text{So, its direction cosines are}
\displaystyle \frac{-2}{\sqrt{(-2)^2+6^2+(-3)^2}},\;\frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}},\;\frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}
\displaystyle =\frac{-2}{7},\;\frac{6}{7},\;\frac{-3}{7}
\displaystyle \text{Thus, the given line passes through the point having position vector }\overrightarrow{a}=4\widehat{i}+\widehat{k}\text{ and is parallel to the vector}
\displaystyle \overrightarrow{b}=-2\widehat{i}+6\widehat{j}-3\widehat{k}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}\text{ Here,}
\displaystyle \overrightarrow{a}=4\widehat{i}+\widehat{k}
\displaystyle \overrightarrow{b}=-2\widehat{i}+6\widehat{j}-3\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(4\widehat{i}+0\widehat{j}+\widehat{k})+\lambda(-2\widehat{i}+6\widehat{j}-3\widehat{k})
\displaystyle \text{Here, }\lambda\text{ is a parameter.}

\displaystyle \textbf{Question 12: }~\text{The cartesian equations of a line are }x=ay+b,\ z=cy+d.\text{ Find its direction ratios and reduce it to vector form.}
\displaystyle \text{Answer:}
\displaystyle \text{The Cartesian equation of the given line is }x=ay+b,\;z=cy+d
\displaystyle \text{It can be re-written as}
\displaystyle \frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}
\displaystyle \text{Thus, the given line passes through the point }(b,0,d)\text{ and its direction ratios are proportional to } \\ a,1,c\text{. It is also parallel to the vector }\overrightarrow{b}=a\widehat{i}+\widehat{j}+c\widehat{k}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(b\widehat{i}+0\widehat{j}+d\widehat{k})+\lambda(a\widehat{i}+\widehat{j}+c\widehat{k})
\displaystyle \text{Here, }\lambda\text{ is a parameter.}

\displaystyle \textbf{Question 13: }~\text{Find the vector equation of a line passing through the point} \\ \text{with position vector }\hat{i}-2\hat{j}-3\hat{k}\text{ and parallel to the line joining the points with} \\ \text{position vectors }\hat{i}-\hat{j}+4\hat{k}\text{ and }2\hat{i}+\hat{j}+2\hat{k}.\text{ Also, find the cartesian equivalent} \\ \text{of this equation.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{b}=(2\widehat{i}+\widehat{j}+2\widehat{k})-(\widehat{i}-\widehat{j}+4\widehat{k})=\widehat{i}+2\widehat{j}-2\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(\widehat{i}-2\widehat{j}-3\widehat{k})+\lambda(\widehat{i}+2\widehat{j}-2\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(\widehat{i}-2\widehat{j}-3\widehat{k})+\lambda(\widehat{i}+2\widehat{j}-2\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(1+\lambda)\widehat{i}+(-2+2\lambda)\widehat{j}+(-3-2\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=1+\lambda,\;y=-2+2\lambda,\;z=-3-2\lambda
\displaystyle \Rightarrow \frac{x-1}{1}=\lambda,\;\frac{y+2}{2}=\lambda,\;\frac{z+3}{-2}=\lambda
\displaystyle \Rightarrow \frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2}

\displaystyle \textbf{Question 14: }~\text{Find the points on the line }\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\text{ at a distance of } \\ 5\text{ units from the point }P(1,3,3). \ [CBSE\ 2010]
\displaystyle \text{Answer:}
\displaystyle \text{The coordinates of any point on the line }\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\text{ are given by}
\displaystyle \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda
\displaystyle \Rightarrow x=3\lambda-2,\;y=2\lambda-1,\;z=2\lambda+3\text{...1}
\displaystyle \text{Let the coordinates of the desired point be }(3\lambda-2,2\lambda-1,2\lambda+3)
\displaystyle \text{The distance between this point and }(1,3,3)\text{ is }5\text{ units.}
\displaystyle \Rightarrow \sqrt{(3\lambda-2-1)^2+(2\lambda-1-3)^2+(2\lambda+3-3)^2}=5
\displaystyle \Rightarrow (3\lambda-3)^2+(2\lambda-4)^2+(2\lambda)^2=25
\displaystyle \Rightarrow 17\lambda^2-34\lambda=0
\displaystyle \Rightarrow \lambda(\lambda-2)=0
\displaystyle \Rightarrow \lambda=0\text{ or }2
\displaystyle \text{Substituting the values of }\lambda\text{ in (1) we get the coordinates of the desired point as } \\ (-2,-1,3)\text{ and }(4,3,7)

\displaystyle \textbf{Question 15: }~\text{Show that the points whose position vectors are }-2\hat{i}+3\hat{j},\\  \hat{i}+2\hat{j}+3\hat{k}\text{ and }7\hat{i}+9\hat{k}\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given points be }P,Q\text{ and }R\text{ and let their position vectors be }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{, respectively.}
\displaystyle \overrightarrow{a}=-2\widehat{i}+3\widehat{j}
\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{c}=7\widehat{i}+9\widehat{k}
\displaystyle \text{Vector equation of the line passing through }P\text{ and }Q\text{ is}
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})
\displaystyle \Rightarrow \overrightarrow{r}=(-2\widehat{i}+3\widehat{j})+\lambda\{(\widehat{i}+2\widehat{j}+3\widehat{k})-(-2\widehat{i}+3\widehat{j})\}
\displaystyle \Rightarrow \overrightarrow{r}=(-2\widehat{i}+3\widehat{j})+\lambda(3\widehat{i}-\widehat{j}+3\widehat{k})\text{...1}
\displaystyle \text{If points }P,Q\text{ and }R\text{ are collinear, then point }R\text{ must satisfy (1).}
\displaystyle \text{Replacing }\overrightarrow{r}\text{ by }\overrightarrow{c}=7\widehat{i}+9\widehat{k}\text{ in (1), we get}
\displaystyle 7\widehat{i}+9\widehat{k}=(-2\widehat{i}+3\widehat{j})+\lambda(3\widehat{i}-\widehat{j}+3\widehat{k})
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle 7=-2+3\lambda,\;0=3-\lambda,\;9=3\lambda
\displaystyle \Rightarrow \lambda=3
\displaystyle \text{These three equations are consistent, i.e., they give the same value of }\lambda \\ \text{Hence, the given three points are collinear.}

\displaystyle \textbf{Question 16: }~\text{Find the cartesian and vector equations of a line which passes} \\ \text{through the point }(1,2,3)\text{ and is parallel to the line }\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}. \ [CBSE\ 2004]
\displaystyle \text{Answer:}
\displaystyle \text{We have }\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}
\displaystyle \text{It can be re-written as}
\displaystyle \frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\frac{3}{2}}
\displaystyle \Rightarrow \frac{x+2}{-2}=\frac{y+3}{14}=\frac{z-3}{3}
\displaystyle \text{This shows that the given line passes through the point }(-2,-3,3) \\ \text{ and its direction ratios are proportional to }-2,14,3
\displaystyle \text{Thus, the parallel vector is }\overrightarrow{b}=-2\widehat{i}+14\widehat{j}+3\widehat{k}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Here,}
\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}=-2\widehat{i}+14\widehat{j}+3\widehat{k}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(-2\widehat{i}+14\widehat{j}+3\widehat{k})\text{...1}
\displaystyle \text{Here, }\lambda\text{ is a parameter.}
\displaystyle \text{Reducing (1) to Cartesian form, we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(-2\widehat{i}+14\widehat{j}+3\widehat{k})
\displaystyle \Rightarrow x\widehat{i}+y\widehat{j}+z\widehat{k}=(1-2\lambda)\widehat{i}+(2+14\lambda)\widehat{j}+(3+3\lambda)\widehat{k}
\displaystyle \text{Comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k}\text{, we get}
\displaystyle x=1-2\lambda,\;y=2+14\lambda,\;z=3+3\lambda
\displaystyle \Rightarrow \frac{x-1}{-2}=\lambda,\;\frac{y-2}{14}=\lambda,\;\frac{z-3}{3}=\lambda
\displaystyle \Rightarrow \frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}=\lambda
\displaystyle \text{Hence, the Cartesian form of (1) is}
\displaystyle \frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}

\displaystyle \textbf{Question 17: }~\text{The cartesian equations of a line are }3x+1=6y-2=1-z. \\ \text{Find the fixed point through which it passes, its direction ratios and also} \\ \text{its vector equation. \ [CBSE\ 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{The Cartesian equation of the given line is }3x+1=6y-2=1-z
\displaystyle \text{It can be re-written as}
\displaystyle \frac{x+\frac{1}{3}}{\frac{1}{3}}=\frac{y-\frac{1}{3}}{\frac{1}{6}}=\frac{z-1}{-1}
\displaystyle \Rightarrow \frac{x-(-\frac{1}{3})}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}
\displaystyle \text{Thus, the given line passes through the point }\left(-\frac{1}{3},\frac{1}{3},1\right)\text{ and its direction ratios are proportional to } \\ 2,1,-6\text{. It is parallel to the vector }\overrightarrow{b}=2\widehat{i}+\widehat{j}-6\widehat{k}
\displaystyle \text{We know that the vector equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Vector equation of the required line is}
\displaystyle \overrightarrow{r}=\left(-\frac{1}{3}\widehat{i}+\frac{1}{3}\widehat{j}+\widehat{k}\right)+\lambda(2\widehat{i}+\widehat{j}-6\widehat{k})
\displaystyle \text{Here, }\lambda\text{ is a parameter.}

\displaystyle \textbf{Question 18: }~\text{Find the vector equation of the line passing through the point } \\ A(1,2,-1)\text{ and parallel to the line }5x-25=14-7y=35z. \ [CBSE\ 2017]
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the line }5x-25=14-7y=35z\text{ can be re-written as}
\displaystyle \frac{x-5}{\frac{1}{5}}=\frac{y-2}{-\frac{1}{7}}=\frac{z}{\frac{1}{35}}
\displaystyle \Rightarrow \frac{x-5}{7}=\frac{y-2}{-5}=\frac{z}{1}
\displaystyle \text{Since the required line is parallel to the given line, the direction ratios of} \\ \text{the required line are proportional to }7,-5,1
\displaystyle \text{The vector equation of the required line passing through the point } \\ (1,2,-1)\text{ and having direction ratios proportional to }7,-5,1\text{ is}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-\widehat{k})+\lambda(7\widehat{i}-5\widehat{j}+\widehat{k})

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