Class 12: Straight Line in Space

$\displaystyle \textbf{Question 1: }~\text{Show that the three lines with direction cosines } \\ \frac{12}{13},\frac{-3}{13},\frac{-4}{13};\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{3}{13},\frac{-4}{13},\frac{12}{13}\text{ are mutually perpendicular. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction cosines of the three lines are}$
$\displaystyle l_1=\frac{12}{13},\;m_1=-\frac{3}{13},\;n_1=-\frac{4}{13}$
$\displaystyle l_2=\frac{4}{13},\;m_2=\frac{12}{13},\;n_2=\frac{3}{13}$
$\displaystyle l_3=\frac{3}{13},\;m_3=-\frac{4}{13},\;n_3=\frac{12}{13}$
$\displaystyle \therefore l_1l_2+m_1m_2+n_1n_2=\frac{48-36-12}{169}=0$
$\displaystyle \text{Also,}$
$\displaystyle l_2l_3+m_2m_3+n_2n_3=\frac{12-48+36}{169}=0$
$\displaystyle l_1l_3+m_1m_3+n_1n_3=\frac{36+12-48}{169}=0$
$\displaystyle \text{Hence, the given lines are perpendicular to each other.}$

$\displaystyle \textbf{Question 2: }~\text{Show that the line through the points }(1,-1,2)\text{ and }(3,4,-2) \\ \text{ is perpendicular to the line through the points }(0,3,2)\text{ and }(3,5,6).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Suppose vector }\overrightarrow{a}\text{ is passing through the points } \\ (1,1,2)\text{ and }(3,4,-2)\text{ and vector }\overrightarrow{b}\text{ is passing through the points }(0,3,2)\text{ and }(3,5,6)$
$\displaystyle \text{Then,}$
$\displaystyle \overrightarrow{a}=2\widehat{i}+5\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{b}=3\widehat{i}+2\widehat{j}+4\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(2\widehat{i}+5\widehat{j}-4\widehat{k})\cdot(3\widehat{i}+2\widehat{j}+4\widehat{k})=0$
$\displaystyle \text{Hence, the given lines are perpendicular to each other.}$

$\displaystyle \textbf{Question 3: }~\text{Show that the line through the points }(4,7,8)\text{ and }(2,3,4) \\ \text{ is parallel to the line through the points }(-1,-2,1)\text{ and }(1,2,5).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equations of lines passing through the points }(x_1,y_1,z_1)\text{ and }(x_2,y_2,z_2)\text{ are given by}$
$\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
$\displaystyle \text{So, the equation of a line passing through }(4,7,8)\text{ and }(2,3,4)\text{ is}$
$\displaystyle \frac{x-4}{2-4}=\frac{y-7}{3-7}=\frac{z-8}{4-8}$
$\displaystyle \Rightarrow \frac{x-4}{-2}=\frac{y-7}{-4}=\frac{z-8}{-4}$
$\displaystyle \text{Also, the equation of the line passing through the points }(-1,-2,1)\text{ and }(1,2,5)\text{ is}$
$\displaystyle \frac{x+1}{1-(-1)}=\frac{y+2}{2-(-2)}=\frac{z-1}{5-1}$
$\displaystyle \Rightarrow \frac{x+1}{2}=\frac{y+2}{4}=\frac{z-1}{4}$
$\displaystyle \text{We know that two lines are parallel if}$
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\displaystyle \text{Cartesian equations of the two lines are given by}$
$\displaystyle \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\text{ and }\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$
$\displaystyle \text{We observe}$
$\displaystyle \frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}=-1$
$\displaystyle \text{Hence, the given lines are parallel to each other.}$

$\displaystyle \textbf{Question 4: }~\text{Find the cartesian equation of the line which passes through the point } \\ (-2,4,-5)\text{ and parallel to the line given by }\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the Cartesian equation of a line passing through a point with position vector } \\ \overrightarrow{a}\text{ and parallel to the vector }\overrightarrow{b}\text{ is } \frac{x-x_1}{a}=\frac{y-y_2}{b}=\frac{z-z_3}{c}$
$\displaystyle \text{Here,}$
$\displaystyle \overrightarrow{a}=-2\widehat{i}+4\widehat{j}-5\widehat{k}$
$\displaystyle \overrightarrow{b}=3\widehat{i}+5\widehat{j}-6\widehat{k}$
$\displaystyle \text{The Cartesian equation of the required line is}$
$\displaystyle \frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6}$
$\displaystyle \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$

$\displaystyle \textbf{Question 5: }~\text{Show that the lines }\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\text{ and } \\ \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\text{ are perpendicular to each other. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have }\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\text{ and }\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
$\displaystyle \text{These equations can be re-written as}$
$\displaystyle \frac{x-5}{7}=\frac{y-(-2)}{-5}=\frac{z-0}{1}\text{...1}$
$\displaystyle \frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3}\text{...2}$
$\displaystyle \therefore \overrightarrow{m_1}=\text{Vector parallel to line (1)}=7\widehat{i}-5\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{m_2}=\text{Vector parallel to line (2)}=\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{m_1}\cdot\overrightarrow{m_2}=(7\widehat{i}-5\widehat{j}+\widehat{k})\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =7-10+3$
$\displaystyle =0$
$\displaystyle \text{Hence, the given two lines are perpendicular to each other.}$

$\displaystyle \textbf{Question 6: }~\text{Show that the line joining the origin to the point }(2,1,1) \\ \text{ is perpendicular to the line determined by the points }(3,5,-1)\text{ and }(4,3,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of the line joining the origin to the point }(2,1,1)\text{ are }2,1,1$
$\displaystyle \text{Let}$
$\displaystyle \overrightarrow{b_1}=2\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{The direction ratios of the line joining the points }(3,5,-1)\text{ and }(4,3,-1)\text{ are }1,-2,0$
$\displaystyle \text{Let}$
$\displaystyle \overrightarrow{b_2}=\widehat{i}-2\widehat{j}+0\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}\cdot\overrightarrow{b_2}=(2\widehat{i}+\widehat{j}+\widehat{k})\cdot(\widehat{i}-2\widehat{j}+0\widehat{k})$
$\displaystyle =2-2+0$
$\displaystyle =0$
$\displaystyle \therefore \overrightarrow{b_1}\perp\overrightarrow{b_2}$
$\displaystyle \text{Hence, the two lines joining the given points are perpendicular to each other.}$

$\displaystyle \textbf{Question 7: }~\text{Find the equation of a line parallel to } \\ x\text{-axis and passing through the origin. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of the line parallel to the }x\text{-axis are proportional to }1,0,0$
$\displaystyle \text{Equation of the line passing through the origin and parallel to the }x\text{-axis is}$
$\displaystyle \frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}$
$\displaystyle \Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

$\displaystyle \textbf{Question 8: }~\text{Find the angle between the following pairs of lines:}$
$\displaystyle (i)\ \overrightarrow{r}=(4\widehat{i}-\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-2\widehat{k})\text{ and }\overrightarrow{r}=\widehat{i}-\widehat{j}+2\widehat{k}-\mu(2\widehat{i}+4\widehat{j}-4\widehat{k})$
$\displaystyle (ii)\ \overrightarrow{r}=(3\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=(5\widehat{j}-2\widehat{k})+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k})$
$\displaystyle (iii)\ \overrightarrow{r}=\lambda(\widehat{i}+\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{j}+\mu[(\sqrt3-1)\widehat{i}-(\sqrt3+1)\widehat{j}+4\widehat{k}]$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \overrightarrow{r}=(4\widehat{i}-\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-2\widehat{k})\text{ and }\overrightarrow{r}=\widehat{i}-\widehat{j}+2\widehat{k}-\mu(2\widehat{i}+4\widehat{j}-4\widehat{k})$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{b_2}=2\widehat{i}+4\widehat{j}-4\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(\widehat{i}+2\widehat{j}-2\widehat{k})\cdot(2\widehat{i}+4\widehat{j}-4\widehat{k})}{\sqrt{1^2+2^2+(-2)^2}\sqrt{2^2+4^2+(-4)^2}}$
$\displaystyle =\frac{2+8+8}{3\times6}$
$\displaystyle =1$
$\displaystyle \textbf{(ii) }$
$\displaystyle \overrightarrow{r}=(3\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=(5\widehat{j}-2\widehat{k})+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k})$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=\widehat{i}+2\widehat{j}+2\widehat{k}$
$\displaystyle \overrightarrow{b_2}=3\widehat{i}+2\widehat{j}+6\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(\widehat{i}+2\widehat{j}+2\widehat{k})\cdot(3\widehat{i}+2\widehat{j}+6\widehat{k})}{\sqrt{1^2+2^2+2^2}\sqrt{3^2+2^2+6^2}}$
$\displaystyle =\frac{3+4+12}{3\times7}$
$\displaystyle =\frac{19}{21}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{19}{21}\right)$
$\displaystyle \textbf{(iii) }$
$\displaystyle \overrightarrow{r}=\lambda(\widehat{i}+\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{j}+\mu\{(\sqrt{3}-1)\widehat{i}-(\sqrt{3}+1)\widehat{j}+4\widehat{k}\}$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \overrightarrow{b_2}=(\sqrt{3}-1)\widehat{i}-(\sqrt{3}+1)\widehat{j}+4\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(\widehat{i}+\widehat{j}+2\widehat{k})\cdot((\sqrt{3}-1)\widehat{i}-(\sqrt{3}+1)\widehat{j}+4\widehat{k})}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2+4^2}}$
$\displaystyle =\frac{(\sqrt{3}-1)-(\sqrt{3}+1)+8}{\sqrt{6}\sqrt{24}}$
$\displaystyle =\frac{6}{12}$
$\displaystyle =\frac{1}{2}$
$\displaystyle \Rightarrow \theta=\frac{\pi}{3}$

$\displaystyle \textbf{Question 9: }~\text{Find the angle between the following pairs of lines:}$
$\displaystyle (i)\ \frac{x+4}{3}=\frac{y-1}{5}=\frac{z+3}{4}\text{ and }\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$
$\displaystyle (ii)\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{-3}\text{ and }\frac{x+3}{-1}=\frac{y-5}{8}=\frac{z-1}{4}$
$\displaystyle (iii)\ \frac{5-x}{-2}=\frac{y+3}{1}=\frac{1-z}{3}\text{ and }\frac{x}{3}=\frac{1-y}{-2}=\frac{z+5}{-1}$
$\displaystyle (iv)\ \frac{x-2}{3}=\frac{y+3}{-2},\ z=5\text{ and }\frac{x+1}{3}=\frac{2y-3}{3}=\frac{z-5}{2}$
$\displaystyle (v)\ \frac{x-5}{1}=\frac{2y+6}{-2}=\frac{z-3}{1}\text{ and }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5}$
$\displaystyle (vi)\ \frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\text{ and }\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}. \ [CBSE\ 2011]$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \frac{x+4}{3}=\frac{y-1}{5}=\frac{z+3}{4}\text{ and }\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$
$\displaystyle \text{Let}$
$\displaystyle \overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \overrightarrow{b_1}=3\widehat{i}+5\widehat{j}+4\widehat{k}$
$\displaystyle \overrightarrow{b_2}=\widehat{i}+\widehat{j}+2\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(3\widehat{i}+5\widehat{j}+4\widehat{k})\cdot(\widehat{i}+\widehat{j}+2\widehat{k})}{\sqrt{3^2+5^2+4^2}\sqrt{1^2+1^2+2^2}}$
$\displaystyle =\frac{3+5+8}{10\sqrt{3}}$
$\displaystyle =\frac{8}{5\sqrt{3}}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)$
$\displaystyle \textbf{(ii) }$
$\displaystyle \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{-3}\text{ and }\frac{x+3}{-1}=\frac{y-5}{8}=\frac{z-1}{4}$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=2\widehat{i}+3\widehat{j}-3\widehat{k}$
$\displaystyle \overrightarrow{b_2}=-\widehat{i}+8\widehat{j}+4\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(2\widehat{i}+3\widehat{j}-3\widehat{k})\cdot(-\widehat{i}+8\widehat{j}+4\widehat{k})}{\sqrt{2^2+3^2+(-3)^2}\sqrt{(-1)^2+8^2+4^2}}$
$\displaystyle =\frac{-2+24-12}{9\sqrt{22}}$
$\displaystyle =\frac{10}{9\sqrt{22}}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{10}{9\sqrt{22}}\right)$
$\displaystyle \textbf{(iii) }$
$\displaystyle \frac{5-x}{-2}=\frac{y+3}{1}=\frac{1-z}{3}\text{ and }\frac{x}{3}=\frac{1-y}{-2}=\frac{z+5}{-1}$
$\displaystyle \text{The equation of the given lines can be re-written as}$
$\displaystyle \frac{x-5}{2}=\frac{y+3}{1}=\frac{z-1}{-3}\text{ and }\frac{x}{3}=\frac{y-1}{2}=\frac{z+5}{-1}$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=2\widehat{i}+\widehat{j}-3\widehat{k}$
$\displaystyle \overrightarrow{b_2}=3\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(2\widehat{i}+\widehat{j}-3\widehat{k})\cdot(3\widehat{i}+2\widehat{j}-\widehat{k})}{\sqrt{2^2+1^2+(-3)^2}\sqrt{3^2+2^2+(-1)^2}}$
$\displaystyle =\frac{6+2+3}{\sqrt{14}\sqrt{14}}$
$\displaystyle =\frac{11}{14}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{11}{14}\right)$
$\displaystyle \textbf{(iv) }$
$\displaystyle \frac{x-2}{3}=\frac{y+3}{-2},\;z=5\text{ and }\frac{x+1}{1}=\frac{2y-3}{3}=\frac{z-5}{2}$
$\displaystyle \text{The equations of the given lines can be re-written as}$
$\displaystyle \frac{x-2}{3}=\frac{y+3}{-2}=\frac{z-5}{0}\text{ and }\frac{x+1}{1}=\frac{y-\frac{3}{2}}{\frac{3}{2}}=\frac{z-5}{2}$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=3\widehat{i}-2\widehat{j}+0\widehat{k}$
$\displaystyle \overrightarrow{b_2}=\widehat{i}+\frac{3}{2}\widehat{j}+2\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(3\widehat{i}-2\widehat{j}+0\widehat{k})\cdot(\widehat{i}+\frac{3}{2}\widehat{j}+2\widehat{k})}{\sqrt{3^2+(-2)^2+0^2}\sqrt{1^2+\left(\frac{3}{2}\right)^2+2^2}}$
$\displaystyle =\frac{3-3+0}{\sqrt{13}\sqrt{\frac{29}{4}}}$
$\displaystyle =0$
$\displaystyle \Rightarrow \theta=\frac{\pi}{2}$
$\displaystyle \textbf{(v) }$
$\displaystyle \frac{x-5}{1}=\frac{2y+6}{-2}=\frac{z-3}{1}\text{ and }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5}$
$\displaystyle \text{The equations of the given lines can be re-written as}$
$\displaystyle \frac{x-5}{1}=\frac{y+3}{-1}=\frac{z-3}{1}\text{ and }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5}$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b_2}=3\widehat{i}+4\widehat{j}+5\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(\widehat{i}-\widehat{j}+\widehat{k})\cdot(3\widehat{i}+4\widehat{j}+5\widehat{k})}{\sqrt{1^2+(-1)^2+1^2}\sqrt{3^2+4^2+5^2}}$
$\displaystyle =\frac{3-4+5}{\sqrt{3}\sqrt{50}}$
$\displaystyle =\frac{4}{5\sqrt{6}}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{4}{5\sqrt{6}}\right)$
$\displaystyle \textbf{(vi) }$
$\displaystyle \frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\text{ and }\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}$
$\displaystyle \text{The equations of the given lines can be re-written as}$
$\displaystyle \frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3}\text{ and }\frac{x+2}{-1}=\frac{y-4}{2}=\frac{z-5}{4}$
$\displaystyle \text{Let }\overrightarrow{b_1}\text{ and }\overrightarrow{b_2}\text{ be vectors parallel to the given lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}=2\widehat{i}+7\widehat{j}-3\widehat{k}$
$\displaystyle \overrightarrow{b_2}=-\widehat{i}+2\widehat{j}+4\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$\displaystyle =\frac{(2\widehat{i}+7\widehat{j}-3\widehat{k})\cdot(-\widehat{i}+2\widehat{j}+4\widehat{k})}{\sqrt{2^2+7^2+(-3)^2}\sqrt{(-1)^2+2^2+4^2}}$
$\displaystyle =\frac{-2+14-12}{\sqrt{62}\sqrt{21}}$
$\displaystyle =0$
$\displaystyle \Rightarrow \theta=\frac{\pi}{2}$

$\displaystyle \textbf{Question 10: }~\text{Find the angle between the pairs of lines with direction ratios } \\ \text{proportional to} (i)\ 5,-12,13\text{ and }-3,4,5;\ (ii)\ 2,2,1\text{ and }4,1,8; \\ (iii)\ 1,2,-2\text{ and }-2,2,1; (iv)\ a,b,c\text{ and }b-c,c-a,a-b.$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle 5,-12,13\text{ and }-3,4,5$
$\displaystyle \text{Let }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}\text{ be vectors parallel to the two given lines.}$
$\displaystyle \text{Then, the angle between the two given lines is the same as the angle between }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{m_1}=\text{Vector parallel to the line having direction ratios proportional to }5,-12,13$
$\displaystyle \overrightarrow{m_2}=\text{Vector parallel to the line having direction ratios proportional to }-3,4,5$
$\displaystyle \therefore \overrightarrow{m_1}=5\widehat{i}-12\widehat{j}+13\widehat{k}$
$\displaystyle \overrightarrow{m_2}=-3\widehat{i}+4\widehat{j}+5\widehat{k}$
$\displaystyle \text{Let }\theta\text{ be the angle between the lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \cos\theta=\frac{\overrightarrow{m_1}\cdot\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|}$
$\displaystyle =\frac{(5\widehat{i}-12\widehat{j}+13\widehat{k})\cdot(-3\widehat{i}+4\widehat{j}+5\widehat{k})}{\sqrt{5^2+(-12)^2+13^2}\sqrt{(-3)^2+4^2+5^2}}$
$\displaystyle =\frac{-15-48+65}{13\sqrt{2}\times5\sqrt{2}}$
$\displaystyle =\frac{1}{65}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{1}{65}\right)$
$\displaystyle \textbf{(ii) }$
$\displaystyle 2,2,1\text{ and }4,1,8$
$\displaystyle \text{Let }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}\text{ be vectors parallel to the given two lines.}$
$\displaystyle \text{Then, the angle between the lines is the same as the angle between }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{m_1}=\text{Vector parallel to the line having direction ratios proportional to }2,2,1$
$\displaystyle \overrightarrow{m_2}=\text{Vector parallel to the line having direction ratios proportional to }4,1,8$
$\displaystyle \therefore \overrightarrow{m_1}=2\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{m_2}=4\widehat{i}+\widehat{j}+8\widehat{k}$
$\displaystyle \text{Let }\theta\text{ be the angle between the lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \cos\theta=\frac{\overrightarrow{m_1}\cdot\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|}$
$\displaystyle =\frac{(2\widehat{i}+2\widehat{j}+\widehat{k})\cdot(4\widehat{i}+\widehat{j}+8\widehat{k})}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}}$
$\displaystyle =\frac{8+2+8}{3\times9}$
$\displaystyle =\frac{2}{3}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{2}{3}\right)$
$\displaystyle \textbf{(iii) }$
$\displaystyle 1,2,-2\text{ and }-2,2,1$
$\displaystyle \text{Let }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}\text{ be vectors parallel to the two given lines.}$
$\displaystyle \text{Then, the angle between the two given lines is the same as the angle between }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{m_1}=\text{Vector parallel to the line having direction ratios proportional to }1,2,-2$
$\displaystyle \overrightarrow{m_2}=\text{Vector parallel to the line having direction ratios proportional to }-2,2,1$
$\displaystyle \therefore \overrightarrow{m_1}=\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{m_2}=-2\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \text{Let }\theta\text{ be the angle between the lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \cos\theta=\frac{\overrightarrow{m_1}\cdot\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|}$
$\displaystyle =\frac{(\widehat{i}+2\widehat{j}-2\widehat{k})\cdot(-2\widehat{i}+2\widehat{j}+\widehat{k})}{\sqrt{1^2+2^2+(-2)^2}\sqrt{(-2)^2+2^2+1^2}}$
$\displaystyle =\frac{-2+4-2}{3\times3}$
$\displaystyle =0$
$\displaystyle \Rightarrow \theta=\frac{\pi}{2}$
$\displaystyle \textbf{(iv) }$
$\displaystyle a,b,c\text{ and }b-c,\;c-a,\;a-b$
$\displaystyle \text{Let }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}\text{ be vectors parallel to the given two lines.}$
$\displaystyle \text{Then, the angle between the two lines is the same as the angle between }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{m_1}=\text{Vector parallel to the line having direction ratios proportional to }a,b,c$
$\displaystyle \overrightarrow{m_2}=\text{Vector parallel to the line having direction ratios proportional to }b-c,\;c-a,\;a-b$
$\displaystyle \therefore \overrightarrow{m_1}=a\widehat{i}+b\widehat{j}+c\widehat{k}\text{ and }\overrightarrow{m_2}=(b-c)\widehat{i}+(c-a)\widehat{j}+(a-b)\widehat{k}$
$\displaystyle \text{Let }\theta\text{ be the angle between the lines.}$
$\displaystyle \text{Now,}$
$\displaystyle \cos\theta=\frac{\overrightarrow{m_1}\cdot\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|}$
$\displaystyle =\frac{(a\widehat{i}+b\widehat{j}+c\widehat{k})\cdot\{(b-c)\widehat{i}+(c-a)\widehat{j}+(a-b)\widehat{k}\}}{\sqrt{a^2+b^2+c^2}\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}$
$\displaystyle =\frac{ab-ac+bc-ba+ca-cb}{\sqrt{a^2+b^2+c^2}\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}$
$\displaystyle =0$
$\displaystyle \Rightarrow \theta=\frac{\pi}{2}$

$\displaystyle \textbf{Question 11: }~\text{Find the angle between two lines, one of which has direction ratios }2,2,1 \\ \text{ while the other one is obtained by joining the points }(3,1,4)\text{ and }(7,2,12).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of the line joining the points }(3,1,4)\text{ and } \\ (7,2,12)\text{ are proportional to }4,1,8$
$\displaystyle \text{Let }\overrightarrow{m_1}\text{ and }\overrightarrow{m_2} \\ \text{ be vectors parallel to the lines having direction ratios proportional to }2,2,1\text{ and }4,1,8$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{m_1}=2\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{m_2}=4\widehat{i}+\widehat{j}+8\widehat{k}$
$\displaystyle \text{If }\theta\text{ is the angle between the given lines, then}$
$\displaystyle \cos\theta=\frac{\overrightarrow{m_1}\cdot\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|}$
$\displaystyle =\frac{(2\widehat{i}+2\widehat{j}+\widehat{k})\cdot(4\widehat{i}+\widehat{j}+8\widehat{k})}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}}$
$\displaystyle =\frac{8+2+8}{3\times9}$
$\displaystyle =\frac{2}{3}$
$\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{2}{3}\right)$

$\displaystyle \textbf{Question 12: }~\text{Find the equation of the line passing through the point }(1,2,-4) \\ \text{ and parallel to the line }\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of the line parallel to the line }\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3} \\ \text{ are proportional to }4,2,3$
$\displaystyle \text{Equation of the required line passing through the point }(1,2,-4) \\ \text{ having direction ratios proportional to }4,2,3\text{ is}$
$\displaystyle \frac{x-1}{4}=\frac{y-2}{2}=\frac{z-(-4)}{3}$
$\displaystyle =\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$

$\displaystyle \textbf{Question 13: }~\text{Find the equations of the line passing through the point } \\ (-1,2,1)\text{ and parallel to the line }\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the line }\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3} \\ \text{ can be re-written as }\frac{x-\frac{1}{2}}{2}=\frac{y+\frac{5}{3}}{\frac{2}{3}}=\frac{z-2}{-3}$
$\displaystyle \text{The direction ratios of the line parallel to the line }\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3} \\ \text{ are proportional to }2,\frac{2}{3},-3$
$\displaystyle \text{Equation of the required line passing through the point }(-1,2,1) \\ \text{ having direction ratios proportional to }2,\frac{2}{3},-3\text{ is}$
$\displaystyle \frac{x-(-1)}{2}=\frac{y-2}{\frac{2}{3}}=\frac{z-1}{-3}$
$\displaystyle \Rightarrow \frac{x+1}{2}=\frac{y-2}{\frac{2}{3}}=\frac{z-1}{-3}$

$\displaystyle \textbf{Question 14: }~\text{Find the equation of the line passing through the point } (2,-1,3) \\ \text{ and parallel to the line }\overrightarrow{r}=(\widehat{i}-2\widehat{j}+\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}-5\widehat{k}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is parallel to the vector }2\widehat{i}+3\widehat{j}-5\widehat{k}\text{ and the required line is parallel to the given line.}$
$\displaystyle \text{So, the required line is parallel to the vector }2\widehat{i}+3\widehat{j}-5\widehat{k}$
$\displaystyle \text{Hence, the equation of the required line passing through the point }(2,-1,3) \\ \text{ and parallel to the vector }2\widehat{i}+3\widehat{j}-5\widehat{k}\text{ is}$
$\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}-5\widehat{k})$

$\displaystyle \textbf{Question 15: }~\text{Find the equations of the line passing through the point }(2,1,3) \\ \text{ and perpendicular to the lines }\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\text{ and }\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}. \ [CBSE\ 2014]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let:}$
$\displaystyle \overrightarrow{b_1}=\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{b_2}=-3\widehat{i}+2\widehat{j}+5\widehat{k}$
$\displaystyle \text{Since the required line is perpendicular to the lines parallel to the vectors }\overrightarrow{b_1}=\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b_2}=-3\widehat{i}+2\widehat{j}+5\widehat{k}\text{ it is parallel to the vector }\overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}\text{ Now,}$
$\displaystyle \overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&3\\-3&2&5\end{vmatrix}$
$\displaystyle =4\widehat{i}-14\widehat{j}+8\widehat{k}$
$\displaystyle =2(2\widehat{i}-7\widehat{j}+4\widehat{k})$
$\displaystyle \text{Thus, the direction ratios of the required line are proportional to }2,-7,4$
$\displaystyle \text{The equation of the required line passing through the point }(2,1,3) \\ \text{ and having direction ratios proportional to }2,-7,4\text{ is}$
$\displaystyle \frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$

$\displaystyle \textbf{Question 16: }~\text{Find the equation of the line passing through the point }\\ \widehat{i}+\widehat{j}-3\widehat{k}\text{ and perpendicular to the lines }\overrightarrow{r}=\widehat{i}+\lambda(2\widehat{i}+\widehat{j}-3\widehat{k}) \\ \text{and }\overrightarrow{r}=(2\widehat{i}+\widehat{j}-\widehat{k})+\mu(\widehat{i}+\widehat{j}+\widehat{k}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The required line is perpendicular to the lines parallel to the vectors }\overrightarrow{b_1}=2\widehat{i}+\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{b_2}=\widehat{i}+\widehat{j}+\widehat{k}\text{ so, the required line is parallel to the vector}$
$\displaystyle \overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&-3\\1&1&1\end{vmatrix}$
$\displaystyle =4\widehat{i}-5\widehat{j}+\widehat{k}$
$\displaystyle \text{Equation of the required line passing through the point }(\widehat{i}+\widehat{j}-3\widehat{k})\text{ and parallel to }(4\widehat{i}-5\widehat{j}+\widehat{k})\text{ is}$
$\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j}-3\widehat{k})+\lambda(4\widehat{i}-5\widehat{j}+\widehat{k})$

$\displaystyle \textbf{Question 17: }~\text{Find the equation of the line passing through the point }(1,-1,1) \\ \text{ and perpendicular to the lines joining the points }(4,3,2),(1,-1,0)\text{ and }(1,2,-1),(2,1,1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of the lines joining the points }(4,3,2),(1,-1,0)\text{ and }(1,2,-1),(2,1,1)\text{ are }-3,-4,-2\text{ and }1,-1,2\text{ respectively.}$
$\displaystyle \text{Let:}$
$\displaystyle \overrightarrow{b_1}=-3\widehat{i}-4\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{b_2}=\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \text{Since the required line is perpendicular to the lines parallel to the vectors }\overrightarrow{b_1}=-3\widehat{i}-4\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{b_2}=\widehat{i}-\widehat{j}+2\widehat{k}\text{ it is parallel to the vector }\overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-3&-4&-2\\1&-1&2\end{vmatrix}$
$\displaystyle =-10\widehat{i}+4\widehat{j}+7\widehat{k}$
$\displaystyle \text{So, the direction ratios of the required line are proportional to }-10,4,7$
$\displaystyle \text{The equation of the required line passing through the point }(1,-1,1) \\ \text{ and having direction ratios proportional to }-10,4,7\text{ is}$
$\displaystyle \frac{x-1}{-10}=\frac{y-(-1)}{4}=\frac{z-1}{7}$
$\displaystyle \Rightarrow \frac{x-1}{-10}=\frac{y+1}{4}=\frac{z-1}{7}$

$\displaystyle \textbf{Question 18: }~\text{Determine the equations of the line passing through the point }(1,2,-4) \\ \text{ and perpendicular to the two lines }\frac{x-8}{8}=\frac{y+9}{-16}=\frac{z-10}{7}\text{ and }\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}. \ [CBSE\ 2012,2016,2017]$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-8}{8}=\frac{y+9}{-16}=\frac{z-10}{7}$
$\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$
$\displaystyle \text{Let:}$
$\displaystyle \overrightarrow{b_1}=8\widehat{i}-16\widehat{j}+7\widehat{k}$
$\displaystyle \overrightarrow{b_2}=3\widehat{i}+8\widehat{j}-5\widehat{k}$
$\displaystyle \text{Since the required line is perpendicular to the lines parallel to the vectors }\overrightarrow{b_1}=8\widehat{i}-16\widehat{j}+7\widehat{k}\text{ and }\overrightarrow{b_2}=3\widehat{i}+8\widehat{j}-5\widehat{k}\text{ it is parallel to the vector }\overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}$
$\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\8&-16&7\\3&8&-5\end{vmatrix}$
$\displaystyle =24\widehat{i}+61\widehat{j}+112\widehat{k}$
$\displaystyle \text{The direction ratios of the required line are proportional to }24,61,112$
$\displaystyle \text{The equation of the required line passing through the point }(1,2,-4) \\ \text{ and having direction ratios proportional to }24,61,112\text{ is}$
$\displaystyle \frac{x-1}{24}=\frac{y-2}{61}=\frac{z-(-4)}{112}$
$\displaystyle \Rightarrow \frac{x-1}{24}=\frac{y-2}{61}=\frac{z+4}{112}$

$\displaystyle \textbf{Question 19: }~\text{Show that the lines }\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\text{ and }\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\text{ are perpendicular to each other.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of the lines }\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\text{ and }\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\text{ are proportional to } \\ 7,-5,1\text{ and }1,2,3\text{ respectively.}$
$\displaystyle \text{Let:}$
$\displaystyle \overrightarrow{b_1}=7\widehat{i}-5\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b_2}=\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{b_1}\cdot\overrightarrow{b_2}=(7\widehat{i}-5\widehat{j}+\widehat{k})\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})$
$\displaystyle =7-10+3$
$\displaystyle =0$
$\displaystyle \therefore \overrightarrow{b_1}\perp\overrightarrow{b_2}$
$\displaystyle \text{Hence, the given lines are perpendicular to each other.}$

$\displaystyle \textbf{Question 20: }~\text{Find the vector equation of the line passing through the point } \\ (2,-1,-1)\text{ which is parallel to the line }6x-2=3y+1=2z-2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the line }6x-2=3y+1=2z-2\text{ can be re-written as}$
$\displaystyle \frac{x-\frac{1}{3}}{\frac{1}{6}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-1}{\frac{1}{2}}$
$\displaystyle \frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3}$
$\displaystyle \text{Since the required line is parallel to the given line, the direction ratios of the required} \\ \text{line are proportional to }1,2,3$
$\displaystyle \text{The vector equation of the required line passing through the point } \\ (2,-1,-1)\text{ and having direction ratios proportional to }1,2,3\text{ is}$
$\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k})$

$\displaystyle \textbf{Question 21: }~\text{If the lines }\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\text{ and }\frac{x-1}{3\lambda}=\frac{y-1}{1}=\frac{z-6}{-5}\text{ are perpendicular, find the value of }\lambda. \ [CBSE\ 2009]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given lines are}$
$\displaystyle \frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}$
$\displaystyle \frac{x-1}{3\lambda}=\frac{y-1}{1}=\frac{z-6}{-5}$
$\displaystyle \text{Since the given lines are perpendicular to each other, we have}$
$\displaystyle (-3)(3\lambda)+(2\lambda)(1)+2(-5)=0$
$\displaystyle -9\lambda+2\lambda-10=0$
$\displaystyle -7\lambda-10=0$
$\displaystyle \lambda=-\frac{10}{7}$

$\displaystyle \textbf{Question 22: }~\text{If the coordinates of the points }A,B,C,D\text{ be }(1,2,3),(4,5,7), \\ (-4,3,-6)\text{ and } (2,9,2)\text{ respectively, then find the angle between the lines }AB\text{ and }CD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The direction ratios of }AB\text{ and }CD\text{ are proportional to }3,3,4\text{ and }6,6,8\text{ respectively.}$
$\displaystyle \text{Let }\theta\text{ be the angle between }AB\text{ and }CD.\text{ Then,}$
$\displaystyle \cos\theta=\frac{3\times6+3\times6+4\times8}{\sqrt{3^{2}+3^{2}+4^{2}}\sqrt{6^{2}+6^{2}+8^{2}}}$
$\displaystyle \cos\theta=\frac{68}{\sqrt{34}\sqrt{136}}$
$\displaystyle \cos\theta=\frac{68}{\sqrt{34\times136}}$
$\displaystyle \cos\theta=1$
$\displaystyle \theta=0^\circ$

$\displaystyle \textbf{Question 23: }~\text{Find the value of }\lambda\text{ so that the following lines are perpendicular} \\ \text{to each other: } \frac{x-5}{5\lambda+2}=\frac{2-y}{5}=\frac{1-z}{-1},\ \frac{x}{1}=\frac{2y+1}{4\lambda}=\frac{1-z}{-3}. \ [CBSE\ 2009]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given lines}$
$\displaystyle \frac{x-5}{5\lambda+2}=\frac{2-y}{5}=\frac{1-z}{-1}\text{ and }\frac{x}{1}=\frac{2y+1}{4\lambda}=\frac{1-z}{-3}$
$\displaystyle \text{can be re-written as}$
$\displaystyle \frac{x-5}{5\lambda+2}=\frac{y-2}{-5}=\frac{z-1}{1}\text{ and }\frac{x}{1}=\frac{y+\frac{1}{2}}{2\lambda}=\frac{z-1}{3}$
$\displaystyle \text{Since the given lines are perpendicular to each other, we have}$
$\displaystyle (5\lambda+2)(1)-5(2\lambda)+1(3)=0$
$\displaystyle 5\lambda+2-10\lambda+3=0$
$\displaystyle -5\lambda+5=0$
$\displaystyle 5\lambda=5$
$\displaystyle \lambda=1$

$\displaystyle \textbf{Question 24: }~\text{Find the direction cosines of the line }\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}. \\ \text{ Also, find the vector equation of the line through the point } A(-1,2,3) \\ \text{ and parallel to the given line. \ [CBSE\ 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the given line is}$
$\displaystyle \frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$
$\displaystyle \text{The given equation can be re-written as}$
$\displaystyle \frac{x+2}{2}=\frac{y-\frac{7}{2}}{3}=\frac{z-5}{-6}$
$\displaystyle \text{This line passes through the point }\left(-2,\frac{7}{2},5\right)\text{ and has direction ratios proportional to }2,3,-6$
$\displaystyle \text{So, its direction cosines are}$
$\displaystyle \frac{2}{\sqrt{2^{2}+3^{2}+(-6)^{2}}},\frac{3}{\sqrt{2^{2}+3^{2}+(-6)^{2}}},\frac{-6}{\sqrt{2^{2}+3^{2}+(-6)^{2}}}$
$\displaystyle \text{Or }\frac{2}{7},\frac{3}{7},\frac{-6}{7}$
$\displaystyle \text{The required line passes through the point having position vector }\overrightarrow{a}=-\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=2\widehat{i}+3\widehat{j}-6\widehat{k}$
$\displaystyle \text{So, its vector equation is}$
$\displaystyle \overrightarrow{r}=(-\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}-6\widehat{k})$