\displaystyle \textbf{Question 1: }~\text{If a line makes angles of }90^\circ,\ 60^\circ\text{ and }30^\circ\text{ with the positive direction of }x,y, \\ \text{and }z\text{-axis respectively, find its direction cosines.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the direction cosines of the line be } l,m,n.
\displaystyle \text{Now, } l=\cos 90^\circ=0.
\displaystyle m=\cos 60^\circ=\frac{1}{2}.
\displaystyle n=\cos 30^\circ=\frac{\sqrt{3}}{2}.
\displaystyle \text{Therefore, the direction cosines of the line are } 0,\ \frac{1}{2},\ \frac{\sqrt{3}}{2}.

\displaystyle \textbf{Question 2: }~\text{If a line has direction ratios }2,-1,-2,\ \text{determine its direction cosines. }
\displaystyle \text{Answer:}
\displaystyle \text{Let the direction cosines of the line be } l,m,n.
\displaystyle \text{Now, } l=\frac{2}{\sqrt{2^2+(-1)^2+(-2)^2}}.
\displaystyle =\frac{2}{\sqrt{4+1+4}}.
\displaystyle =\frac{2}{3}.
\displaystyle m=\frac{-1}{\sqrt{2^2+(-1)^2+(-2)^2}}.
\displaystyle =\frac{-1}{\sqrt{4+1+4}}.
\displaystyle =\frac{-1}{3}.
\displaystyle n=\frac{-2}{\sqrt{2^2+(-1)^2+(-2)^2}}.
\displaystyle =\frac{-2}{\sqrt{4+1+4}}.
\displaystyle =\frac{-2}{3}.
\displaystyle \text{Therefore, the direction cosines of the line are } \frac{2}{3},\ \frac{-1}{3},\ \frac{-2}{3}.

\displaystyle \textbf{Question 3: }~\text{Find the direction cosines of the line passing through two points } \\ (-2,4,-5)\text{ and }(1,2,3). 
\displaystyle \text{Answer:}
\displaystyle \text{The direction cosines of the line passing through two points }  \\ P(x_1,y_1,z_1) \text{ and } Q(x_2,y_2,z_2) \text{ are } \frac{x_2-x_1}{PQ},\ \frac{y_2-y_1}{PQ},\ \frac{z_2-z_1}{PQ}.
\displaystyle \text{Here, } PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.
\displaystyle P=(-2,4,-5).
\displaystyle Q=(1,2,3).
\displaystyle PQ=\sqrt{[1-(-2)]^2+(2-4)^2+[3-(-5)]^2}.
\displaystyle =\sqrt{9+4+64}.
\displaystyle =\sqrt{77}.
\displaystyle \text{Thus, the direction cosines of the line joining the two points are } \frac{1-(-2)}{\sqrt{77}},\ \frac{2-4}{\sqrt{77}},\ \frac{3-(-5)}{\sqrt{77}}.
\displaystyle \text{i.e. } \frac{3}{\sqrt{77}},\ \frac{-2}{\sqrt{77}},\ \frac{8}{\sqrt{77}}.

\displaystyle \textbf{Question 4: }~\text{Using direction ratios show that the points }A(2,3,-4),\ B(1,-2,3) \\ \text{ and }C(3,8,-11)\text{ are collinear.  }
\displaystyle \text{Answer:}
\displaystyle \text{The given points are } A(2,3,-4),\ B(1,-2,3) \text{ and } C(3,8,-11).
\displaystyle \text{We know that the direction ratios of the line joining the points } (x_1,y_1,z_1) \\ \text{ and } (x_2,y_2,z_2) \text{ are } x_2-x_1,\ y_2-y_1,\ z_2-z_1.
\displaystyle \text{The direction ratios of the line joining } A \text{ and } B \text{ are } 1-2,\ -2-3,\ 3-(-4).
\displaystyle =-1,\ -5,\ 7.
\displaystyle \text{The direction ratios of the line joining } B \text{ and } C \text{ are } 3-1,\ 8-(-2),\ -11-3.
\displaystyle =2,\ 10,\ -14.
\displaystyle \text{It is clear that the direction ratios of } BC \text{ are } -2 \text{ times those of } AB.
\displaystyle \text{Therefore, } AB \parallel BC.
\displaystyle \text{Also, point } B \text{ is common to both } AB \text{ and } BC.
\displaystyle \text{Therefore, points } A,\ B \text{ and } C \text{ are collinear.}

\displaystyle \textbf{Question 5: }~\text{Find the direction cosines of the sides of the triangle whose vertices are } \\ (3,5,-4),\ (-1,1,2)\text{ and }(-5,-5,-2). 
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of } \triangle ABC \text{ are } A(3,5,-4),\ B(-1,1,2) \text{ and } C(-5,-5,-2).
\displaystyle \text{The direction ratios of } AB \text{ are } (-1-3),\ (1-5),\ [2-(-4)].
\displaystyle =-4,\ -4,\ 6.
\displaystyle \text{Therefore, the direction cosines of } AB \text{ are } \frac{-4}{\sqrt{(-4)^2+(-4)^2+6^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+6^2}}, \\ \frac{6}{\sqrt{(-4)^2+(-4)^2+6^2}}.
\displaystyle =\frac{-4}{\sqrt{16+16+36}},\ \frac{-4}{\sqrt{16+16+36}},\ \frac{6}{\sqrt{16+16+36}}.
\displaystyle =\frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}}.
\displaystyle =\frac{-2}{\sqrt{17}},\ \frac{-2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}.
\displaystyle \text{The direction ratios of } BC \text{ are } [-5-(-1)],\ (-5-1),\ (-2-2).
\displaystyle =-4,\ -6,\ -4.
\displaystyle \text{Therefore, the direction cosines of } BC \text{ are } \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}, \\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}.
\displaystyle =\frac{-4}{\sqrt{16+36+16}},\ \frac{-6}{\sqrt{16+36+16}},\ \frac{-4}{\sqrt{16+36+16}}.
\displaystyle =\frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}}.
\displaystyle =\frac{-2}{\sqrt{17}},\ \frac{-3}{\sqrt{17}},\ \frac{-2}{\sqrt{17}}.
\displaystyle \text{The direction ratios of } CA \text{ are } [3-(-5)],\ [5-(-5)],\ [-4-(-2)].
\displaystyle =8,\ 10,\ -2.
\displaystyle \text{Therefore, the direction cosines of } CA \text{ are } \frac{8}{\sqrt{8^2+10^2+(-2)^2}},\\ \frac{10}{\sqrt{8^2+10^2+(-2)^2}},\ \frac{-2}{\sqrt{8^2+10^2+(-2)^2}}.
\displaystyle =\frac{8}{\sqrt{64+100+4}},\ \frac{10}{\sqrt{64+100+4}},\ \frac{-2}{\sqrt{64+100+4}}.
\displaystyle =\frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}}.
\displaystyle =\frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ \frac{-1}{\sqrt{42}}.

\displaystyle \textbf{Question 6: }~\text{Find the angle between the vectors with direction ratios proportional} \\ \text{to }1,-2,1\text{ and }4,3,2. \ 
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{a} \text{ be a vector with direction ratios } 1,-2,1.
\displaystyle \Rightarrow \overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k}.
\displaystyle \text{Let } \overrightarrow{b} \text{ be a vector with direction ratios } 4,3,2.
\displaystyle \Rightarrow \overrightarrow{b}=4\widehat{i}+3\widehat{j}+2\widehat{k}.
\displaystyle \text{Let } \theta \text{ be the angle between the given vectors.}
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(\widehat{i}-2\widehat{j}+\widehat{k})\cdot(4\widehat{i}+3\widehat{j}+2\widehat{k})}{|\widehat{i}-2\widehat{j}+\widehat{k}|\ |4\widehat{i}+3\widehat{j}+2\widehat{k}|}.
\displaystyle =\frac{4-6+2}{\sqrt{1^2+(-2)^2+1^2}\ \sqrt{4^2+3^2+2^2}}.
\displaystyle =\frac{0}{\sqrt{6}\sqrt{29}}.
\displaystyle =0.
\displaystyle \therefore\ \theta=\frac{\pi}{2}.
\displaystyle \text{Thus, the angle between the given vectors measures } \frac{\pi}{2}.

\displaystyle \textbf{Question 7: }~\text{Find the angle between the vectors whose direction cosines are} \\ \text{proportional to }2,3,-6\text{ and }3,-4,5.
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{a} \text{ be a vector with direction ratios } 2,3,-6.
\displaystyle \Rightarrow \overrightarrow{a}=2\widehat{i}+3\widehat{j}-6\widehat{k}.
\displaystyle \text{Let } \overrightarrow{b} \text{ be a vector with direction ratios } 3,-4,5.
\displaystyle \Rightarrow \overrightarrow{b}=3\widehat{i}-4\widehat{j}+5\widehat{k}.
\displaystyle \text{Let } \theta \text{ be the angle between the given vectors.}
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(2\widehat{i}+3\widehat{j}-6\widehat{k})\cdot(3\widehat{i}-4\widehat{j}+5\widehat{k})}{|2\widehat{i}+3\widehat{j}-6\widehat{k}|\ |3\widehat{i}-4\widehat{j}+5\widehat{k}|}.
\displaystyle =\frac{6-12-30}{\sqrt{2^2+3^2+(-6)^2}\ \sqrt{3^2+(-4)^2+5^2}}.
\displaystyle =\frac{-36}{\sqrt{49}\sqrt{50}}.
\displaystyle =\frac{-36}{7\sqrt{50}}.
\displaystyle =\frac{-36}{35\sqrt{2}}.
\displaystyle \text{Rationalising the result, we get } \cos\theta=\frac{-18\sqrt{2}}{35}.
\displaystyle \therefore\ \theta=\cos^{-1}\left(\frac{-18\sqrt{2}}{35}\right).
\displaystyle \text{Thus, the angle between the given vectors measures } \cos^{-1}\left(\frac{-18\sqrt{2}}{35}\right).

\displaystyle \textbf{Question 8: }~\text{Find the acute angle between the lines whose direction ratios are} \\ \text{proportional to }2:3:6\text{ and }1:2:2.
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{a} \text{ be a vector parallel to the vector with direction ratios } 2,3,6.
\displaystyle \Rightarrow \overrightarrow{a}=2\widehat{i}+3\widehat{j}+6\widehat{k}.
\displaystyle \text{Let } \overrightarrow{b} \text{ be a vector parallel to the vector with direction ratios } 1,2,2.
\displaystyle \Rightarrow \overrightarrow{b}=\widehat{i}+2\widehat{j}+2\widehat{k}.
\displaystyle \text{Let } \theta \text{ be the angle between the given vectors.}
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(2\widehat{i}+3\widehat{j}+6\widehat{k})\cdot(\widehat{i}+2\widehat{j}+2\widehat{k})}{|2\widehat{i}+3\widehat{j}+6\widehat{k}|\ |\widehat{i}+2\widehat{j}+2\widehat{k}|}.
\displaystyle =\frac{2+6+12}{\sqrt{2^2+3^2+6^2}\ \sqrt{1^2+2^2+2^2}}.
\displaystyle =\frac{20}{\sqrt{49}\sqrt{9}}.
\displaystyle =\frac{20}{21}.
\displaystyle \therefore\ \theta=\cos^{-1}\left(\frac{20}{21}\right).

\displaystyle \textbf{Question 9: }~\text{Show that the points }(2,3,4),\ (-1,-2,1),\ (5,8,7)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose the points are } A(2,3,4),\ B(-1,-2,1) \text{ and } C(5,8,7).
\displaystyle \text{We know that the direction ratios of the line joining the points } (x_1,y_1,z_1)  \\ \text{ and } (x_2,y_2,z_2) \text{ are } x_2-x_1,\ y_2-y_1,\ z_2-z_1.
\displaystyle \text{The direction ratios of } AB \text{ are } (-1-2),\ (-2-3),\ (1-4).
\displaystyle =-3,\ -5,\ -3.
\displaystyle \text{The direction ratios of } BC \text{ are } (5-(-1)),\ (8-(-2)),\ (7-1).
\displaystyle =6,\ 10,\ 6.
\displaystyle \text{It can be seen that the direction ratios of } BC \text{ are } -2 \text{ times those of } AB.
\displaystyle \text{Therefore, } AB \parallel BC.
\displaystyle \text{Since point } B \text{ is common to both } AB \text{ and } BC,\ \text{points } A,\ B \text{ and } C \text{ are collinear.}

\displaystyle \textbf{Question 10: }~\text{Show that the line through points }(4,7,8) \\ \text{ and }(2,3,4)\text{ is parallel to the} \\ \text{line through the points }(-1,-2,1)\text{ and }(1,2,5).
\displaystyle \text{Answer:}
\displaystyle \text{We know that the direction ratios of the line passing through the points } (x_1,y_1,z_1) \text{ and } (x_2,y_2,z_2) \text{ are } x_2-x_1,\ y_2-y_1,\ z_2-z_1.
\displaystyle \text{Let the first two points be } A(4,7,8) \text{ and } B(2,3,4).
\displaystyle \text{Thus, the direction ratios of } AB \text{ are } (2-4),\ (3-7),\ (4-8).
\displaystyle =-2,\ -4,\ -4.
\displaystyle \text{Similarly, let the other two points be } C(-1,-2,1) \text{ and } D(1,2,5).
\displaystyle \text{Thus, the direction ratios of } CD \text{ are } [1-(-1)],\ [2-(-2)],\ (5-1).
\displaystyle =2,\ 4,\ 4.
\displaystyle \text{It can be seen that the direction ratios of } CD \text{ are } -1 \text{ times those of } AB.
\displaystyle \text{Therefore, } AB \parallel CD.

\displaystyle \textbf{Question 11: }~\text{Show that the line through the points }(1,-1,2)\text{ and }(3,4,-2)\text{ is perpendicular to the} \\ \text{line through the points }(0,3,2)\text{ and }(3,5,6).
\displaystyle \text{Answer:}
\displaystyle \text{We know that two lines with direction ratios } a_1,b_1,c_1 \text{ and } a_2,b_2,c_2  \\ \text{ are perpendicular if } a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \text{The direction ratios of the line passing through the points } (1,-1,2) \text{ and } (3,4,-2) \text{ are } (3-1),\ [4-(-1)],\ (-2-2).
\displaystyle =2,\ 5,\ -4.
\displaystyle \Rightarrow a_1=2,\ b_1=5,\ c_1=-4.
\displaystyle \text{Similarly, the direction ratios of the line passing through the points } (0,3,2) \\ \text{ and } (3,5,6) \text{ are } (3-0),\ (5-3),\ (6-2).
\displaystyle =3,\ 2,\ 4.
\displaystyle \Rightarrow a_2=3,\ b_2=2,\ c_2=4.
\displaystyle \therefore\ a_1a_2+b_1b_2+c_1c_2=2\times3+5\times2+(-4)\times4.
\displaystyle =6+10-16.
\displaystyle =0.
\displaystyle \text{Thus, the line through the points } (1,-1,2) \text{ and } (3,4,-2) \text{ is perpendicular to the}\\ \text{line through the points } (0,3,2) \text{ and } (3,5,6).

\displaystyle \textbf{Question 12: }~\text{Show that the line joining the origin to the point }(2,1,1) \\ \text{ is perpendicular to the} \text{line determined by the points }(3,5,-1)\text{ and }(4,3,-1).
\displaystyle \text{Answer:}
\displaystyle \text{We know that two lines with direction ratios } a_1,b_1,c_1 \text{ and } a_2,b_2,c_2  \\ \text{ are perpendicular if } a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \text{The direction ratios of the line joining the origin } (0,0,0) \text{ to the point } (2,1,1) \text{ are } (2-0),\ (1-0),\ (1-0).
\displaystyle =2,\ 1,\ 1.
\displaystyle \Rightarrow a_1=2,\ b_1=1,\ c_1=1.
\displaystyle \text{Similarly, the direction ratios of the line joining the points } (3,5,-1) \text{ and } (4,3,-1) \text{ are } (4-3),\ (3-5),\ [-1-(-1)].
\displaystyle =1,\ -2,\ 0.
\displaystyle \Rightarrow a_2=1,\ b_2=-2,\ c_2=0.
\displaystyle \therefore\ a_1a_2+b_1b_2+c_1c_2=2\times1+1\times(-2)+1\times0.
\displaystyle =2-2+0.
\displaystyle =0.
\displaystyle \text{Therefore, the line joining the origin to the point } (2,1,1) \text{ is}\\ \text{perpendicular to the line determined by the points } (3,5,-1) \text{ and } (4,3,-1).

\displaystyle \textbf{Question 13: }~\text{Find the angle between the lines whose direction ratios are} \\ \text{proportional to }a,b,c\text{ and }b-c,\ c-a,\ a-b.
\displaystyle \text{Answer:}
\displaystyle \text{Let } \theta \text{ be the angle between the given lines.}
\displaystyle \text{We have } a_1=a,\ b_1=b,\ c_1=c.
\displaystyle a_2=b-c,\ b_2=c-a,\ c_2=a-b.
\displaystyle \text{Now, } \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\ \sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle =\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}\ \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}.
\displaystyle =\frac{ab-ac+bc-ab+ac-bc}{\sqrt{a^2+b^2+c^2}\ \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}.
\displaystyle =0.
\displaystyle \therefore\ \theta=\frac{\pi}{2}.
\displaystyle \text{Thus, the angle between the given lines measures } 90^\circ.

\displaystyle \textbf{Question 14: }~\text{If the coordinates of the points }A,B,C,D\text{ are }(1,2,3),\ (4,5,7),\ (-4,3,-6) \\ \text{ and }(2,9,2),\text{ then find the angle between }AB\text{ and }CD.
\displaystyle \text{Answer:}
\displaystyle \text{The given points are } A(1,2,3),\ B(4,5,7),\ C(-4,3,-6) \text{ and } D(2,9,2).
\displaystyle \text{We know that the direction ratios of the line joining the points } (x_1,y_1,z_1)  \\ \text{ and } (x_2,y_2,z_2) \text{ are } x_2-x_1,\ y_2-y_1,\ z_2-z_1.
\displaystyle \text{The direction ratios of } AB \text{ are } (4-1),\ (5-2),\ (7-3).
\displaystyle =3,\ 3,\ 4.
\displaystyle \text{The direction ratios of } CD \text{ are } [2-(-4)],\ (9-3),\ [2-(-6)].
\displaystyle =6,\ 6,\ 8.
\displaystyle \text{Let } \theta \text{ be the angle between } AB \text{ and } CD.
\displaystyle \text{We have } a_1=3,\ b_1=3,\ c_1=4.
\displaystyle a_2=6,\ b_2=6,\ c_2=8.
\displaystyle \text{Now, } \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\ \sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle =\frac{3\times6+3\times6+4\times8}{\sqrt{3^2+3^2+4^2}\ \sqrt{6^2+6^2+8^2}}.
\displaystyle =\frac{18+18+32}{\sqrt{9+9+16}\ \sqrt{36+36+64}}.
\displaystyle =\frac{68}{\sqrt{34}\ \sqrt{136}}.
\displaystyle =\frac{68}{68}.
\displaystyle =1.
\displaystyle \therefore\ \theta=0^\circ.
\displaystyle \text{Thus, the angle between } AB \text{ and } CD \text{ measures } 0^\circ.

\displaystyle \textbf{Question 15: }~\text{Find the direction cosines of the lines, connected by the relations: } \\ l+m+n=0\ \text{and }2lm+2ln-mn=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle l+m+n=0.\qquad(1)
\displaystyle 2lm+2ln-mn=0.\qquad(2)
\displaystyle \text{From (1), we get } l=-m-n.
\displaystyle \text{Substituting } l=-m-n \text{ in (2), we get}
\displaystyle 2(-m-n)m+2(-m-n)n-mn=0.
\displaystyle \Rightarrow -2m^2-2mn-2mn-2n^2-mn=0.
\displaystyle \Rightarrow 2m^2+2n^2+5mn=0.
\displaystyle \Rightarrow (m+2n)(2m+n)=0.
\displaystyle \Rightarrow m=-2n \text{ or } m=-\frac{n}{2}.
\displaystyle \text{If } m=-2n,\ \text{then from (1), we get } l=n.
\displaystyle \text{If } m=-\frac{n}{2},\ \text{then from (1), we get } l=-\frac{n}{2}.
\displaystyle \text{Thus, the direction ratios of the two lines are proportional to } (n,-2n,n) \text{ and } \left(-\frac{n}{2},-\frac{n}{2},n\right).
\displaystyle \text{i.e. } (1,-2,1) \text{ and } \left(-\frac{1}{2},-\frac{1}{2},1\right).
\displaystyle \text{Hence, their direction cosines are } \pm\frac{1}{\sqrt{6}},\ \pm\frac{-2}{\sqrt{6}},\ \pm\frac{1}{\sqrt{6}}.
\displaystyle \text{and } \pm\frac{-1}{\sqrt{6}},\ \pm\frac{-1}{\sqrt{6}},\ \pm\frac{2}{\sqrt{6}}.

\displaystyle \textbf{Question 16: }~\text{Find the angle between the lines whose direction cosines are given by the} \\ \text{equations:}
\displaystyle \text{(i) }l+m+n=0\ \text{and }l^{2}+m^{2}-n^{2}=0
\displaystyle \text{(ii) }2l-m+2n=0\ \text{and }mn+nl+lm=0
\displaystyle \text{(iii) }l+2m+3n=0\ \text{and }3lm-4ln+mn=0
\displaystyle \text{(iv) }2l+2m-n=0,\ mn+ln+lm=0.\qquad 
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given:}
\displaystyle l+m+n=0.\qquad(1)
\displaystyle l^2+m^2-n^2=0.\qquad(2)
\displaystyle \text{From (1), we get } m=-l-n.
\displaystyle \text{Substituting } m=-l-n \text{ in (2), we get}
\displaystyle l^2+(-l-n)^2-n^2=0.
\displaystyle \Rightarrow l^2+l^2+n^2+2ln-n^2=0.
\displaystyle \Rightarrow 2l^2+2ln=0.
\displaystyle \Rightarrow 2l(l+n)=0.
\displaystyle \Rightarrow l=0 \text{ or } l=-n.
\displaystyle \text{If } l=0,\ \text{then substituting in (1), we get } m=-n.
\displaystyle \text{If } l=-n,\ \text{then substituting in (1), we get } m=0.
\displaystyle \text{Thus, the direction ratios of the two lines are proportional to } (0,-n,n) \text{ and } (-n,0,n).
\displaystyle \text{i.e. } (0,-1,1) \text{ and } (-1,0,1).
\displaystyle \text{Vectors parallel to these lines are } \overrightarrow{a}=0\widehat{i}-\widehat{j}+\widehat{k} \text{ and } \overrightarrow{b}=-\widehat{i}+0\widehat{j}+\widehat{k}.
\displaystyle \text{If } \theta \text{ is the angle between the lines, then } \theta \text{ is also the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}.
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{0^2+(-1)^2+1^2}\ \sqrt{(-1)^2+0^2+1^2}}.
\displaystyle =\frac{1}{\sqrt{2}\sqrt{2}}.
\displaystyle =\frac{1}{2}.
\displaystyle \therefore\ \theta=\frac{\pi}{3}.
\displaystyle \text{(ii) }
\displaystyle \text{Given:}
\displaystyle 2l-m+2n=0.\qquad(1)
\displaystyle mn+nl+lm=0.\qquad(2)
\displaystyle \text{From (1), we get } m=2l+2n.
\displaystyle \text{Substituting } m=2l+2n \text{ in (2), we get}
\displaystyle (2l+2n)n+nl+l(2l+2n)=0.
\displaystyle \Rightarrow 2ln+2n^2+nl+2l^2+2ln=0.
\displaystyle \Rightarrow 2l^2+5ln+2n^2=0.
\displaystyle \Rightarrow (l+2n)(2l+n)=0.
\displaystyle \Rightarrow l=-2n \text{ or } l=-\frac{n}{2}.
\displaystyle \text{If } l=-2n,\ \text{then substituting in (1), we get } m=-2n.
\displaystyle \text{If } l=-\frac{n}{2},\ \text{then substituting in (1), we get } m=n.
\displaystyle \text{Thus, the direction ratios of the two lines are proportional to } (-2n,-2n,n) \text{ and } \left(-\frac{n}{2},n,n\right).
\displaystyle \text{i.e. } (-2,-2,1) \text{ and } \left(-\frac{1}{2},1,1\right).
\displaystyle \text{Vectors parallel to these lines are } \overrightarrow{a}=-2\widehat{i}-2\widehat{j}+\widehat{k} \text{ and } \overrightarrow{b}=-\frac{1}{2}\widehat{i}+\widehat{j}+\widehat{k}.
\displaystyle \text{If } \theta \text{ is the angle between the lines, then } \theta \text{ is also the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}.
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(-2)\left(-\frac{1}{2}\right)+(-2)(1)+(1)(1)}{\sqrt{(-2)^2+(-2)^2+1^2}\ \sqrt{\left(-\frac{1}{2}\right)^2+1^2+1^2}}.
\displaystyle =\frac{1-2+1}{\sqrt{4+4+1}\ \sqrt{\frac{1}{4}+1+1}}.
\displaystyle =0.
\displaystyle \therefore\ \theta=\frac{\pi}{2}.
\displaystyle \text{(iii) }
\displaystyle \text{Given:}
\displaystyle l+2m+3n=0.\qquad(1)
\displaystyle 3lm-4ln+mn=0.\qquad(2)
\displaystyle \text{From (1), we get } l=-2m-3n.
\displaystyle \text{Substituting } l=-2m-3n \text{ in (2), we get}
\displaystyle 3(-2m-3n)m-4(-2m-3n)n+mn=0.
\displaystyle \Rightarrow -6m^2-9mn+8mn+12n^2+mn=0.
\displaystyle \Rightarrow 12n^2-6m^2=0.
\displaystyle \Rightarrow m^2=2n^2.
\displaystyle \Rightarrow m=\sqrt{2}\,n \text{ or } m=-\sqrt{2}\,n.
\displaystyle \text{If } m=\sqrt{2}\,n,\ \text{then from (1), } l=-2\sqrt{2}\,n-3n.
\displaystyle \text{If } m=-\sqrt{2}\,n,\ \text{then from (1), } l=2\sqrt{2}\,n-3n.
\displaystyle \text{Thus, the direction ratios of the two lines are proportional to } (-2\sqrt{2}-3,\ \sqrt{2},\ 1) \text{ and } (2\sqrt{2}-3,\ -\sqrt{2},\ 1).
\displaystyle \text{Vectors parallel to these lines are } \overrightarrow{a}=(-2\sqrt{2}-3)\widehat{i}+\sqrt{2}\widehat{j}+\widehat{k} \text{ and } \overrightarrow{b}=(2\sqrt{2}-3)\widehat{i}-\sqrt{2}\widehat{j}+\widehat{k}.
\displaystyle \text{If } \theta \text{ is the angle between the lines, then } \theta \text{ is also the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}.
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(-2\sqrt{2}-3)(2\sqrt{2}-3)+(\sqrt{2})(-\sqrt{2})+(1)(1)}{\sqrt{(-2\sqrt{2}-3)^2+(\sqrt{2})^2+1^2}\ \sqrt{(2\sqrt{2}-3)^2+(\sqrt{2})^2+1^2}}.
\displaystyle =\frac{(9-8)-2+1}{\sqrt{20+12\sqrt{2}}\ \sqrt{20-12\sqrt{2}}}.
\displaystyle =\frac{0}{\sqrt{20+12\sqrt{2}}\ \sqrt{20-12\sqrt{2}}}.
\displaystyle =0.
\displaystyle \therefore\ \theta=\frac{\pi}{2}.
\displaystyle \text{(iv) }
\displaystyle \text{The given relations are}
\displaystyle 2l+2m-n=0.\qquad(1)
\displaystyle mn+ln+lm=0.\qquad(2)
\displaystyle \text{From (1), we have } n=2l+2m.
\displaystyle \text{Putting this value of } n \text{ in (2), we get}
\displaystyle m(2l+2m)+l(2l+2m)+lm=0.
\displaystyle \Rightarrow 2lm+2m^2+2l^2+2lm+lm=0.
\displaystyle \Rightarrow 2m^2+5lm+2l^2=0.
\displaystyle \Rightarrow (2m+l)(m+2l)=0.
\displaystyle \Rightarrow 2m+l=0 \text{ or } m+2l=0.
\displaystyle \Rightarrow l=-2m \text{ or } l=-\frac{m}{2}.
\displaystyle \text{When } l=-2m,\ \text{we have } n=2(-2m)+2m=-4m+2m=-2m.
\displaystyle \text{When } l=-\frac{m}{2},\ \text{we have } n=2\left(-\frac{m}{2}\right)+2m=-m+2m=m.
\displaystyle \text{Thus, the direction ratios of the two lines are proportional to } (-2m,m,-2m) \text{ and } \left(-\frac{m}{2},m,m\right).
\displaystyle \text{Or } (-2,1,-2) \text{ and } (-1,2,2).
\displaystyle \text{So, vectors parallel to these lines are } \overrightarrow{a}=-2\widehat{i}+\widehat{j}-2\widehat{k} \text{ and } \overrightarrow{b}=-\widehat{i}+2\widehat{j}+2\widehat{k}.
\displaystyle \text{Let } \theta \text{ be the angle between these lines; then } \theta \text{ is also the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}.
\displaystyle \text{Now, } \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}.
\displaystyle =\frac{(-2)(-1)+(1)(2)+(-2)(2)}{\sqrt{(-2)^2+1^2+(-2)^2}\ \sqrt{(-1)^2+2^2+2^2}}.
\displaystyle =\frac{2+2-4}{3\times3}.
\displaystyle =0.
\displaystyle \therefore\ \theta=\frac{\pi}{2}.
\displaystyle \text{Thus, the angle between the two lines whose direction cosines are given by the relations is } \frac{\pi}{2}.

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