NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.2) Exercise 11.2

Q.1

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{three}\mathrm{lines}\mathrm{with}\mathrm{direction}\mathrm{cosines}\\ \frac{\mathrm{12}}{\mathrm{13}}\mathrm{,}\frac{–3}{\mathrm{13}}\mathrm{,}\frac{–4}{\mathrm{13}};\frac{\mathrm{4}}{\mathrm{13}}\mathrm{,}\frac{\mathrm{12}}{\mathrm{13}}\mathrm{,}\frac{\mathrm{3}}{\mathrm{13}};\frac{\mathrm{3}}{\mathrm{13}}\mathrm{,}\frac{–4}{\mathrm{13}}\mathrm{,}\frac{\mathrm{12}}{\mathrm{13}}\mathrm{are}\mathrm{mutually}\mathrm{perpendicular}\mathrm{.}\end{array}$

Ans

\begin{array}{l}Two\text{lines with direction cosines}{l}_{\text{1}}{\text{,m}}_{\text{1}}{\text{,n}}_{\text{1}}\text{and}{l}_{\text{2}}{\text{,m}}_{\text{2}}{\text{,n}}_{\text{2}}\text{are}\\ \text{perpendicular to each other if}{l}_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=0\\ \left(i\right)\text{For the lines with direction cosines,}\frac{12}{13},\frac{-3}{13},\frac{-4}{13}\text{and}\frac{4}{13},\frac{12}{13},\frac{3}{13},\\ we\text{have}\\ l_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=\frac{12}{13}.\frac{4}{13}+\frac{-3}{13}.\frac{12}{13}+\frac{-4}{13}.\frac{3}{13}\\ =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=0\\ Thus,\text{lines are perpendicular}\text{.}\\ \left(ii\right)\text{For the lines with direction cosines}\frac{4}{13},\frac{12}{13},\frac{3}{13}\text{and}\frac{3}{13},\frac{-4}{13},\frac{12}{13},\\ we\text{have}\\ l_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=\frac{4}{13}.\frac{3}{13}+\frac{12}{13}.\left(\frac{-4}{13}\right)+\frac{3}{13}.\frac{12}{13}\\ =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}=0\\ Thus,\text{lines are perpendicular}\text{.}\\ \left(iii\right)\text{For the lines with direction cosines}\frac{3}{13},\frac{-4}{13},\frac{12}{13}\text{and}\frac{12}{13},\frac{-3}{13},\frac{-4}{13},\\ we\text{have}\\ l_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=\frac{3}{13}.\frac{12}{13}+\frac{-4}{13}.\left(\frac{-3}{13}\right)+\frac{12}{13}.\left(\frac{-4}{13}\right)\\ =\frac{36}{169}+\frac{12}{169}-\frac{48}{169}=0\\ Thus,\text{lines are perpendicular}\text{.}\\ \text{Therefore, all lines are mutually perpendicular}\text{.}\end{array}

Q.2 Show that the line through the points (1,–1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Ans

Let line AB has end points A(1,–1, 2) and B(3, 4, –2). And line CD has end points C (0, 3, 2) and D(3, 5, 6).

The direction ratios a₁, a₂ and a₃ of AB are (3 – 1), {4 –(–1)} and (–2–2) i.e., 2, 5, – 4.The direction ratios b₁, b₂ and b₃ of CD are (3 – 0), (5 –3) and (6 – 2) i.e., 3, 2, 4.AB and CD will be perpendicular to each other, if a₁b₁+ a₂b₂ + a₃b₃ = 0.Then,
a₁b₁+ a₂b₂ + a₃b₃ = (2)(3) + (5)(2) + (– 4)(4)

= 6 + 10 – 16
= 0

Therefore, AB and CD are perpendicular to each other.

Q.3 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

Ans

Let line AB has end points A (4, 7, 8) and B (2, 3, 4). And line CD has end points C (–1, –2, 1) and D (1, 2, 5).

The direction ratios a₁, a₂ and a₃ of AB are (2 – 4), (3 –7) and (4 – 8) i.e., –2, – 4, – 4.The direction ratios b₁, b₂ and b₃ of CD are {1–(–1)}, {2 –(–2)} and (5 – 1) i.e., 2, 4, 4.AB will be parallel to CD, if(a₁/b₁) = (a₂/b₂) = (a₃/b₃). Then,
(a₁ / b₁) = (–2/2) = –1

(a₂ / b₂) = (– 4/4) = – 1

(a₃ / b₃) = (– 4/4) = – 1

So, (a₁/b₁) = (a₂/b₂) = (a₃/b₃)

Therefore, AB is parallel to CD.

Q.4 Find the equation of the line which passes through the point ( 1 , 2 , 3 ) and is parallel to the vector 3 i ^ + 2 j ^ − 2 k ^ .

Ans

$\begin{array}{l}\mathrm{The}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{point}\mathrm{A}\left(1,2,3\right)\mathrm{is}\overrightarrow{\mathrm{a}}\mathrm{=}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\mathrm{and}\mathrm{it}\mathrm{is}\\ \mathrm{parallelt}\mathrm{to}\mathrm{vector}\overrightarrow{\mathrm{b}}=3\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}–2\widehat{\mathrm{k}}\mathrm{.}\\ \mathrm{Since},\mathrm{the}\mathrm{line}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{A}\left(1,2,3\right)\mathrm{and}\mathrm{parallel}\mathrm{to}\\ \mathrm{vector}\overrightarrow{\mathrm{b}}\mathrm{is}\mathrm{given}\mathrm{by}\\ \overrightarrow{\mathrm{r}}\mathrm{=}\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{b}},\mathrm{where}\mathrm{\lambda }\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{.}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\overrightarrow{\mathrm{r}}\mathrm{=}\left(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\right)+\mathrm{\lambda }\left(\mathrm{3}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}–2\widehat{\mathrm{k}}\right)\\ \hspace{0.17em}\hspace{0.17em}=\left(1+3\mathrm{\lambda }\right)\widehat{\mathrm{i}}\mathrm{+}\left(2+2\mathrm{\lambda }\right)\widehat{\mathrm{j}}\mathrm{+}\left(3-2\mathrm{\lambda }\right)\widehat{\mathrm{k}}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{.}\end{array}$

Q.5 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 i ^ − j ^ + 4 k ^ and is in the direction i ^ + 2 j ^ − k ^ .

Ans

\begin{array}{l}\text{Let}\overrightarrow{\text{a}}\text{=2}\widehat{\text{i}}-\widehat{\text{j}}\text{+4}\widehat{\text{k}}\\ \overrightarrow{\text{b}}\text{=}\widehat{\text{i}}\text{+2}\widehat{\text{j}}\text{–}\widehat{\text{k}}\\ \text{Since, the line passes through the point with position vector}\overrightarrow{\text{a}}\\ \text{and parallel to vector}\overrightarrow{\text{b}}\text{is given by}\\ \overrightarrow{\text{r}}\text{=}\overrightarrow{\text{a}}\text{+λ}\overrightarrow{\text{b}}\text{, where λ is a constant}\text{.}\\ \Rightarrow \overrightarrow{\text{r}}\text{=}\left(\text{2}\widehat{\text{i}}-\widehat{\text{j}}\text{+4}\widehat{\text{k}}\right)\text{+λ}\left(\widehat{\text{i}}\text{+2}\widehat{\text{j}}-\widehat{\text{k}}\right)\mathrm{\dots }\left(\text{i}\right)\\ \text{This is the required equation of the line in vector form}\text{.}\\ \text{Putting}\overrightarrow{\text{r}}\text{=x}\widehat{\text{i}}\text{+y}\widehat{\text{j}}\text{+z}\widehat{\text{k}}\text{in equation}\left(\text{i}\right)\text{, we get}\\ \text{x}\widehat{\text{i}}\text{+y}\widehat{\text{j}}\text{+z}\widehat{\text{k}}\text{=}\left(\text{2}\widehat{\text{i}}-\widehat{\text{j}}\text{+4}\widehat{\text{k}}\right)\text{+λ}\left(\widehat{\text{i}}\text{+2}\widehat{\text{j}}-\widehat{\text{k}}\right)\\ \text{=}\left(\text{2+λ}\right)\widehat{\text{i}}-\left(\text{1}-\text{2λ}\right)\widehat{\text{j}}\text{+}\left(\text{4}-\text{λ}\right)\widehat{\text{k}}\\ \Rightarrow \text{x=2+λ,}\text{y=2λ-1 and z=4-λ}\\ \Rightarrow \text{λ=x-2,}\text{​}\text{λ=}\frac{\text{y+1}}{\text{2}}\text{and λ=4-z}\\ \Rightarrow \frac{\text{x}-\text{2}}{\text{1}}\text{=}\frac{\text{y+1}}{\text{2}}\text{=}\frac{\text{z}-\text{4}}{-\text{1}}\\ \text{This is the required equation of the given line in cartesian form}\text{.}\end{array}

Q.6 Find the cartesian equation of the line which passes through the point −2, 4, −5 and parallel to the line given by x+3 3 = y−4 5 = z+8 6 .

Ans

\begin{array}{l}\text{The direction ratios of}\frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y}-\text{4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}\text{are 3,5 and 6}\text{.}\\ \text{Let the direction ratios of required line parallel to}\\ \frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y}-\text{4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}\text{are 3k, 5k and 6k}\text{.}\\ \text{Since, equation of the line through the point}\left({\text{x}}_{\text{1}}\text{,}{\text{y}}_{\text{1}}\text{,}{\text{z}}_{\text{1}}\right)\text{and with}\\ \text{the direction ratios a, b and c is}\\ \frac{\text{x}-{\text{x}}_{\text{1}}}{\text{a}}\text{=}\frac{\text{y}-{\text{y}}_{\text{1}}}{\text{b}}\text{=}\frac{\text{z}-{\text{z}}_{\text{1}}}{\text{c}}\\ \text{Therefore, the equation of required line}\text{through the point}\\ \left(-\text{2,}\text{4,}-\text{5}\right)\text{is}\\ \frac{\text{x+2}}{\text{3k}}\text{=}\frac{\text{y}-\text{4}}{\text{5k}}\text{=}\frac{\text{z+5}}{\text{6k}}\\ \text{or}\frac{\text{x+2}}{\text{3}}\text{=}\frac{\text{y}-\text{4}}{\text{5}}\text{=}\frac{\text{z+5}}{\text{6}}\\ \text{which is the required equation of line}\text{.}\end{array}

Q.7

$\begin{array}{l}\mathrm{The}\mathrm{cartesian}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{is}\frac{\mathrm{x}-5}{3}=\frac{\mathrm{y}+4}{7}=\frac{\mathrm{z}-6}{2}.\\ \mathrm{Write}\mathrm{its}\mathrm{vector}\mathrm{form}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{cartesian}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}\\ \frac{\mathrm{x}-\mathrm{5}}{\mathrm{3}}\mathrm{=}\frac{\mathrm{y}+4}{\mathrm{7}}\mathrm{=}\frac{\mathrm{z}-\mathrm{6}}{\mathrm{2}}...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{given}\mathrm{lines}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(5,-\mathrm{\hspace{0.17em}}4,6\right).\mathrm{The}\mathrm{position}\\ \mathrm{vector}\mathrm{of}\mathrm{this}\mathrm{point}\mathrm{is}\overrightarrow{\mathrm{a}}=5\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\mathrm{4}\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}\mathrm{.}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{are}3,\; 7\mathrm{and}2\mathrm{.}\\ \mathrm{The}\mathrm{direction}\mathrm{vector}\mathrm{of}\mathrm{given}\mathrm{line}\mathrm{is}\overrightarrow{\mathrm{b}}=3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\mathrm{.}\\ \mathrm{So},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{through}\mathrm{the}\mathrm{point}\overrightarrow{\mathrm{a}}\mathrm{and}\mathrm{in}\mathrm{the}\mathrm{direction}\\ \mathrm{of}\mathrm{the}\mathrm{vector}\overrightarrow{\mathrm{b}}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{equation},\\ \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{b}},\mathrm{where}\mathrm{\lambda }\mathrm{is}\mathrm{a}\mathrm{contant}\mathrm{.}\\ \mathrm{So},\overrightarrow{\mathrm{r}}=\left(\mathrm{5\hspace{0.17em}}\widehat{\mathrm{i}}–4\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}\right)+\mathrm{\lambda }\left(\mathrm{3\hspace{0.17em}}\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{in}\mathrm{the}\mathrm{vector}\mathrm{form}\mathrm{.}\end{array}$

Q.8 Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2 , 3).

Ans

\begin{array}{l}\text{The required line passes through origin}\text{.Therefore its position}\\ \text{vector is}\overrightarrow{\text{a}}\text{=}\text{0}\text{.}\\ \text{The direction ratios of line through origin and the point}\left(\text{5,}-\text{2,3}\right)\\ \text{are:}\left(\text{5-0}\right)\text{,}\left(-\text{2}-\text{0}\right)\text{,}\left(\text{3}-\text{0}\right)\text{i}\text{.e}\text{., 5,-2,3}\\ \text{The line is parallel to the position vector}\overrightarrow{\text{b}}\text{=5}\widehat{\text{i}}\text{-2}\widehat{\text{j}}\text{+3}\widehat{\text{k}}\text{.}\\ \text{So, the equation of line passes through origin and parallel to}\overrightarrow{\text{b}}\text{is}\\ \overrightarrow{\text{r}}\text{=}\overrightarrow{\text{a}}\text{+λ}\overrightarrow{\text{b}}\text{,}\text{λ}\in \text{R}\\ \Rightarrow \overrightarrow{\text{r}}\text{=0+λ}\left(\text{5}\widehat{\text{i}}-\text{2}\widehat{\text{j}}\text{+3}\widehat{\text{k}}\right)\\ \Rightarrow \overrightarrow{\text{r}}\text{=λ}\left(\text{5}\widehat{\text{i}}-\text{2}\widehat{\text{j}}\text{+3}\widehat{\text{k}}\right)\\ \text{The equation of line through the point}\left({\text{x}}_{\text{1}}{\text{,y}}_{\text{1}}{\text{,z}}_{\text{1}}\right)\text{and direction}\\ \text{ratios a,b,c is given by}\frac{\text{x}-{\text{x}}_{\text{1}}}{\text{a}}\text{=}\frac{\text{y}-{\text{y}}_{\text{1}}}{\text{b}}\text{=}\frac{\text{z}-{\text{z}}_{\text{1}}}{\text{c}}\\ \text{Therefore, equation of the required line in the cartesian form is}\\ \frac{\text{x-0}}{\text{5}}\text{=}\frac{\text{y-0}}{-\text{2}}\text{=}\frac{\text{z-0}}{\text{3}}\\ \Rightarrow \frac{\text{x}}{\text{5}}\text{=}\frac{\text{y}}{-\text{2}}\text{=}\frac{\text{z}}{\text{3}}\end{array}

Q.9 Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{given}\mathrm{points}\mathrm{are}\mathrm{A}(3,–2,-\mathrm{5})\mathrm{and}\mathrm{B}(3,-2,\; 6)\mathrm{through}\\ \mathrm{which}\mathrm{line}\mathrm{AB}\mathrm{passes}\mathrm{.}\\ \mathrm{So},\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{point}(3,–2,-\mathrm{5})\left(\overrightarrow{\mathrm{a}}\right)=3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-5\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{AB}\mathrm{are}\mathrm{given}\mathrm{as}\\ (3-3),(-2+2),(6+5)\mathrm{i}.\mathrm{e}.,\; 0,0,11\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{vector}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{AB}\mathrm{is}\\ \overrightarrow{\mathrm{b}}=0\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+0\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+11\widehat{\mathrm{k}}=11\widehat{\mathrm{k}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \overrightarrow{\mathrm{r}}=(3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}-5\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}})+\mathrm{\lambda }\left(11\widehat{\mathrm{k}}\right)\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-3}{0}=\frac{\mathrm{y}+2}{0}=\frac{\mathrm{z}+5}{11}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{following}\mathrm{pairs}\mathrm{of}\mathrm{lines}:\\ \left(\mathrm{i}\right)\overrightarrow{\mathrm{r}}=2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-5\widehat{\mathrm{j}}+\widehat{\mathrm{k}}+\mathrm{\lambda }\left(3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)\mathrm{and}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\overrightarrow{\mathrm{r}}=7\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-6\widehat{\mathrm{k}}+\mathrm{\mu }\left(\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\right)\\ \left(\mathrm{ii}\right)\overrightarrow{\mathrm{r}}=3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+\widehat{\mathrm{j}}-2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}+\mathrm{\lambda }\left(\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)\mathrm{and}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\overrightarrow{\mathrm{r}}=2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-56\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}+\mathrm{\mu }\left(3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-5\widehat{\mathrm{j}}-4\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{}\mathrm{Here},\mathrm{}\overrightarrow{{\mathrm{b}}_1}=3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\mathrm{}\mathrm{and}\mathrm{}\overrightarrow{{\mathrm{b}}_2}=\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\\ \mathrm{Then}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{\theta }\mathrm{}\mathrm{between}\mathrm{}\mathrm{the}\mathrm{}\mathrm{two}\mathrm{}\mathrm{lines}\mathrm{}\mathrm{is}\mathrm{}\mathrm{given}\mathrm{}\mathrm{by}\\ \mathrm{cos\theta }=\left|\frac{\overrightarrow{{\mathrm{b}}_1}.\overrightarrow{{\mathrm{b}}_2}}{\left|\overrightarrow{{\mathrm{b}}_1}\right|\left|\overrightarrow{{\mathrm{b}}_1}\right|}\right|\\ =\left|\frac{\left(3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right).\left(\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\right)}{\left|3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right|\left|\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\right|}\right|\\ =\left|\frac{3+4+12}{\sqrt{3^2+2^2+6^2}\sqrt{1^2+2^2+2^2}}\right|\\ =\left|\frac{19}{\sqrt{9+4+36}\sqrt{1+4+4}}\right|\\ =\frac{19}{7\times 3}\\ =\frac{19}{21}\\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{\theta }={\cos }^{–1}\left(\frac{19}{21}\right)\\ \left(\mathrm{ii}\right)\mathrm{}\mathrm{Here},\mathrm{}\overrightarrow{{\mathrm{b}}_1}=\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\mathrm{}\mathrm{and}\mathrm{}\overrightarrow{{\mathrm{b}}_2}=3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-5\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}-4\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \mathrm{Then}\mathrm{angle}\mathrm{\theta }\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{cos\theta }=\left|\frac{\overrightarrow{{\mathrm{b}}_1}.\overrightarrow{{\mathrm{b}}_2}}{\left|\overrightarrow{{\mathrm{b}}_1}\right|\left|\overrightarrow{{\mathrm{b}}_1}\right|}\right|\\ =\left|\frac{\left(\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right).\left(3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-5\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}-4\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)}{\left|\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right|\left|3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-5\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}-4\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right|}\right|\\ =\left|\frac{3+5+8}{\sqrt{1^2+{\left(-1\right)}^2+{\left(-2\right)}^2}\sqrt{3^2+{\left(-5\right)}^2+{\left(-4\right)}^2}}\right|\\ =\left|\frac{16}{\sqrt{\mathrm{\hspace{0.17em}}1+1+4}\sqrt{9+25+16}}\right|\\ =\frac{16}{\sqrt{6}\times \sqrt{50}}\\ =\frac{16}{10\sqrt{3}}\\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{\theta }={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)\end{array}$

Q.11 Find the angle between the following pair of lines : i  x−22=y−15=z+3−3 and x+2−1=y−48=z−54ii  x2=y2=z1 and x−54=y−21=z−38

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-1}{5}=\frac{\mathrm{z}+3}{-3}\mathrm{and}\frac{\mathrm{x}+2}{-1}=\frac{\mathrm{y}-4}{8}=\frac{\mathrm{z}-5}{4}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}2,5,-3\mathrm{and}\mathrm{the} \\ direction\; ratios \mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}-1,8,4.\mathrm{If}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\\ \mathrm{them},\mathrm{then}\\ \mathrm{cos\theta }=\left|\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{{\mathrm{a}}_1}^2+{{\mathrm{b}}_1}^2+{{\mathrm{c}}_1}^2}.\sqrt{{{\mathrm{a}}_2}^2+{{\mathrm{b}}_2}^2+{{\mathrm{c}}_2}^2}}\right|\\ \mathrm{\hspace{0.17em}}=\left|\frac{2\times -1+5\times 8+(-3)\times 4}{\sqrt{{\left(2\right)}^2+{\left(5\right)}^2+{(-3)}^2}.\sqrt{{(-1)}^2+8^2+4^2}}\right|\\ \mathrm{\hspace{0.17em}}=\left|\frac{-2+40-12}{\sqrt{4+25+9}.\sqrt{1+64+16}}\right|\\ \mathrm{\hspace{0.17em}}=\left|\frac{-2+40-12}{\sqrt{38}.\sqrt{81}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{\theta }={\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}\mathrm{and}\frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-3}{8}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}2,\mathrm{\hspace{0.17em}}2,\mathrm{\hspace{0.17em}}1\mathrm{and}\mathrm{the} \\ direction\; ratios \mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}4,\hspace{0.17em}1,\hspace{0.17em}8.\mathrm{If}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\\ \mathrm{them},\mathrm{then}\\ \mathrm{cos\theta }=\left|\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{{\mathrm{a}}_1}^2+{{\mathrm{b}}_1}^2+{{\mathrm{c}}_1}^2}.\sqrt{{{\mathrm{a}}_2}^2+{{\mathrm{b}}_2}^2+{{\mathrm{c}}_2}^2}}\right|\\ \mathrm{\hspace{0.17em}}=\left|\frac{2\times 4+2\times 1+1\times 8}{\sqrt{{\left(2\right)}^2+{\left(2\right)}^2+{\left(1\right)}^2}.\sqrt{{\left(4\right)}^2+1^2+8^2}}\right|\\ \mathrm{\hspace{0.17em}}=\left|\frac{8+2+8}{\sqrt{4+4+1}.\sqrt{16+1+64}}\right|\\ \mathrm{\hspace{0.17em}}=\left|\frac{18}{\sqrt{9}.\sqrt{81}}\right|=\left|\frac{18}{3\times 9}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{\theta }={\cos }^{-1}\left(\frac{2}{3}\right)\end{array}$

Q.12 Find the values of p so that the lines 1−x3=7y−142p=z−32and  7−7x3p=y−51=6−z5 are at right angles.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{eqution}\mathrm{of}\mathrm{given}\mathrm{lines}\mathrm{are}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1-\mathrm{x}}{3}=\frac{7\mathrm{y}-14}{2\mathrm{p}}=\frac{\mathrm{z}-3}{2}\mathrm{and}\frac{7-7\mathrm{x}}{3\mathrm{p}}=\frac{\mathrm{y}-5}{1}=\frac{6-\mathrm{z}}{5}\\ \Rightarrow \frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-2}{\left(\frac{2\mathrm{p}}{7}\right)}=\frac{\mathrm{z}-3}{2}\mathrm{and}\frac{\mathrm{x}-1}{(-\frac{3\mathrm{p}}{7})}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-6}{-5}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}-3,\mathrm{\hspace{0.17em}}\frac{2\mathrm{p}}{7},\hspace{0.17em}2\mathrm{and}\mathrm{the} \\ direction\; ratios \mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}-\frac{3\mathrm{p}}{7},\hspace{0.17em}1,\mathrm{\hspace{0.17em}}-\mathrm{5}.\mathrm{Since},\mathrm{both}\mathrm{lines}\mathrm{are}\\ \mathrm{perpendicular},\mathrm{so}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=0\\ \Rightarrow (-3).(-\frac{3\mathrm{p}}{7})+\frac{2\mathrm{p}}{7}.1+2.(-5)=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{9\mathrm{p}}{7}+\frac{2\mathrm{p}}{7}-10=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}11\mathrm{p}-70=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{p}=\frac{70}{11}\\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{p}\mathrm{is}\frac{70}{11}.\end{array}$

Q.13 Show that the lines x−57=x+2−5=z1 and x1=y2=z3 are perpendicular to each other.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{given}\mathrm{lines}\mathrm{are}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{x}-5}{7}=\frac{\mathrm{x}+2}{-5}=\frac{\mathrm{z}}{1}\mathrm{and}\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}7,\hspace{0.17em}-5,\hspace{0.17em}1\mathrm{and}\mathrm{the}\mathrm{direction}\\ \mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}1,\hspace{0.17em}2,\mathrm{\hspace{0.17em}}3.\mathrm{}\mathrm{So},\\ {\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=7.1+\left(-5\right).2+1.3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=7-10+3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\\ \mathrm{Thus},\mathrm{both}\mathrm{lines}\mathrm{are}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}\mathrm{.}\end{array}$

Q.14 Find the shortest distance between the lines r→=i^+2j^+k^+λi^−j^+k^ and r→=2 i^−j^+k^+μ2 i^+j^+2 k^

Ans

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{equations}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{given}\mathrm{}\mathrm{lines}\mathrm{}\mathrm{are}:\\ \mathrm{\hspace{0.17em}}\overrightarrow{\mathrm{r}}=\left(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right)+\mathrm{\lambda }\left(\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{\hspace{0.17em}}\overrightarrow{\mathrm{r}}=\left(2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\right)+\mathrm{\mu }\left(2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)...\left(\mathrm{ii}\right)\\ \mathrm{Comparing}\mathrm{equatin}\mathrm{}\left(\mathrm{i}\right)\mathrm{with}\overrightarrow{\mathrm{r}}=\overrightarrow{{\mathrm{a}}_1}+\mathrm{\lambda }\overrightarrow{{\mathrm{b}}_1}\mathrm{and}\mathrm{equation}\mathrm{}\left(\mathrm{ii}\right)\\ \mathrm{with}\hspace{0.17em}\hspace{0.17em}\overrightarrow{\mathrm{r}}=\overrightarrow{{\mathrm{a}}_2}+\mathrm{\mu }\overrightarrow{{\mathrm{b}}_2},\mathrm{we}\mathrm{get}\\ \overrightarrow{{\mathrm{a}}_1}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\overrightarrow{{\mathrm{b}}_1}=\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\overrightarrow{{\mathrm{a}}_2}=2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-\widehat{\mathrm{k}},\overrightarrow{{\mathrm{b}}_2}=2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \mathrm{The}\mathrm{shortest}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{lines}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{d}=\left|\frac{\left(\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right).\left(\overrightarrow{{\mathrm{a}}_2}-\overrightarrow{{\mathrm{a}}_1}\right)}{\left|\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right|}\right|...\left(\mathrm{iii}\right)\\ \mathrm{So},\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|\\ \mathrm{\hspace{0.17em}}=\left(-2-1\right)\widehat{\mathrm{i}}-\left(2-2\right)\widehat{\mathrm{j}}+\left(1+2\right)\widehat{\mathrm{k}}\\ \mathrm{\hspace{0.17em}}=-3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-0.\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right|=\left|-3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-0.\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right|\\ \mathrm{\hspace{0.17em}}=\sqrt{{\left(-3\right)}^2+3^2}\\ \mathrm{\hspace{0.17em}}=\sqrt{9+9}\\ \mathrm{\hspace{0.17em}}=3\sqrt{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\overrightarrow{{\mathrm{a}}_2}-\overrightarrow{{\mathrm{a}}_1}=\left(2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\right)-\left(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right)\\ =\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{d}=\left|\frac{\left(-3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right).\left(\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}\right)}{3\sqrt{2}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\frac{-3-6}{3\sqrt{2}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{3}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{3\sqrt{2}}{2}\\ \mathrm{Thus},\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{two}\mathrm{lines}\mathrm{is}\frac{3\sqrt{2}}{2}\mathrm{units}\mathrm{.}\end{array}$

Q.15 Find the shortest distance between the linesx+17=y+1− 6=z−11 and x−31=y−5−2=z−71

Ans

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{equations}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{given}\mathrm{\hspace{0.17em}}\mathrm{lines}\mathrm{}\mathrm{are}:\\ \frac{\mathrm{X}+1}{7}=\frac{\mathrm{Y}+1}{-6}=\frac{\mathrm{Z}-1}{1}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}\frac{\mathrm{X}-3}{1}=\frac{\mathrm{Y}-5}{-2}=\frac{\mathrm{Z}-7}{1}\\ \mathrm{Comparing}\mathrm{}\mathrm{both}\hspace{0.17em}\mathrm{the}\hspace{0.17em}\mathrm{equations}\hspace{0.17em}\mathrm{with}\\ \frac{\mathrm{X}-{\mathrm{X}}_1}{{\mathrm{a}}_1}=\frac{\mathrm{Y}-{\mathrm{Y}}_1}{{\mathrm{b}}_1}=\frac{\mathrm{Z}-{\mathrm{Z}}_1}{{\mathrm{C}}_1}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\frac{\mathrm{X}-{\mathrm{X}}_2}{{\mathrm{a}}_2}=\frac{\mathrm{Y}-5}{{\mathrm{b}}_2}=\frac{\mathrm{Z}-7}{{\mathrm{C}}_2}\hspace{0.17em}\mathrm{respectively}.\\ \\ {\mathrm{X}}_1=-1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{X}}_2=3\\ {\mathrm{Y}}_1=-1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{Y}}_2=5\\ {\mathrm{Z}}_1=1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{Z}}_2=7\\ {\mathrm{a}}_1=7,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{a}}_2=1\\ {\mathrm{b}}_1=-6,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{b}}_2=-2\\ {\mathrm{c}}_1=1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{c}}_2=1\\ \mathrm{Distance}\hspace{0.17em}\mathrm{between}\hspace{0.17em}\mathrm{two}\hspace{0.17em}\mathrm{lines}\hspace{0.17em}\mathrm{is}​\hspace{0.17em}\mathrm{given}\hspace{0.17em}\mathrm{by}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{d}=\left|\frac{\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\end{array}\right|}{\sqrt{{\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)}^2+{\left({\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{c}}_2{\mathrm{a}}_1\right)}^2+{\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)}^2}}\right|\\ =\left|\frac{\left|\begin{array}{ccc}3-\left(-1\right)& 5-\left(-1\right)& 7-\left(-1\right)\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|}{\sqrt{{\left\{-6\times 1-\left(-2\right).1\right\}}^2+{\left(1\times 1-1\times 7\right)}^2+{\left(7\times -2-1\times -6\right)}^2}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{d}=\left|\frac{\left|\begin{array}{ccc}4& 6& 8\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|}{\sqrt{{\left(-4\right)}^2+{\left(-6\right)}^2+{\left(-8\right)}^2}}\right|\\ =\left|\frac{4\left(-6+2\right)-6\left(7-1\right)+8\left(-14+6\right)}{\sqrt{16+36+64}}\right|\\ =\left|\frac{-16-36-64}{\sqrt{116}}\right|\\ =\left|\frac{-116}{2\sqrt{29}}\right|\\ =\left|-2\sqrt{29}\right|\\ \mathrm{So},\mathrm{}\mathrm{the}\mathrm{}\mathrm{distance}\mathrm{}\mathrm{between}\mathrm{}\mathrm{two}\mathrm{}\mathrm{given}\mathrm{}\mathrm{lines}\mathrm{}\mathrm{is}2\sqrt{29}\mathrm{}\mathrm{units}.\end{array}$

Q.16 Find the shortest distance between the lines whose vector equations are r→=i^+2j^+3k^+λi^− j^+2 k^ and r→=4 i^+ j^+6 k^+μ2 i^+3 j^+ k^

Ans

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{equations}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{given}\mathrm{}\mathrm{lines}\mathrm{}\mathrm{are}:\\ \overrightarrow{\mathrm{r}}=\left(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\right)+\mathrm{\lambda }\left(\widehat{\mathrm{i}}-3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)...\left(\mathrm{i}\right)\\ \overrightarrow{\mathrm{r}}=\left(4\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+5\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)+\mathrm{\mu }\left(2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{Comparing}\mathrm{}\mathrm{equatin}\mathrm{}\left(\mathrm{i}\right)\mathrm{}\mathrm{with}\mathrm{}\overrightarrow{\mathrm{r}}=\overrightarrow{{\mathrm{a}}_1}+\mathrm{\lambda }\overrightarrow{{\mathrm{b}}_1}\mathrm{}\mathrm{and}\mathrm{}\mathrm{equation}\mathrm{}\left(\mathrm{ii}\right)\\ \mathrm{with}\hspace{0.17em}\mathrm{\hspace{0.17em}}\overrightarrow{\mathrm{r}}=\overrightarrow{{\mathrm{a}}_2}+\mathrm{\mu }\overrightarrow{{\mathrm{b}}_2},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\overrightarrow{{\mathrm{a}}_1}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}},\overrightarrow{{\mathrm{b}}_1}=\widehat{\mathrm{i}}-3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+2\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\overrightarrow{{\mathrm{a}}_2}=4\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+5\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}},\overrightarrow{{\mathrm{b}}_2}=2\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \mathrm{The}\mathrm{}\mathrm{shortest}\mathrm{}\mathrm{distance}\mathrm{}\mathrm{between}\mathrm{}\mathrm{the}\mathrm{}\mathrm{lines}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{d}=\left|\frac{\left(\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right).\left(\overrightarrow{{\mathrm{a}}_2}-\overrightarrow{{\mathrm{a}}_1}\right)}{\left|\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right|}\right|...\left(\mathrm{iii}\right)\\ \mathrm{So},\mathrm{}\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 1& -3& 2\\ 2& 3& 1\end{array}\right|\\ \mathrm{\hspace{0.17em}}=\left(-3-6\right)\widehat{\mathrm{i}}-\left(1-4\right)\widehat{\mathrm{j}}+\left(3+6\right)\widehat{\mathrm{k}}\\ \mathrm{\hspace{0.17em}}=-9\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+9\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\right|=\left|-9\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+9\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right|\\ \mathrm{\hspace{0.17em}}=\sqrt{{\left(-9\right)}^2+3^2+9^2}\\ \mathrm{\hspace{0.17em}}=\sqrt{81+9+81}\\ \mathrm{\hspace{0.17em}}=\sqrt{171}=3\sqrt{19}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\overrightarrow{{\mathrm{a}}_2}-\overrightarrow{{\mathrm{a}}_1}=\left(4\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+5\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+6\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)-\left(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)\\ =3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\\ \mathrm{From}\mathrm{}\mathrm{equation}\mathrm{}\left(\mathrm{i}\right),\mathrm{}\mathrm{we}\mathrm{}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{d}=\left|\frac{\left(-9\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+9\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right).\left(3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)}{3\sqrt{19}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\frac{3\left(-3\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+\mathrm{\hspace{0.17em}}\widehat{\mathrm{j}}+3\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right).\left(\mathrm{\hspace{0.17em}}\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\mathrm{\hspace{0.17em}}\widehat{\mathrm{k}}\right)}{\sqrt{19}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\frac{3\left(-3+1+3\right)}{\sqrt{19}}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{3}{\sqrt{19}}\times \frac{\sqrt{19}}{\sqrt{19}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{3\sqrt{19}}{19}\\ \mathrm{So},\mathrm{}\mathrm{the}\mathrm{}\mathrm{distance}\mathrm{}\mathrm{between}\mathrm{}\mathrm{two}\mathrm{}\mathrm{given}\mathrm{}\mathrm{lines}\mathrm{}\mathrm{is}\mathrm{}\frac{3\sqrt{19}}{19}\mathrm{}\mathrm{units}.\end{array}$