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

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-11 (Three Dimensional Geometry)Exercise 11.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 11.2

Q1. Show that the three lines with direction cosines $\frac{12}{13},\frac{-3}{13},\frac{-4}{13};\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{3}{13},\frac{-4}{13},\frac{12}{13}$ are mutually perpendicular.

Answer. $\begin{array}{c}\text{Two lines with direction cosines,}{l}_1,m_1,n_1\text{and}{l}_2,m_2,n_2,\text{are perpendicular to each}\\ \text{other, if}{I}_1{I}_2+m_1{m}_2+n_1{n}_2=0\end{array}$ (i) 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 obtain $\begin{array}{cc}{l}_1{l}_2+m_1{m}_2+n_1{n}_2& =\frac{12}{13}\times \frac{4}{13}+\left(\frac{-3}{13}\right)\times \frac{12}{13}+\left(\frac{-4}{13}\right)\times \frac{3}{13}\\ & =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\\ & =0\end{array}$ Therefore, the lines are perpendicular. (ii) 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 obtain $\begin{array}{cc}{l}_1{l}_2+m_1{m}_2+n_1{n}_2& =\frac{4}{13}\times \frac{3}{13}+\frac{12}{13}\times \left(\frac{-4}{13}\right)+\frac{3}{13}\times \frac{12}{13}\\ & =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\\ & =0\end{array}$ Therefore, the lines are perpendicular. Thus, all the lines are mutually perpendicular.

Q2. 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).

Answer. $\begin{array}{c}\text{Let}AB\text{be the line joining the points,}(1,-1,2)\text{and}(3,4,-2),\text{and}CD\text{be the line}\\ \text{joining the points,}(0,3,2)\text{and}(3,5,6)\text{.}\\ \text{The direction ratios,}{a}_1,b_1,c_1,\text{of}AB\text{are}(3-1),(4-(-1\left)\right),\text{and}(-2-2)\text{i.e.t}2,5,\text{and}\\ -4\text{.}\end{array}$ $\begin{array}{c}\text{The direction ratios,}{a}_2,b_2,c_2\text{of co are}(3-0),(5-3),\text{and}(6-2)\text{i.e.,}3,2,\text{and}4.\\ \text{AB and}CD\text{will be perpendicular to each other, if}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 3+5\times 2+(-4)\times 4\\ =6+10-16\\ =0\end{array}$ Therefore, AB and CD are perpendicular to each other.

Q3. 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).

Answer. Let AB be the line through the points (4, 7, 8) and (2, 3, 4) adn CD be the through the points (-1, -2, 1) and (1, 2, 5). $\begin{array}{c}\text{The directions ratios,}{a}_1,b_1,c_1,\text{of}AB\text{are}(2-4),(3-7),\text{and}(4-8)\text{i.e.,}-2,-4,\text{and}\\ -4\text{.}\end{array}$ $\begin{array}{c}\text{The direction ratios,}{a}_2,b_2,c_2,\text{of}CD\text{are}(1-(-1\left)\right),(2-(-2\left)\right),\text{and}(5-1)\text{l.e.,}2,4,\\ \text{and}4.\end{array}$ AB will be parallel to CD, if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ $\begin{array}{c}\frac{a_1}{a_2}=\frac{-2}{2}=-1\\ \frac{b_1}{b_2}=\frac{-4}{4}=-1\\ \frac{c_1}{c_2}=\frac{-4}{4}=-1\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$ Thus, AB is parallel to CD.

Q4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector $3\hat{i}+2\hat{j}-2\hat{k}$.

Answer. It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$ $\overrightarrow{b}=3\hat{i}+2\hat{j}-2\hat{k}$ It is known the the line which passes through point A and parallel to $0\overrightarrow{b}$ is given by $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\text{where}\lambda$ is a constant. $\Rightarrow \overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (3\hat{i}+2\hat{j}-2\hat{k})$ This is the required equation of the line.

Q5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{i}-j+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.

Answer. $\begin{array}{c}\text{It is given that the line passes through the point with position vector}\\ \overline{a}=2\hat{i}+\hat{j}+4\hat{k}\\ \overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}\\ \text{It is known that a line through a point with position vector}\overrightarrow{a}\text{and parallel to}\overline{b}\text{is given by}\end{array}$ $\begin{array}{c}\text{the equation,}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda (\hat{i}+2\hat{j}-\hat{k})\\ \text{This is the required equation of the line in vector form.}\\ \overrightarrow{r}=x\hat{i}-y\hat{j}+z\hat{k}\\ \Rightarrow x\hat{i}-y\hat{j}+z\dot{k}=(\lambda +2)\hat{i}+(2\lambda -1)\hat{j}+(-\lambda +4)\hat{k}\end{array}$ $\begin{array}{c}\text{Eliminating}{\lambda }_r\text{we obtain the Cartesian form equation as}\\ \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\\ \text{This is the required equation of the given line in Cartesian form.}\end{array}$

Q6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$.

Answer. $\begin{array}{c}\text{It is given that the line passes through the point}(-2,4,-5)\text{and is parallel to}\\ \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\end{array}$ $\begin{array}{c}\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\\ \text{The direction ratios of the line,}\frac{y-4}{3}=\frac{y-4}{5}=\frac{z+8}{6},\text{are}3,5,\text{and}6.\\ \text{The required line is parallel to}\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\end{array}$ $\begin{array}{c}\text{Therefore, its direction ratios are}3k,5k,\text{and}6k,\text{where}k\ne 0\\ \text{It is known that the equation of the line through the point}(x_1,y_1,z_1)\text{and with direction}\\ \text{ratios,}a,b,c,\text{is given by}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \text{Therefore the equation of the required line is}\end{array}$ $\begin{array}{c}\frac{x+2}{3k}=\frac{y-4}{5k}=\frac{z+5}{6k}\\ \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k\end{array}$

Q7. The cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$. Write its vector form.

Answer. $\begin{array}{c}\text{The Cartesian equation of the line is}\\ \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\\ \text{The given line passes through the point}(5,-4,6)\text{. The position vector of this point is}\\ \overline{a}=5\hat{i}-4\hat{j}+6\hat{k}\end{array}$ Also, the direction ratios of the given line are 3, 7 and 2. This means that the line is in the direction of vector $\overrightarrow{b}=3\hat{i}+7\hat{j}+2\hat{k}$ It is known that the line through position vector $\overrightarrow{b}$ and in the direction of the vector $\overrightarrow{a}$ is given by the equation , $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda \in R$ $\Rightarrow \overrightarrow{r}=(5\hat{i}-4\hat{j}+6\dot{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$ This is the required equation of the given line in vector form.

Q8. Find the vector and the Cartesian equations of the lines that pass through the origin and (5, -2, 3).

Answer. The required line passes through the origin. Therefore, its position vector is given by, $\overrightarrow{a}=\overrightarrow{0}$ The direction ratios of the line through origin and (5, -2, 3) are (5-0)=5,(-2-0)=-2,(3-0)=3 The line is parallel to the vector given by the equation $\overrightarrow{b}=5\hat{i}-2\hat{j}+3\hat{k}$ The equation of the line in vector through a point with position vector and parallel to ${\overrightarrow{b}}_{\text{is},}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda \in R$ $\begin{array}{c}\Rightarrow \overrightarrow{r}=\overrightarrow{0}+\lambda (5\hat{i}-2\hat{j}+3\hat{k})\\ \Rightarrow \overrightarrow{r}=\lambda (5\hat{i}-2\hat{j}+3\hat{k})\end{array}$ The equation of the line through the point $(x_1,y_1,z_1)$ and direction ratios a, b, c is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ Therefore, the equation of the required line in the Cartesian form is $\begin{array}{c}\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\\ \Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\end{array}$.

Q9. Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

Answer. Let the line passing through the points, $P(3,-2,-5)\text{and}Q(3,-2,6),\text{be}PQ$. Since PQ passes through P (3,-2,-5), its position vector is given by, $\overrightarrow{a}=3\hat{i}-2\hat{j}-5\hat{k}$ The direction ratios of PQ are given by, (3-3)=0,(-2+2)=0,(6+5)=11 The equation of the vector in the direction of PQ is $\begin{array}{c}\overrightarrow{b}=0\hat{i}-0,\hat{j}+11\hat{k}=11\hat{k}\\ \text{The equation of}PQ\text{in vector form is given by,}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda \in R\\ \Rightarrow \overrightarrow{r}=(3\hat{i}-2\hat{j}-5\hat{k})+11\lambda \hat{k}\end{array}$ $\begin{array}{c}\text{The equation of}PQ\text{in Cartesian form is}\\ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\end{array}$

Q10. Find the angle between the following pairs of lines: $\begin{array}{cc}\text{(i)}& \overrightarrow{r}=2\hat{i}-5\hat{j}+\hat{k}+\lambda (3\hat{i}+2\hat{j}+6\hat{k})\text{and}\\ & \overrightarrow{r}=7\hat{i}-6\hat{k}+\mu (\hat{i}+2\hat{j}+2\hat{k})\end{array}$ $\begin{array}{cc}\text{(ii)}& \overrightarrow{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda (\hat{i}-\hat{j}-2\hat{k})\text{and}\\ & \overrightarrow{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu (3\hat{i}-5\hat{j}-4\hat{k})\end{array}$

Answer. (i) Let Q be the angle between the given lines. The angle between the given pairs of lines is given by, $\cos Q=\left|\frac{{\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2}{{\overrightarrow{b}}_1||{\overrightarrow{b}}_2|}\right|$ The given lines are parallel to the vectors, ${\overrightarrow{b}}_1=3\hat{i}+2\hat{j}+6\hat{k}\text{and}{\overrightarrow{b}}_2=\hat{i}+2\hat{j}+2\hat{k}$ respectively. $\begin{array}{cc}\therefore \left|{\overrightarrow{b}}_1\right|& =\sqrt{3^2+2^2+6^2}=7\\ \left|{\overrightarrow{b}}_2\right|=& \sqrt{(1{)}^2+(2{)}^2+(2{)}^2}=3\\ {\overline{b}}_1\cdot {\overrightarrow{b}}_2& =(3\hat{i}+2\hat{j}+6\hat{k})\cdot (\hat{i}+2\hat{j}+2\hat{k})\\ & =3\times 1+2\times 2+6\times 2\\ & =3+4+12\\ & =19\end{array}$ $\begin{array}{c}\Rightarrow \cos Q=\frac{19}{7\times 3}\\ \Rightarrow Q={\cos }^{-1}\left(\frac{19}{21}\right)\end{array}$ (ii) The given lines are parallel to the vectors, ${\overrightarrow{b}}_1=\hat{i}-\hat{j}-2\hat{k}\text{and}{\overrightarrow{b}}_2=3\hat{i}-5\hat{j}-4\hat{k}$ respectively. $\begin{array}{cc}\therefore \left|{\overrightarrow{b}}_1\right|& =\sqrt{(1{)}^2+(-1{)}^2+(-2{)}^2}=\sqrt{6}\\ \left|{\overrightarrow{b}}_2\right|=& \sqrt{(3{)}^2+(-5{)}^2+(-4{)}^2}=\sqrt{50}=5\sqrt{2}\\ {\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2& =(\hat{i}-\hat{j}-2\hat{k})\cdot (3\hat{i}-5\hat{j}-4\hat{k})\\ & =1\cdot 3-1(-5)-2(-4)\\ & =3+5+8\\ & =16\end{array}$ $\begin{array}{c}\cos Q=\frac{|\overrightarrow{b}\cdot {\overrightarrow{b}}_2|}{|\overrightarrow{b}|\left|{\overrightarrow{b}}_2\right|}\\ \Rightarrow \cos Q=\frac{16}{\sqrt{6}\cdot 5\sqrt{2}}=\frac{16}{\sqrt{2}\cdot \sqrt{3}\cdot 5\sqrt{2}}=\frac{16}{10\sqrt{3}}\\ \Rightarrow \cos Q=\frac{8}{5\sqrt{3}}\\ \Rightarrow Q={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)\end{array}$

Q11. Find the angle between the following pair of lines: $\begin{array}{c}\text{(i)}\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\text{and}\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\\ \text{(ii)}\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\text{and}\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\end{array}$

Answer. (i) $\begin{array}{c}\text{Let}{\overrightarrow{b}}_1\text{and}{\overrightarrow{b}}_2\text{be the vectors parallel to the pair of lines,}\\ \frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\text{and}\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\text{, respectively.}\end{array}$ $\begin{array}{cc}\therefore {\overrightarrow{b}}_1& =2\hat{i}+5\hat{j}-3\hat{k}\text{and}{\overrightarrow{b}}_2=-\hat{i}+8\hat{j}+4\hat{k}\\ \left|{\overrightarrow{b}}_1\right|=& \sqrt{(2{)}^2+(5{)}^2+(-3{)}^2}=\sqrt{38}\\ \left|{\overrightarrow{b}}_2\right|=& \sqrt{(-1{)}^2+(8{)}^2+(4{)}^2}=\sqrt{81}=9\\ {\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2& =(2\hat{i}+5\hat{j}-3\hat{k})\cdot (-\hat{i}+8\hat{j}+4\hat{k})\\ & =-2+40-12\\ & =26\end{array}$ The angle, Q, between the given pair of lines is given by the relation, $\begin{array}{c}\cos Q=|{\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2|\\ {\overrightarrow{b}}_1\left|{\overrightarrow{b}}_2\right|\\ \Rightarrow \cos Q=\frac{26}{9\sqrt{38}}\\ \Rightarrow Q={\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)\end{array}$ (ii) Let ${\overrightarrow{b}}_1,{\overrightarrow{b}}_2\text{be the vectors parallel to the given pair of lines,}\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ $\frac{x-5}{4}=\frac{y-5}{1}=\frac{z-3}{8},$ respectively. $\begin{array}{cc}{\overrightarrow{b}}_1& =2\hat{i}+2\hat{j}+\hat{k}\\ {\overrightarrow{b}}_2& =4\hat{i}+\hat{j}+8\hat{k}\end{array}$ $\begin{array}{cc}\therefore \left|{\overrightarrow{b}}_1\right|& =\sqrt{(2{)}^2+(2{)}^2+(1{)}^2}=\sqrt{9}=3\\ \left|{\overrightarrow{b}}_2\right|=& \sqrt{4^2+1^2+8^2}=\sqrt{81}=9\\ {\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2& =(2\hat{i}+2\hat{j}+\hat{k})\cdot (4\hat{i}+\hat{j}+8\hat{k})\\ & =2\times 4+2\times 1+1\times 8\\ & =8+2+8\\ & =18\end{array}$ If Q is the angle between the given pair of lines, then $\cos Q=\left|\frac{|\overrightarrow{b}\cdot {\overrightarrow{b}}_2|}{|\overrightarrow{b}||{\overrightarrow{b}}_2|}\right|$ $\begin{array}{c}\Rightarrow \cos Q=\frac{18}{3\times 9}=\frac{2}{3}\\ \Rightarrow Q={\cos }^{-1}\left(\frac{2}{3}\right)\end{array}$

Q12. Find the values of p so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

Answer. The given equations can be written in the standard form as $\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}$ and $\frac{x-1}{-3p}=\frac{y-5}{1}=\frac{z-6}{-5}$ The direction ratios of the lines are $-3,\frac{2p}{7},2\text{and}\frac{-3p}{7},1,-5$ respectively. Two lines with direction ratios, $a_1,b_1,c_1\text{and}{a}_2,b_2,c_2$ are perpendicular to each other, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ $\begin{array}{c}\therefore (-3)\cdot \left(\frac{-3p}{7}\right)+\left(\frac{2p}{7}\right)\cdot \left(1\right)+2\cdot (-5)=0\\ \Rightarrow \frac{9p}{7}+\frac{2p}{7}=10\\ \Rightarrow 11p=70\\ \Rightarrow p=\frac{70}{11}\end{array}$ Thus, the value of p is $\frac{70}{11}$.

Q13. Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\text{and}\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

Answer. The equations of the given lines are $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\text{and}\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ The direction ratios of the given lines are (7,-5,1) and ( 1,2,3 ) respectively. Two lines with direction ratios , $a_1,b_1,c_1\text{and}{a}_2,b_2,c_2$ are perpendicular to each other, if $\begin{array}{c}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ \therefore 7\times 1+(-5)\times 2+1\times 3\\ =7-10+3\\ =0\end{array}$ Therefore, the given lines are perpendicular to each other.

Q14. Find the shortest distance between the lines $\begin{array}{c}\overrightarrow{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})\text{and}\\ \overline{r}=2\hat{i}-\hat{j}-\hat{k}+\mu (2\hat{i}+\hat{j}+2\hat{k})\end{array}$

Answer. The equations of the given lines are $\begin{array}{c}\overrightarrow{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})\text{and}\\ \overline{r}=2\hat{i}-\hat{j}-\hat{k}+\mu (2\hat{i}+\hat{j}+2\hat{k})\end{array}$ It is known that the shortest distance between the lines, $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1\text{and}\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ is given by , $d=\left|\frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_2)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}\right|$ Comparing the given equations, we obtain $\begin{array}{c}{\overrightarrow{a}}_1=\hat{i}+2\hat{j}+\hat{k}\\ {\overrightarrow{b}}_1=\hat{i}-\hat{j}+\hat{k}\\ {\overrightarrow{a}}_2=2\hat{i}-\hat{j}-\hat{k}\\ {\overrightarrow{b}}_2=2\hat{i}+\hat{j}+2\hat{k}\\ {\overrightarrow{a}}_2-{\overrightarrow{a}}_1=(2\hat{i}-\hat{j}-\hat{k})-(\hat{i}+2\hat{j}+\hat{k})=\hat{i}-3\hat{j}-2\hat{k}\end{array}$ ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \dot{k}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|$ $\begin{array}{c}{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=(-2-1)\hat{i}-(2-2)\hat{j}+(1+2)\hat{k}=-3\hat{i}+3\hat{k}\\ \Rightarrow |{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{(-3{)}^2+(3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\end{array}$ Substituting all the values in equation (1), we obtain $\begin{array}{c}d=\left|\frac{(-3\hat{i}+3\hat{k})\cdot (\hat{i}-3\hat{j}-2\hat{k})}{3\sqrt{2}}\right|\\ \Rightarrow d=\left|\frac{-3.1+3(-2)}{3\sqrt{2}}\right|\\ \Rightarrow d=\left|\frac{-9}{3\sqrt{2}}\right|\\ \Rightarrow d=\frac{3}{\sqrt{2}}=\frac{3\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{3\sqrt{2}}{2}\end{array}$ Therefore, the shortest distance between the two lines is $\frac{3\sqrt{2}}{2}$ units.

Q15. Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\text{and}\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Answer. The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\text{and}\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ It is known that the shortest distance between the two lines, $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\text{and}\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2},$ is given by, $d=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{(b{c}_2-b_2{c}_1)}^2+{(c_1{a}_2-c_2{a}_1)}^2+{(a_1{b}_2-a_2{b}_1)}^2}}$ Comparing the given equations, we obtain $\begin{array}{cc}{x}_1=-1,& y_1=-1,z_1=-1\\ a_1& =7,b_1=-6,c_1=1\\ x_2& =3,y_2=5,z_2=7\\ a_2& =1,b_2=-2,c_2=1\end{array}$ Then, $\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=\left|\begin{array}{ccc}4& 6& 8\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|$ $\begin{array}{c}=4(-6+2)-6(7-1)+8(-14+6)\\ =-16-36-64\\ =-116\end{array}$ $\begin{array}{c}\Rightarrow \sqrt{{(b_1{c}_2-b_2{c}_1)}^2+{(c_1{a}_2-c_2{a}_1)}^2+{(a_1{b}_2-a_2{b}_1)}^2}\\ =\sqrt{(-6+2{)}^2+(1+7{)}^2+(-14+6{)}^2}\\ =\sqrt{16+36+64}\\ =\sqrt{116}\\ =2\sqrt{29}\end{array}$ Substituting distance is always non-negative, the distance between the given lines is $2\sqrt{29}$ units.

Q16. Find the shortest distance between the lines whose vector equations are $\begin{array}{c}\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}-3\hat{j}+2\hat{k})\\ \text{and}\overline{r}=4\hat{i}+5\hat{j}+6\hat{k}+\mu (2\hat{i}+3\hat{j}+\hat{k}\end{array}$

Answer. The given lines are $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}-3\hat{j}+2\hat{k}{)}_{\text{and}}\overrightarrow{r}=4\hat{i}+5\hat{j}+6\hat{k}+\mu (2\hat{i}+3\hat{j}+\hat{k})$ It is known that the shortest distance between the lines, $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_{\text{and}}\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2,$ is given by $d=\left|\frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_2)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}\right|$ Comparing the given equations with $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_{1\text{and}}\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$, we obtain $\begin{array}{cc}{\overrightarrow{a}}_1& =\hat{i}+2\hat{j}+3\hat{k}\\ {\overrightarrow{b}}_1& =\hat{i}-3\hat{j}+2\hat{k}\\ {\overrightarrow{a}}_2& =4\hat{i}+5\hat{j}+6\hat{k}\\ {\overrightarrow{b}}_2& =2\hat{i}+3\hat{j}+\hat{k}\end{array}$ ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=(4\hat{i}+5\hat{j}+6\dot{k})-(\hat{i}+2\hat{j}+3\dot{k})=3\hat{i}+3\hat{j}+3\hat{k}$ ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \dot{k}\\ 1& -3& 2\\ 2& 3& 1\end{array}\right|=(-3-6)\hat{i}-(1-4)\hat{j}+(3+6)\hat{k}=-9\hat{i}+3\hat{j}+9\hat{k}$ $\begin{array}{c}\Rightarrow |{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{(-9{)}^2+(3{)}^2+(9{)}^2}=\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}\\ ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(-9\hat{i}+3\hat{j}+9\hat{k})\cdot (3\hat{i}+3\hat{j}+3\hat{k})\end{array}$ $\begin{array}{cc}& =-9\times 3+3\times 3+9\times 3\\ & =9\end{array}$ Substituting all the values in equation (1), we obtain $d=\left|\frac{9}{3\sqrt{19}}\right|=\frac{3}{\sqrt{19}}$ Therefore, the shortest distance between the two given lines is $\frac{3}{\sqrt{19}}$ units.

Q17. Find the shortest distance between the lines whose vector equations are $\begin{array}{cc}\overrightarrow{r}& =(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}\text{and}\\ \overrightarrow{r}& =(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}\end{array}$

Answer. $\begin{array}{c}\text{The given lines are}\\ \overrightarrow{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}\\ \Rightarrow \overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k})\\ \overrightarrow{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}\\ \Rightarrow \overrightarrow{r}=(\hat{i}-\hat{j}+\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k})\end{array}$ It is known that the shortest distance between the lines, $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_{\text{and}}\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$, is given by, $d=\left|\frac{(\overrightarrow{b}1\times \overrightarrow{b}2)\cdot (\overrightarrow{a}2-\overrightarrow{a}2)}{|\overrightarrow{b}1\times \overrightarrow{b}2|}\right|$ For the given equations, $\begin{array}{c}{\overrightarrow{a}}_1=\hat{i}-\hat{j}+3\hat{k}\\ {\overrightarrow{b}}_1=-\hat{i}+\hat{j}-2\hat{k}\\ {\overrightarrow{a}}_2=\hat{i}-\hat{j}-\hat{k}\\ {\overrightarrow{b}}_2=\hat{i}+2\hat{j}-2\hat{k}\\ {\overrightarrow{a}}_2-{\overrightarrow{a}}_1=(\hat{i}-\hat{j}-\hat{k})-(i-2\hat{j}+3\hat{k})=\hat{j}-4\hat{k}\end{array}$ ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\begin{array}{ccc}\hat{i}& \hat{j}& \dot{k}\\ -1& 1& -2\\ 1& 2& -2\end{array}|=(-2+4)\hat{i}-(2+2)\hat{j}+(-2-1)\hat{k}=2\hat{i}-4\hat{j}-3\hat{k}$ $\begin{array}{c}\Rightarrow |{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{(2{)}^2+(-4{)}^2+(-3{)}^2}=\sqrt{4+16+9}=\sqrt{29}\\ \therefore ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(2\hat{i}-4\hat{j}-3\hat{k})\cdot (\hat{j}-4\hat{k})=-4+12=8\end{array}$ Substituting all the values in equation (3), we obtain $d=\left|\frac{8}{\sqrt{29}}\right|=\frac{8}{\sqrt{29}}$ Therefore, the shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units.

Chapter-11 (Three Dimensional Geometry)

• Exercise 11.1
• Exercise 11.3