# 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.  (i) For the lines with direction cosines, , we obtain $\begin{array}{rl}{l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2}& =\frac{12}{13}×\frac{4}{13}+\left(\frac{-3}{13}\right)×\frac{12}{13}+\left(\frac{-4}{13}\right)×\frac{3}{13}\\ & =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\\ & =0\end{array}$ Therefore, the lines are perpendicular. (ii) For the lines with direction cosines, , we obtain $\begin{array}{rl}{l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2}& =\frac{4}{13}×\frac{3}{13}+\frac{12}{13}×\left(\frac{-4}{13}\right)+\frac{3}{13}×\frac{12}{13}\\ & =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\\ & =0\end{array}$ 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.   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).   AB will be parallel to CD, if $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ $\begin{array}{l}\frac{{a}_{1}}{{a}_{2}}=\frac{-2}{2}=-1\\ \frac{{b}_{1}}{{b}_{2}}=\frac{-4}{4}=-1\\ \frac{{c}_{1}}{{c}_{2}}=\frac{-4}{4}=-1\\ \therefore \frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\end{array}$ 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 $3\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}$.

Answer. It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is $\stackrel{\to }{a}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ $\stackrel{\to }{b}=3\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}$ It is known the the line which passes through point A and parallel to $0\stackrel{\to }{b}$ is given by  is a constant. $⇒\stackrel{\to }{r}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}\right)$ 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 $2\stackrel{^}{i}-j+4\stackrel{^}{k}$ and is in the direction $\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}$.

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

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.  Also, the direction ratios of the given line are 3, 7 and 2. This means that the line is in the direction of vector $\stackrel{\to }{b}=3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}$ It is known that the line through position vector $\stackrel{\to }{b}$ and in the direction of the vector $\stackrel{\to }{a}$ is given by the equation , $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b},\lambda \in R$ $⇒\stackrel{\to }{r}=\left(5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{˙}{k}\right)+\lambda \left(3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}\right)$ This is the required equation of the given line in vector form.
Answer. The required line passes through the origin. Therefore, its position vector is given by, $\stackrel{\to }{a}=\stackrel{\to }{0}$ 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 $\stackrel{\to }{b}=5\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}$ The equation of the line in vector through a point with position vector and parallel to  $\begin{array}{l}⇒\stackrel{\to }{r}=\stackrel{\to }{0}+\lambda \left(5\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}\right)\\ ⇒\stackrel{\to }{r}=\lambda \left(5\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}\right)\end{array}$ The equation of the line through the point $\left({x}_{1},{y}_{1},{z}_{1}\right)$ 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 $\begin{array}{l}\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\\ ⇒\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\end{array}$.
Answer. Let the line passing through the points, . Since PQ passes through P (3,-2,-5), its position vector is given by, $\stackrel{\to }{a}=3\stackrel{^}{i}-2\stackrel{^}{j}-5\stackrel{^}{k}$ 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