NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

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January 30, 2026

Ex 11.2 class 12 Maths is a set of math problems about lines in 3D space. It teaches you how to find the path of a line and the space between two different lines. By solving these questions, you learn how lines lean, how they cross, and how to measure the gap between them.

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NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Solve The Following Questions of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Question 1. Show that the three lines with direction cosines are mutually perpendicular.Solution : Two lines with direction cosines, l ₁, m ₁, n ₁and l ₂, m ₂, n ₂, are perpendicular to each other, if l ₁l ₂+ m ₁m ₂+ n ₁n ₂= 0 NCERT Solutions class 12 Maths Three Dimensional Geometry Therefore, the lines are perpendicular. Thus, all the lines are mutually perpendicular.

Question 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).

Solution : Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6). The direction ratios, a ₁, b ₁, c ₁, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4. The direction ratios, a ₂, b ₂, c ₂, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4. AB and CD will be perpendicular to each other, if a ₁a ₂+ b ₁b ₂+ c ₁c ₂= 0 a ₁a ₂+ b ₁b ₂+ c ₁c ₂= 2 × 3 + 5 × 2 + (− 4) × 4 = 6 + 10 − 16 = 0 Therefore, AB and CD are perpendicular to each other.

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

Solution : Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5). The directions ratios, a ₁, b ₁, c ₁, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4. The direction ratios, a ₂, b ₂, c ₂, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4. AB will be parallel to CD, if a1/a2 = b1/b2 = c1/c2 a1/a2 = -2/2 = -1 b1/b2 = -4/4 = -1 c1/c2 = -4/4 = -1 ∴a1/a2 = b1/b2 = c1/c2 Thus, AB is parallel to CD.

Question 4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector

Solution : It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is chapter 11-Three Dimensional Geometry Exercise 11.2/image037.png

It is known that the line which passes through point A and parallel to is given by λ is a constant. ncert solution chapter 11-Three Dimensional Geometry Exercise 11.2/image037.png

This is the required equation of the line.

Question 5. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector and is in the direction

Solution : It is given that the line passes through the point with position vector

Three Dimension Geometry

It is known that a line through a point with position vector and parallel to is given by the equation, ncert solution chapter 11-Three Dimensional Geometry Exercise 11.2/image037.png Three Dimension Geometry This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as Three Dimension Geometry

Question 6. Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by chapter 11-Three Dimensional Geometry Exercise 11.2/image057.png

Solution : It is given that the line passes through the point (−2, 4, −5) and is parallel to

chapter 11-Three Dimensional Geometry Exercise 11.2/image057.png

The direction ratios of the line,

are 3, 5, and 6. The required line is parallel to

Therefore, its direction ratios are 3 k , 5 k , and 6 k , where k ≠ 0 It is known that the equation of the line through the point ( x ₁, y ₁, z ₁) and with direction ratios, a , b , c , is given by Three Dimension Geometry Therefore the equation of the required line is

Question 7. The Cartesian equation of a line is Write its vector form.

Solution : The Cartesian equation of the line is ....(1)

The given line passes through the point (5, −4, 6). The position vector of this point is chapter 11-Three Dimensional Geometry Exercise 11.2/image068.png

Also, the direction ratios of the given line are 3, 7, and 2. This means that the line is in the direction of vector, chapter 11-Three Dimensional Geometry Exercise 11.2/image068.png It is known that the line through position vector and in the direction of the vector is given by the equation, ncert solution

This is the required equation of the given line in vector form.

Question 8. Find the vector and Cartesian equations of the line that passes through the origin and (5, −2, 3). Solution :

The required line passes through the origin. Therefore, its position vector is given by, = 0

.....(1) ncert solution 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, chapter 11-Three Dimensional Geometry Exercise 11.2/image041.png

The equation of the line in vector form through a point with position vector and parallel to is,

ncert solution

chapter 11-Three Dimensional Geometry Exercise 11.2/image068.png

chapter 11-Three Dimensional Geometry Exercise 11.2/image041.png

The equation of the line through the point ( x ₁, y ₁, z ₁) and direction ratios a , b , c is given by,

Therefore, the equation of the required linein the Cartesian form is

chapter 11-Three Dimensional Geometry Exercise 11.2/image041.png

Question 9. Find vector and Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Solution : Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by, chapter 11-Three Dimensional Geometry Exercise 11.2/image092.png

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 chapter 11-Three Dimensional Geometry Exercise 11.2/image092.png

The equation of PQ in vector form is given by,

chapter 11-Three Dimensional Geometry Exercise 11.2/image068.png

The equation of PQ in Cartesian form is i.e.,

chapter 11-Three Dimensional Geometry Exercise 11.2/image092.png

Question 10. Find the angle between the following pairs of lines: (i)(ii)Solution : (i) Let Q be the angle between the given lines. The angle

between the given pairs of lines is given by

chapter 11-Three Dimensional Geometry Exercise 11.2/image073.png

The given lines are parallel to the vectors,

, respectively.

chapter 11-Three Dimensional Geometry Exercise 11.2/image073.png

(ii) The given lines are parallel to the vectors, respectively. chapter 11-Three Dimensional Geometry Exercise 11.2/image073.png

Question 11. Find the angle between the following pair of lines

(i)(ii)Solution : Let ₁and ₂be the vectors parallel to the pair of lines,, respectively. ncert solution chapter 11-Three Dimensional Geometry Exercise 11.2/image131.png (ii) Let ₁₂be the vectors parallel to the given pair of lines,, respectively.

Question 12. Find the values of p so that the lines are at right angles.

Solution : The given equations can be written in the standard form asThe direction ratios of the lines are respectively NCERT Solutions class 12 Maths Three Dimensional Geometry Two lines with direction ratios , a ₁, b ₁, c ₁and a ₂, b ₂, c ₂, are perpendicular to each other, if a ₁a ₂+ b ₁b ₂+ c ₁c ₂= 0 chapter 11-Three Dimensional Geometry Exercise 11.2/image152.png

Question 13. Show that the lines are perpendicular to each other.

Solution : The equations of the given lines are

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 ₁and a ₂, b ₂, c ₂,

are perpendicular to each other,

if a ₁a ₂+ b ₁b ₂+ c ₁c ₂= 0 ∴ 7 × 1 + (−5) × 2 + 1 × 3 = 7 − 10 + 3 = 0

Therefore, the given lines are perpendicular to each other.

Question 14. Find the shortest distance between the lines

Solution : The equations of the given lines areSince, the shortest distance between the two skew lines is given by chapter 11-Three Dimensional Geometry Exercise 11.2 Therefore, the shortest distance between the two lines is 3√2/2 units.

Question 15. Find the shortest distance between the lines .

Solution : The given lines are

It is known that the shortest distance between the two lines, ,

is given by Comparing this equation we have chapter 11-Three Dimensional Geometry Exercise 11.2/image184.png Since distance is always non-negative, the distance between the given lines is 2√29 units.

Question 16. Find the shortest distance between the lines whose vector equations are

Solution : The given lines areIt is known that the shortest distance between the lines, is given by, chapter 11-Three Dimensional Geometry Exercise 11.2/image201.png .....

(1)

chapter 11-Three Dimensional Geometry Exercise 11.2/image201.png

Comparing the given equations with , we obtain chapter 11-Three Dimensional Geometry Exercise 11.2/image201.png Substituting all the values in equation

(1), we obtain chapter 11-Three Dimensional Geometry Exercise 11.2/image205.png

Therefore, the shortest distance between the two given lines is 3/√19 units.

Question 17. Find the shortest distance between the lines whose vector equations are

Solution : The given lines are chapter 11-Three Dimensional Geometry Exercise 11.2/image216.png It is known that the shortest distance between the lines, is given by,For the given equations,Substituting all the values in equation (3), we obtainTherefore, the shortest distance between the lines is 8/√29 units.

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Ex 11.2 Class 12 Maths

When you study ex 11.2 class 12 Maths, you are looking at how lines move in a world with three sides: length, width, and height. This is called 3D geometry. In this exercise, we use special numbers called direction ratios and cosines to talk about where a line is going. It's like giving a line its own home address in space.

Why is this exercise helpful

It shows you how to write the "name" of a line using vectors.
It teaches you how to find if two lines are standing straight against each other.
You learn the way to find the shortest gap between two lines that never touch.
It helps you get ready for big tests by practicing 17 different types of questions.

Most students use the ex 11.2 class 12 Maths NCERT solutions to check their steps. These solutions show you how to move from a vector form to a Cartesian form. This sounds hard, but it’s just two different ways to write the same math story. One uses arrows (vectors) and the other uses letters like x , y , and z .

Class 12 Maths Chapter 11 Exercise 11.2

The class 12 Maths chapter 11 exercise 11.2 focus is all about the "Equation of a Line." Imagine a line as a walking path. To know this path, you only need two things: a point where it starts and the way it is pointing. If you have these, you can write the whole equation.

Concept Type	What it tells us	Key Formula
Vector Form	The path using arrows	\vec{r} = \vec{a} + \lambda\vec{b}
Cartesian Form	The path using coordinates	\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}
Shortest Distance	The smallest gap between lines	Uses the "Cross Product" rule

Tips for Class 12 Maths Chapter 11 Exercise 11.2 Solutions

If you are stuck on a problem, try to draw it! Even a simple drawing helps you see if the lines are parallel (running side-by-side) or perpendicular (crossing like a "+" sign). Many ex 11.2 class 12 maths solutions suggest finding the "direction ratios" first because they make the rest of the work much easier.

Ex 11.2 Class 12 Maths Ncert Solutions PDF

Using an ex 11.2 class 12 maths NCERT solutions pdf is very smart because you can carry it on your phone. You can look at it whenever you have a doubt about a formula. The PDF versions usually have every step written out clearly so you don't miss a single thing.

Important things in the PDF:

Angle between two lines: You use a formula with "cos" to find the corner where two lines meet.
Shortest Distance: This is the most famous part of ex 11.2 class 12 Maths. It's used for skew lines—lines that are in different planes and never meet.
Proving Perpendicularity: You multiply the direction numbers. If the final answer is 0, the lines are perpendicular!

To find the ex 11.2 class 12 maths NCERT answers, always look for the most updated version. The board sometimes changes the numbers or the syllabus, so staying updated is a good habit.

Ex 11.2 Class 12 Maths Solutions

To do well in Ex 11.2 class 12 Maths, you should follow a set of steps. Don't jump to the end! Math is like building a house; you need a strong base first.

Step 1: Write down what the question gives you (points or vectors).
Step 2: Decide which formula to use. Are you finding a distance or an angle?
Step 3: Put the numbers into the formula carefully.
Step 4: Check your plus and minus signs. This is where most kids make mistakes!

The class 12 Maths chapter 11 exercise 11.2 solutions usually highlight Question 12 and Question 15. These are the "star" questions that often show up in exams. Question 12 asks you to find a missing value (like p ) so that lines are at right angles. Question 15 asks for the distance between two lines. If you can do these two, you can do anything in this chapter!

Summary of Ex 11.2 Class 12 Maths NCERT

When we look at ex 11.2 class 12 Maths NCERT, we see that it's the heart of 3D geometry. It links what we know about points in 2D to the real world of 3D. Every line has its own unique direction. If you know how to find that direction, you've won half the battle.

Key Takeaways

Direction Cosines: These are the cosine angles a line makes with the x , y , and z axes.
Skew Lines: These are special lines that are not parallel but still never touch because they are at different heights.
Cartesian Equation: A way to write a line's path using a set of ratios.

Using ex 11.2 class 12 Maths helps you think in 3D. This is useful for many jobs like making video games, building bridges, or even flying planes! It is all about knowing where things are and how they move in space.