# NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.1) Exercise 11.1

## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.1) Exercise 11.1 Free PDF download of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.1 (Ex 11.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

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1. If a line makes angles $90^\circ ,135^\circ ,45^\circ$ with the $x,y{\text{ }}$and $z$axis respectively, find its direction cosines.

Ans: Let the direction of cosines of the given line be $l,m$ and $n$ .

Therefore,

$l = \cos 90^\circ$

$l = 0$

$m = \cos 135^\circ$

$m = - \dfrac{1}{{\sqrt 2 }}$

$n = \cos 45^\circ$

$n = \dfrac{1}{{\sqrt 2 }}$

Therefore, the direction of cosines are $0, - \dfrac{1}{{\sqrt 2 }}$ and $\dfrac{1}{{\sqrt 2 }}$ .

2. Find the direction cosines of a line which makes equal angles with the coordinate axes.

Ans: Let the direction of the line that makes an angle $\alpha$ with each of the coordinate axes.

Therefore,

$l = \cos \alpha$

$m = \cos \alpha$

$n = \cos \alpha$

As, we know that,

${l^2} + {m^2} + {n^2} = 1$

So,

${\cos ^2}\alpha + {\cos ^2}\alpha + {\cos ^2}\alpha = 1$

$3{\cos ^2}\alpha = 1$

${\cos ^2}\alpha = \dfrac{1}{3}$

$\cos \alpha = \pm \dfrac{1}{{\sqrt 3 }}$

Therefore, the direction of cosines of the line, which is equally inclined to the coordinate axes are $\pm \dfrac{1}{{\sqrt 3 }}, \pm \dfrac{1}{{\sqrt 3 }}$ and $\pm \dfrac{1}{{\sqrt 3 }}$ .

3. If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

Ans: In this question it is given the direction ratio a, b and c which is $- 18,12$and $- 4$ respectively.

So,

$a = - 18$

$b = 12$

$c = - 4$

The direction cosines is given as,

$l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$l = \dfrac{{ - 18}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }}$

$l = - \dfrac{{18}}{{22}}$

$m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$m = \dfrac{{12}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }}$

$m = \dfrac{{12}}{{22}}$

$n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$n = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }}$

$n = - \dfrac{2}{{22}}$

Therefore, the direction cosines are $- \dfrac{{18}}{{22}},\dfrac{{12}}{{22}}$ and $- \dfrac{2}{{22}}$ .

4. Show that the points $\left( {2,3,4} \right),\left( { - 1, - 2,1} \right),\left( {5,8,7} \right)$ are collinear.

Ans: The given points are $A\left( {2,3,4} \right),B\left( { - 1, - 2,1} \right)$ and $C\left( {5,8,7} \right)$ .

The direction ratio of line joining the two coordinates $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ is given as $\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)$ and $\left( {{z_1} - {z_2}} \right)$ .

The direction ratio of line $AB$ is given as $\left( { - 1 - 2} \right),\left( { - 2 - 3} \right)$ and $\left( {1 - 4} \right)$ .

So, the direction ratio of line $AB$ is $- 3, - 5$ and $- 3$ .

The direction ratio of line $BC$ is given as $\left( {5 - \left( { - 1} \right)} \right),\left( {8 - \left( { - 2} \right)} \right)$ and $\left( {7 - 1} \right)$ .

So, the direction ratio of line $BC$ is 6, 10 and 6 .

On comparing the direction ratio of $AB$ and $BC$, it can be seen that the direction ratio of $BC$ is $- 2$ times of $AB$ i.e. they are proportional.

$AB = \lambda \left( {BC} \right)$

Therefore, $AB\parallel BC$. As point B is common to both $AB$ and $BC$ .

Therefore, given points $A\left( {2,3,4} \right),B\left( { - 1, - 2,1} \right)$ and $C\left( {5,8,7} \right)$ are collinear.

5. Find the direction cosines of the sides of the triangle whose vertices are

(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Ans: The given vertices of $\Delta ABC$ are $A\left( {3,5, - 4} \right),B\left( { - 1,1,2} \right)$ and $C\left( { - 5, - 5, - 2} \right)$ .

Calculating direction cosines of side $AB$ .

The direction ratio of line joining the two coordinates $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ is given as $\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)$ and $\left( {{z_1} - {z_2}} \right)$ .

The direction ratio of $AB$ is given as $\left( { - 1 - 3} \right),\left( {1 - 5} \right)$ and $\left( {2 - \left( { - 4} \right)} \right)$ .

The direction ratio of $AB$ is $- 4, - 4$ and $6$ .

The direction cosines of side $AB$ using direction ratio is given as,

$l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$l = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }}$

$l = - \dfrac{4}{{2\sqrt {17} }}$

$l = - \dfrac{2}{{\sqrt {17} }}$

$m = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$m = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }}$

$m = - \dfrac{4}{{2\sqrt {17} }}$

$m = - \dfrac{2}{{\sqrt {17} }}$

$n = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$n = \dfrac{6}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }}$

$n = \dfrac{6}{{2\sqrt {17} }}$

$n = \dfrac{3}{{\sqrt {17} }}$

So, the direction cosines of $AB$ is $- \dfrac{2}{{\sqrt {17} }}, - \dfrac{2}{{\sqrt {17} }}$ and $\dfrac{3}{{\sqrt {17} }}$ .

Calculating the direction cosines of side $BC$

The direction ratio of line joining the two coordinates $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ is given as $\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)$ and $\left( {{z_1} - {z_2}} \right)$ .

The direction ratio of $BC$ is given as $\left( { - 1 - 3} \right),\left( {1 - 5} \right)$ and $\left( {2 - \left( { - 4} \right)} \right)$ .

The direction ratio of $BC$  is $- 4, - 4$ and $6$ .

The direction cosines of side $BC$ using direction ratio is given as,

$l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$l = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }}$

$l = - \dfrac{4}{{2\sqrt {17} }}$

$l = - \dfrac{2}{{\sqrt {17} }}$

$m = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$m = \dfrac{{ - 6}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }}$

$m = - \dfrac{6}{{2\sqrt {17} }}$

$m = - \dfrac{3}{{\sqrt {17} }}$

$n = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$n = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }}$

$n = - \dfrac{4}{{2\sqrt {17} }}$

$n = - \dfrac{2}{{\sqrt {17} }}$

So, the direction cosines of $BC$ is $- \dfrac{2}{{\sqrt {17} }}, - \dfrac{3}{{\sqrt {17} }}$ and $- \dfrac{2}{{\sqrt {17} }}$ .

Calculating the direction cosines of side $AC$

The direction ratio of line joining the two coordinates $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ is given as $\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)$ and $\left( {{z_1} - {z_2}} \right)$ .

The direction ratio of $AC$ is given as $\left( { - 5 - 3} \right),\left( { - 5 - 5} \right)$ and $\left( { - 2 - \left( { - 4} \right)} \right)$ .

The direction ratio of $AC$  is $- 8, - 10$ and 2 .

The direction cosines of side $AC$ using direction ratio is given as,

$l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$l = \dfrac{{ - 8}}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( 2 \right)}^2}} }}$

$l = - \dfrac{8}{{2\sqrt {42} }}$

$l = - \dfrac{4}{{\sqrt {42} }}$

$m = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$m = \dfrac{{ - 10}}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( 2 \right)}^2}} }}$

$m = - \dfrac{{10}}{{2\sqrt {42} }}$

$m = - \dfrac{5}{{\sqrt {42} }}$

$n = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$n = \dfrac{2}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( 2 \right)}^2}} }}$

$n = \dfrac{2}{{2\sqrt {42} }}$

$n = \dfrac{1}{{\sqrt {42} }}$

So, the direction cosines of $BC$ is $- \dfrac{4}{{\sqrt {42} }}, - \dfrac{5}{{\sqrt {42} }}$ and $\dfrac{1}{{\sqrt {42} }}$ .

Therefore,

Direction cosines of $AB$ is $- \dfrac{2}{{\sqrt {17} }}, - \dfrac{2}{{\sqrt {17} }}$ and $\dfrac{3}{{\sqrt {17} }}$ .

Direction cosines of $BC$ is $- \dfrac{2}{{\sqrt {17} }}, - \dfrac{3}{{\sqrt {17} }}$ and $- \dfrac{2}{{\sqrt {17} }}$ .

Direction cosines of $BC$ is $- \dfrac{4}{{\sqrt {42} }}, - \dfrac{5}{{\sqrt {42} }}$ and $\dfrac{1}{{\sqrt {42} }}$ .

## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.1

Opting for the NCERT solutions for Ex 11.1 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 11 Exercise 11.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 11 Exercise 11.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 11 Exercise 11.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

1. What is the weightage of Exercise 11.1 of class 12 Maths?

The weightage of any exercise is decided randomly based on the chapter and the format. For board exams, the sums of any chapter can be randomly selected and hence, it is important to practice all the sums to be ready for any question that is set. Apart from this students are advised to check the chapter-wise weightage to get a clearer picture.

2. Which question is considered an important one from Class 12 Maths exercise 11.1?

No such thing as the most significant question exists! You see, the easier ones help you remember your principles, while the more complex ones evaluate your problem-solving abilities and how you apply your knowledge in practice. To understand the topics in Chapter 11 3D Geometry you need to pay close attention. As a result, please practice all amounts, no matter how tough they are. You can visit the page NCERT Solutions Class 12 Maths Chapter 11 on the official website of  Vedantu or the Vedantu app to access Chapter related resources at free of cost.

3. What exactly is 3D Geometry?

Shapes in 3D space may be calculated using 3 coordinates on the XYZ plane: the x-coordinate, y-coordinate, and z-coordinate. Three-dimensional forms are those that occupy space. As a solid form with three dimensions (length, width, and height), 3D shapes may also be described in this way. It is an integral part of Mathematics and students must learn its concepts very clearly and practice all the problems related to the same for better understanding.

4. What is the need for 3D Geometry?

As the name suggests, 3D geometry is the study of three-dimensional forms in space. A z-coordinate may be determined by subtracting the x-coordinate from the value of the y-coordinate. If you want to discover the precise position of a point in three-dimensional space, you need three parameters. It is a vast field and also has multiple applications in the domain of mathematics.

5. Where can we apply Geometry?

A few examples of how geometry is used in the actual world include computer-aided design (CAD) for construction plans and industrial assembly systems. These include nanotechnology, computer graphics, visual graphs, video game programming and virtual reality development (VR). These are just a few examples of many of Geometry’s applications in our life and the domain of Mathematics. SHARE TWEET SHARE SUBSCRIBE