Download Free PDF of Three Dimensional Geometry Exercise 11.1 for Class 12 Maths
You’re about to build a strong foundation in 3D Geometry with these NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.1. This section demystifies direction cosines, direction ratios, and the equations of a line in space—all key CBSE concepts essential for approaching spatial problems with confidence.
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Many students look for step-by-step help on topics like "11.1 class 12" because visualizing three dimensions and applying formulas such as the cartesian and vector forms of a line can feel tricky at first. Clear explanations and 3D geometry solved examples here are designed to support your revision and clarify common sticking points.
With a CBSE weightage of nearly 14 marks for Three Dimensional Geometry (combined with Vectors), mastering these techniques can boost your exam performance. Vedantu’s expert solutions ensure you get precise, board-aligned answers, making your exam practice efficient and dependable.
Exercise 11.1
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 \[\dfrac{1}{{\sqrt {42} }}\] .
Conclusion
In Ex 11.1 Class 12 on Three Dimensional Geometry, students explore the fundamental concepts of Direction Cosines and Direction Ratios of a line. These concepts are essential for understanding the orientation and direction of lines in three-dimensional space. By understanding the use of direction cosines and ratios, students can accurately describe and analyze the spatial relationships of lines. Class 12 Ex 11.1 provides a solid foundation for more advanced topics in three-dimensional geometry and its applications in fields such as physics, engineering, and computer graphics. Through this exercise, students develop critical problem-solving skills and gain confidence in working with three-dimensional geometric concepts.
Class 12 Maths Chapter 11: Exercises Breakdown
CBSE Class 12 Maths Chapter 11 Other Study Materials
NCERT Solutions for Class 12 Maths | Chapter-wise List
Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.
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FAQs on CBSE Class 12 Mathematics Chapter 11 Three Dimensional Geometry – NCERT Solutions 2025-26
1. How are direction cosines and direction ratios calculated in Three Dimensional Geometry as per NCERT Class 12 Maths Chapter 11 solutions?
Direction cosines (l, m, n) of a line are calculated by finding the cosine values of the angles the line makes with the x, y, and z-axes. For a line joining points A(x1, y1, z1) and B(x2, y2, z2), you first determine direction ratios as (x2 – x1), (y2 – y1), (z2 – z1). Then normalize these ratios to get direction cosines by dividing each by the distance between the points:
- l = (x2 – x1) / d
- m = (y2 – y1) / d
- n = (z2 – z1) / d
2. What is the step-by-step method to solve NCERT Class 12 Maths Chapter 11 Exercise 11.1 questions according to board standards?
To solve Exercise 11.1 as per NCERT Solutions and CBSE requirements, use this order:
- Write down all given coordinates or line equations clearly.
- Calculate the direction ratios (DRs) or differences between relevant points.
- Apply the direction cosine (DC) formulas to normalize DRs, if asked.
- Set up the equation of a line in either vector or cartesian form as required by the question.
- Complete all calculations step by step, showing intermediate steps and final answer.
- Box or highlight the final answer following exam norms for clarity.
3. Why is it important to differentiate between direction ratios and direction cosines when solving NCERT Maths Class 12 Chapter 11 problems?
Direction ratios are any set of three numbers in proportion to the difference in coordinates (indicating the direction of a line), while direction cosines are always normalized so that their squares add up to 1.
- Using DRs alone in place of DCs can lead to calculation errors, especially in questions involving angles with axes or normalization steps.
- NCERT Solutions emphasize distinguishing them, as CBSE marks are awarded for precise terminology and correct method use.
4. How can errors in calculating direction cosines affect the accuracy of answers in Class 12 board exams?
Mistakes in finding direction cosines, such as not normalizing direction ratios, using incorrect signs, or calculation slips, can result in completely wrong answers and loss of marks. In CBSE and competitive exams, accuracy in each step is critical, and NCERT Solutions highlight these errors to help students avoid common pitfalls and improve exam performance.
5. What is the process to prove collinearity of three points in NCERT Class 12 Maths Chapter 11 using solutions methodology?
To prove three points A, B, and C are collinear in 3D geometry:
- Calculate the direction ratios of AB and BC.
- If the ratios are proportional (one is a scalar multiple of the other), the points are collinear.
- Show the scalar relationship clearly and mention the common point for stepwise NCERT/CBSE method compliance.
6. How does mastering 3D Geometry concepts in Chapter 11 help in higher-level maths and exams like JEE?
Building a strong foundation in direction cosines, ratios, and line equations from Class 12 Chapter 11 develops spatial reasoning skills required in advanced mathematics, physics, and entrance exams like JEE. These concepts reappear in vectors, calculus, and engineering problems, so mastering their step-wise solutions early boosts long-term accuracy and confidence.
7. In what ways do NCERT Solutions for Class 12 Maths Chapter 11 align with the official CBSE 2025–26 syllabus?
All stepwise solutions and methods in the NCERT Solutions for Chapter 11 are mapped directly with the latest CBSE Class 12 Maths syllabus for 2025–26. They follow prescribed formulas, include all required intermediate steps, and address relevant types of exam questions to ensure students are fully prepared for current CBSE board assessment patterns.
8. What are the most common misconceptions students have while using NCERT Solutions for 3D Geometry, and how can these be avoided?
Students often confuse direction cosines and direction ratios, neglect normalization, use incorrect signs, or skip steps outlined in the NCERT textbook. Following the NCERT Solutions stepwise, double-checking formulas, and practicing error analysis after each problem are key habits to avoid these misconceptions and secure full marks in board exams.
9. How does the equation of a line in 3D geometry differ in vector and cartesian form, and when should each be used?
The vector form of a line is best when using position vectors and parameters: r = a + λb, where a is a point on the line, b is the direction vector, and λ is a scalar. The cartesian form equates ratios: (x – x1)/(l) = (y – y1)/(m) = (z – z1)/(n), where (l, m, n) are direction ratios. Use vector form for questions involving operations between vectors, and cartesian form where coordinate-based comparison is needed.
10. What higher-order thinking strategies should students use when tackling challenging 3D Geometry problems in Class 12?
Apply logical analysis to identify given information and what is required, break problems into smaller steps, cross-check each formula's applicability, visualize geometric relationships in 3D, and clearly differentiate between direction ratios and cosines. Use these strategies as guided in the NCERT Solutions for improved problem-solving and to handle unexpected twists in CBSE or entrance exam questions.
