NCERT Solutions for Class 11 Maths Chapter 11 Miscellaneous Exercise - Free PDF Download
NCERT Solutions for Miscellaneous Exercise Class 11 Chapter 11, Introduction to Three Dimensional Geometry, explains the basics of points, lines, and planes in 3D space. This chapter focuses on finding the coordinates of points, understanding the direction ratios and cosines of lines, and determining the equations of planes. These concepts are essential for solving complex geometry problems and are widely used in fields like physics and engineering.
- 2.1Miscellaneous Exercise
The Miscellaneous Exercise in this chapter provides various problems to practice and apply these concepts. It is important to focus on these exercises to gain a better understanding of three-dimensional geometry and prepare well for exams. Downloading the NCERT Solutions of Maths class 11 in PDF format allows for offline practice and helps in improving problem-solving skills. Find the latest CBSE Class 11 Maths Syllabus here.
Miscellaneous Exercise
1. Three vertices of a parallelogram $\text{ABCD}$ are $\text{A}\left( \text{3,-1,2} \right)$, $\text{B}\left( \text{1,2,-4} \right)$ and $\text{C}\left( \text{-1,1,2} \right)$. Find the coordinates of the fourth vertex.
Ans: We are given the three vertices of a parallelogram $\text{ABCD}$ are $\text{A}\left( \text{3,-1,2} \right)$, $\text{B}\left( \text{1,2,-4} \right)$ and $\text{C}\left( \text{-1,1,2} \right)$.
Let the coordinates of the fourth vertex of the parallelogram $\text{ABCD}$ be $\text{D}\left( \text{x,y,z} \right)$.
According to the property of a parallelogram, the diagonals of the parallelogram bisect each other.
In this parallelogram $\text{ABCD}$, $\text{AC}$ and $\text{BD}$ at point $\text{O}$.
So,
$\text{Mid-point of AC = Mid-point of BD}$
$\left( \frac{\text{3-1}}{\text{2}},\text{ }\frac{\text{-1+1}}{\text{2}},\text{ }\frac{\text{2+2}}{\text{2}} \right)\text{ = }\left( \frac{\text{x+1}}{\text{2}},\text{ }\frac{\text{y+1}}{\text{2}},\text{ }\frac{\text{z-4}}{\text{2}} \right)$
$\left( \text{1,0,2} \right)\text{ = }\left( \frac{\text{x+1}}{\text{2}},\text{ }\frac{\text{y+1}}{\text{2}},\text{ }\frac{\text{z-4}}{\text{2}} \right)$
$\frac{\text{x+1}}{\text{2}}\text{ = 1}$
$\frac{\text{y+2}}{\text{2}}\text{ = 0}$
$\frac{\text{z-4}}{\text{2}}\text{ = 2}$
We get, $\text{x = 1}$, $\text{y = 2}$ and $\text{z = 8}$
Therefore, the coordinates of the fourth vertex of the parallelogram $\text{ABCD}$ are $\text{D}\left( \text{1,-2,8} \right)$.
2. Find the lengths of the medians of the triangle with $\text{A}\left( \text{0,0,6} \right)$, $\text{B}\left( \text{0,4,0} \right)$ and $\text{C}\left( \text{6,0,0} \right)$.
Ans: For the given triangle $\text{ABC}$. Let $\text{AD}$, $\text{BE}$ and $\text{CF}$ are the medians:
We know that the median divides the line segment into two equal parts, so $\text{D}$ is the midpoint of $\text{BC}$, therefore,
$\text{Coordinates of point D = }\left( \frac{\text{0+6}}{\text{2}},\text{ }\frac{\text{4+0}}{\text{2}},\text{ }\frac{\text{0+0}}{\text{2}} \right)$
$\text{Coordinates of point D = }\left( \text{3,2,0} \right)$
$\text{AD = }\sqrt{{{\left( \text{0-3} \right)}^{\text{2}}}\text{+}{{\left( \text{0-2} \right)}^{\text{2}}}\text{+}{{\left( \text{6-0} \right)}^{\text{2}}}}$
$\text{AD = }\sqrt{\text{9+4+36}}$
$\text{AD = }\sqrt{\text{49}}$
$\text{AD = 7}$
Similarly, $\text{E}$ is the midpoint of $\text{AC}$,
$\text{Coordinates of point E = }\left( \frac{\text{0+6}}{\text{2}},\text{ }\frac{\text{0+0}}{\text{2}},\text{ }\frac{\text{0+6}}{\text{2}} \right)$
$\text{Coordinates of point E = }\left( \text{3,0,3} \right)$
$\text{AC = }\sqrt{{{\left( \text{3-0} \right)}^{\text{2}}}\text{+}{{\left( \text{0-4} \right)}^{\text{2}}}\text{+}{{\left( \text{3-0} \right)}^{\text{2}}}}$
$\text{AC = }\sqrt{\text{9+16+9}}$
$\text{AC = }\sqrt{\text{34}}$
Similarly, $\text{F}$ is the midpoint of $\text{AB}$,
$\text{Coordinates of point F = }\left( \frac{\text{0+0}}{\text{2}},\text{ }\frac{\text{0+4}}{\text{2}},\text{ }\frac{\text{6+0}}{\text{2}} \right)$
$\text{Coordinates of point F = }\left( \text{0,2,3} \right)$
$\text{CF = }\sqrt{{{\left( \text{6-0} \right)}^{\text{2}}}\text{+}{{\left( \text{0-2} \right)}^{\text{2}}}\text{+}{{\left( \text{0-3} \right)}^{\text{2}}}}$
$\text{CF = }\sqrt{\text{36+4+9}}$
$\text{CF = }\sqrt{\text{49}}$
$\text{CF = 7}$
Therefore, the lengths of the medians of the triangle $\text{ABC}$ we obtain are, $\text{7, }\sqrt{\text{34}}\text{, 7}$
3. If the origin is the centroid of the triangle $\text{PQR}$ with vertices $\text{P}\left( \text{2a,2,6} \right)$, $\text{Q}\left( \text{-4,3b,-10} \right)$ and $\text{R}\left( \text{8,14,2c} \right)$, then find the values of $\text{a}$, $\text{b}$ and $\text{c}$
Ans: The given triangle $\text{PQR}$
We know that the coordinates of the centroid of triangle with the vertices $\left( {{\text{x}}_{1}},{{\text{y}}_{1}},{{\text{z}}_{1}} \right)$, $\left( {{\text{x}}_{2}},{{\text{y}}_{2}},{{\text{z}}_{2}} \right)$ and $\left( {{\text{x}}_{3}},{{\text{y}}_{3}},{{\text{z}}_{3}} \right)$ are,
$\frac{{{\text{x}}_{\text{1}}}\text{+}{{\text{x}}_{\text{2}}}\text{+}{{\text{x}}_{\text{3}}}}{\text{3}}=\frac{{{\text{y}}_{\text{1}}}\text{+}{{\text{y}}_{\text{2}}}\text{+}{{\text{y}}_{\text{3}}}}{\text{3}}=\frac{{{\text{z}}_{\text{1}}}\text{+}{{\text{z}}_{\text{2}}}\text{+}{{\text{z}}_{\text{3}}}}{\text{3}}$
For triangle $\text{PQR}$, the coordinates will be,
$\text{ }\!\!\Delta\!\!\text{ PQR = }\frac{\text{2a-4+8}}{\text{3}}=\frac{\text{2+3b+14}}{\text{3}}=\frac{\text{6-10+2c}}{\text{3}}$
$\text{ }\!\!\Delta\!\!\text{ PQR = }\frac{\text{2a+4}}{\text{3}}=\frac{\text{3b+16}}{\text{3}}=\frac{\text{2c-4}}{\text{3}}$
Now, we are given that centroid is the origin,
$\frac{\text{2a+4}}{\text{3}}\text{ = 0}$
$\frac{\text{3b+16}}{\text{3}}\text{ = 0}$
$\frac{\text{2c-4}}{\text{3}}\text{ = 0}$
$\text{a = -2}$,
$\text{b = -}\frac{\text{16}}{\text{3}}$
$\text{c = 2}$
Therefore, we obtain the values as $\text{a = -2}$, $\text{b = -}\frac{\text{16}}{\text{3}}$ and $\text{c = 2}$.
4. If $\text{A}$ and $\text{B}$ be the points $\left( \text{3,4,5} \right)$ and $\left( \text{-1,3,-7} \right)$ respectively, find the equation of the set of points $\text{P}$ such that $\text{P}{{\text{A}}^{\text{2}}}\text{+P}{{\text{B}}^{\text{2}}}\text{ = }{{\text{k}}^{\text{2}}}$, where $\text{k}$ is a constant.
Ans: Let the coordinates of point $\text{P}$ be $\left( \text{x,y,z} \right)$.
Using the distance formula we get,
$\text{P}{{\text{A}}^{\text{2}}}\text{=}{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-4} \right)}^{\text{2}}}\text{+}{{\left( \text{z-5} \right)}^{\text{2}}}$
$\text{P}{{\text{A}}^{\text{2}}}\text{ = }{{\text{x}}^{\text{2}}}\text{+9-6x+}{{\text{y}}^{\text{2}}}\text{+16-8y+}{{\text{z}}^{\text{2}}}\text{+25-10z}$
$\text{P}{{\text{A}}^{\text{2}}}\text{ = }{{\text{x}}^{\text{2}}}\text{-6x+}{{\text{y}}^{\text{2}}}\text{-8y+}{{\text{z}}^{\text{2}}}\text{-10z+50}$
Similarly,
$\text{P}{{\text{B}}^{\text{2}}}\text{ = }{{\left( \text{x-1} \right)}^{\text{2}}}\text{+}{{\left( \text{y-3} \right)}^{\text{2}}}\text{+}{{\left( \text{z-7} \right)}^{\text{2}}}$
$\text{P}{{\text{A}}^{\text{2}}}\text{ = }{{\text{x}}^{\text{2}}}\text{-2x+}{{\text{y}}^{\text{2}}}\text{-6y+}{{\text{z}}^{\text{2}}}\text{-14z+59}$
We are given that,
$\text{P}{{\text{A}}^{\text{2}}}\text{+P}{{\text{B}}^{\text{2}}}\text{ = }{{\text{k}}^{\text{2}}}$
So,
$\left( {{\text{x}}^{\text{2}}}\text{-6x+}{{\text{y}}^{\text{2}}}\text{-8y+}{{\text{z}}^{\text{2}}}\text{-10z+50} \right)\text{+}\left( {{\text{x}}^{\text{2}}}\text{-2x+}{{\text{y}}^{\text{2}}}\text{-6y+}{{\text{z}}^{\text{2}}}\text{-14z+59} \right)\text{ = }{{\text{k}}^{2}}$
$\text{2}{{\text{x}}^{\text{2}}}\text{+2}{{\text{y}}^{\text{2}}}\text{+2}{{\text{z}}^{\text{2}}}\text{-4x-14y+14z+109 =}\ {{\text{k}}^{\text{2}}}$
$\text{2}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}}\text{-2x-7y+7z} \right)\text{ = }{{\text{k}}^{\text{2}}}\text{-109}$
${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}}\text{-2x-7y+7z = }\frac{{{\text{k}}^{\text{2}}}\text{-109}}{\text{2}}$
Therefore, the equation is as follows ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}}\text{-2x-7y+2z = }\frac{{{\text{k}}^{\text{2}}}\text{-109}}{\text{2}}$
Conclusion
NCERT Solutions for Miscellaneous Exercise Class 11 Chapter 11 helps in understanding three-dimensional geometry. The Miscellaneous Exercise includes different types of problems that improve skills in 3D geometry. Practising these problems is important for doing well in exams and for future studies in Maths. These exercises help students gain confidence and become better at solving complex geometry questions.
Class 11 Maths Chapter 11: Exercises Breakdown
CBSE Class 11 Maths Chapter 11 Other Study Materials
Chapter-Specific NCERT Solutions for Class 11 Maths
Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
FAQs on NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 11 - Introduction to Three Dimensional Geometry
1. What are the key concepts covered in NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 11?
In NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 11 covers important topics like the distance formula, section formula, and coordinates of points in space. It includes problems that require calculating distances between points and finding coordinates of dividing points. Understanding these fundamental concepts is essential for solving the problems correctly. Focus on practising these concepts as they form the base for more complex issues in three-dimensional geometry.
2. How do students approach solving problems in NCERT Class 11 Maths Chapter 11 Miscellaneous Solutions?
Start by reading each problem carefully to understand what is being asked. Identify the known and unknown variables. Choose the appropriate formula based on the problem type, such as the distance or section formula. Substitute the values correctly and perform the calculations step by step. Practice regularly to improve both accuracy and speed in solving problems.
3. Why is understanding the coordinate axes in 3D geometry in NCERT Class 11 Maths Chapter 11 Miscellaneous Solutions?
The coordinate axes (x, y, z) are the foundation of three-dimensional geometry in Class 11 Maths Chapter 11 Miscellaneous Solutions. They help in locating points in space accurately. Understanding these axes is crucial for visualising problems and applying the correct formulas. It aids in interpreting the spatial relationships between points and lines, which is essential for solving geometry problems.
4. What common mistakes should I avoid while solving NCERT Class 11 Maths Chapter 11 Miscellaneous Solutions?
Avoid mistakes such as incorrect substitution of values into formulas, misunderstanding direction ratios, and making algebraic errors. Double-check your values and calculations to ensure accuracy. Pay attention to the signs of the coordinates and ratios. Regular practice and careful review of your work can help in minimizing these common mistakes.
5. How can I effectively prepare for exams using the NCERT Solutions of Miscellaneous Ch 11 Class 11?
Thoroughly practice all the problems in the Miscellaneous Ch 11 Class 11. Focus on understanding and mastering the key concepts like the distance and section formulas. Review any mistakes you make to learn from them. Regular revision and solving sample papers can help reinforce your understanding and improve your problem-solving speed, making you well-prepared for exams.
6. What types of problems are included in the Miscellaneous Ch 11 Class 11?
The Miscellaneous Ch 11 Class 11 includes problems related to points, lines, and planes in three-dimensional space. These problems test understanding of 3D coordinates, direction ratios, and equations of planes.
7. How can I find the coordinates of a point in 3D space in Miscellaneous Ch 11 Class 11?
To find the coordinates of a point in 3D space, you need to know its position along the x, y, and z axes. The exercise will guide you through different methods to determine these coordinates.
8. What are direction ratios and how are they used in the 3D Miscellaneous Exercise Class 11?
Direction ratios are numbers that describe the direction of a line in 3D space. The problems will require using these ratios to find the equation of a line and to solve related questions.
9. How do I find the equation of a plane in three-dimensional space in 3D Miscellaneous Exercise Class 11?
The equation of a plane can be found using a point on the plane and the normal vector to the plane. This exercise will include problems that help practice these calculations.
10. What should I focus on while solving the 3D Miscellaneous Exercise Class 11 problems?
Focus on understanding how to apply the formulas for points, lines, and planes in 3D space. Practice solving different types of problems to gain confidence and accuracy.
11. Why is practising the 3D Miscellaneous Exercise Class 11 important for exams?
Practicing these exercises helps solidify the understanding of three-dimensional geometry concepts, which are crucial for doing well in board exams and competitive tests. It also improves problem-solving skills and analytical thinking.
