Three Dimensional Geometry: Chapter 11 Links
Current page: Three Dimensional Geometry Miscellaneous NCERT Solutions. Go to other exercises and content from this chapter :
Three Dimensional Geometry Miscellaneous Exercise is a comprehensive revision of the entire chapter. It combines concepts such as direction ratios and direction cosines, equations of a line, the angle between two lines, and the shortest distance between two lines in a variety of application-based problems. The questions test your understanding of all the important ideas covered in this chapter.
\displaystyle \vec r=\vec a+\lambda\vec b;\qquad \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c};\qquad \cos\theta=\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|
While solving these questions, you will revise how to convert between vector, parametric and Cartesian forms of the equation of a line, use direction ratios and direction cosines to find angles, determine the shortest distance between skew or parallel lines, and apply vector operations in three dimensions. The final question is especially important because it combines multiple concepts from the chapter into a single problem.
By practising this exercise, you will strengthen your problem-solving skills and gain confidence in applying the complete chapter. These mixed-concept questions are highly useful for CBSE board examinations as well as CUET, making this exercise an excellent final revision of Three Dimensional Geometry.
Key Concepts
Before solving the Miscellaneous Exercise, quickly revise the important concepts from this chapter. These formulas and results will help you solve questions on direction ratios, equations of a line, the angle between two lines, and the shortest distance between two lines.
1. Direction Ratios (DRs) and Direction Cosines (DCs)
Direction Ratios (DRs) are any three numbers proportional to the direction of a line.
and Direction Cosines (DCs) are the cosines of the angles made by the line with the positive coordinate axes.
\displaystyle l^2+m^2+n^2=1
If the direction ratios are a,b,c, then the corresponding direction cosines are
\displaystyle l=\frac{a}{\sqrt{a^2+b^2+c^2}},\qquad m=\frac{b}{\sqrt{a^2+b^2+c^2}},\qquad n=\frac{c}{\sqrt{a^2+b^2+c^2}}
2. Equation of a Line
If a line passes through the point A(x_1,y_1,z_1) and has direction ratios a,b,c, then
Vector form
\displaystyle \vec r=\vec a+\lambda\vec b
Parametric form
\displaystyle x=x_1+\lambda a,\qquad y=y_1+\lambda b,\qquad z=z_1+\lambda c
Cartesian (Symmetric) form
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
If two points are given, first find the direction ratios by subtracting the corresponding coordinates, then write the required form of the equation.
3. Angle Between Two Lines
If two lines have direction ratios a_1,b_1,c_1 and a_2,b_2,c_2, then
\displaystyle \cos\theta=\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|
If the direction cosines are known, then
\displaystyle \cos\theta=l_1l_2+m_1m_2+n_1n_2
- Two lines are parallel if their direction ratios are proportional.
- Two lines are perpendicular if a_1a_2+b_1b_2+c_1c_2=0.
4. Shortest Distance Between Two Lines
Skew lines are non-parallel, non-intersecting lines that lie in different planes. The shortest distance between them is the length of the common perpendicular.
If \displaystyle \vec r=\vec a_1+\lambda\vec b_1,\qquad \vec r=\vec a_2+\mu\vec b_2
then \displaystyle d=\frac{\left|(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)\right|}{\left|\vec b_1\times\vec b_2\right|}
For parallel lines,
\displaystyle d=\frac{\left|\vec b\times(\vec a_2-\vec a_1)\right|}{|\vec b|}
5. Quick Revision Points
- Any proportional set of numbers represents the same direction ratios.
- Direction cosines always satisfy l^2+m^2+n^2=1.
- A line can be written in vector, parametric or Cartesian form.
- For the acute angle between two lines, always use the absolute value in the cosine formula.
- Parallel lines have proportional direction ratios.
- Perpendicular lines satisfy a_1a_2+b_1b_2+c_1c_2=0.
- Skew lines are neither parallel nor intersecting.
- The shortest distance between skew lines is the common perpendicular.
- Use the parallel-line distance formula only when both lines have the same direction vector.
Let us now solve all the NCERT questions of Three Dimensional Geometry Miscellaneous Exercise step by step.
Question 1: Three Dimensional Geometry Miscellaneous Exercise
1. Find the angle between the lines whose direction ratios are a,\;b,\;c and b-c,\;c-a,\;a-b.
Solution
Think First… Use the formula for the angle between two lines based on their direction ratios. First find the dot product of the two direction vectors.
The direction ratios of the two given lines are
\displaystyle a,\;b,\;c\quad\text{and}\quad b-c,\;c-a,\;a-b
Using the formula
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}
We will first find the numerator by substituting the values, we get
\displaystyle a(b-c)+b(c-a)+c(a-b)
\displaystyle =ab-ac+bc-ab+ac-bc=0
Therefore,
\displaystyle \cos\theta=\frac{0}{\sqrt{a^2+b^2+c^2}\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}=0
Hence,
\boxed{\theta=90^\circ}
Question 2: Three Dimensional Geometry Miscellaneous
2. Find the equation of a line parallel to the x-axis and passing through the origin.
Solution
Think First… A line parallel to the x-axis has direction ratios (1,0,0). Since it passes through the origin, use the Cartesian form of the equation of a line.
The given line passes through the origin, whose coordinates are
\displaystyle (0,0,0)
Since the line is parallel to the x-axis, its direction ratios are
\displaystyle 1,0,0
Using the Cartesian form of the equation of a line,
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
Substituting (x_1,y_1,z_1)=(0,0,0) and the direction ratios as 1,0,0, we get
\displaystyle \frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}
Therefore, the required equation of the line is
\boxed{\frac{x}{1}=\frac{y}{0}=\frac{z}{0}}
Equivalently, the line can also be written in the parametric form as
\displaystyle x=\lambda,\qquad y=0,\qquad z=0,\qquad \lambda\in\mathbb{R}
You may already be following Maths Better for the NCERT Solutions of the earlier Class 12 Maths chapters. If not, do check them out after completing this exercise.
Likewise, Three Dimensional Geometry extends the ideas of Vector Algebra to study lines in space. In this chapter, you learned about direction ratios, direction cosines, the equation of a line, the angle between two lines, and the shortest distance between lines. These concepts are closely connected, so identifying the correct concept first and then applying the appropriate formula will make solving problems much easier. Keep practising these questions to build speed and accuracy for the CBSE Board Examination as well as CUET. Now, let’s proceed to the next question.
Question 3: Three Dimensional Geometry Miscellaneous Ex.
3. If the lines \displaystyle \frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2} and \displaystyle \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5} are perpendicular, find the value of k.
Solution
Think First… For two perpendicular lines, the dot product of their direction ratios is zero.
The given equations of lines are already in standard form. So, clearly the direction ratios of the given lines are
\displaystyle -3,\;2k,\;2\quad\text{and}\quad 3k,\;1,\;-5
Since the lines are perpendicular, the dot product of their direction ratios is zero, i.e.
\displaystyle a_1a_2+b_1b_2+c_1c_2=0
Substituting the corresponding values,
We get, \displaystyle (-3)(3k)+(2k)(1)+(2)(-5)=0
\displaystyle -9k+2k-10=0
\displaystyle -7k=10
Therefore,
\boxed{k=-\frac{10}{7}}
Question 4: Shortest Distance
4. Find the shortest distance between the lines \displaystyle \vec r=6\hat i+2\hat j+2\hat k+\lambda(\hat i-2\hat j+2\hat k) and \displaystyle \vec r=-4\hat i-\hat k+\mu(3\hat i-2\hat j-2\hat k).
Solution
Think First… Since the lines are not parallel, use the shortest distance formula for two skew lines.
We have,
\displaystyle \vec a_1=6\hat i+2\hat j+2\hat k,\qquad \vec b_1=\hat i-2\hat j+2\hat k
\displaystyle \vec a_2=-4\hat i-\hat k,\qquad \vec b_2=3\hat i-2\hat j-2\hat k
Using the shortest distance formula
\displaystyle d=\frac{\left|(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)\right|}{\left|\vec b_1\times\vec b_2\right|}
First, find the cross product:
\displaystyle \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&2\\3&-2&-2\end{vmatrix}=8\hat i+8\hat j+4\hat k
Also,
\displaystyle \vec a_2-\vec a_1=(-4-6)\hat i+(0-2)\hat j+(-1-2)\hat k
\displaystyle =-10\hat i-2\hat j-3\hat k
Now,
\displaystyle (\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=8(-10)+8(-2)+4(-3)=-108
and
\displaystyle \left|\vec b_1\times\vec b_2\right|=\sqrt{8^2+8^2+4^2}=\sqrt{144}=12
Therefore, we get
\displaystyle d=\frac{|-108|}{12}=9
Hence, the shortest distance between the given lines is
\boxed{9}
By now, you have already completed the NCERT Solutions for Matrices, Determinants, Relations and Functions, Inverse Trigonometric Functions, Continuity and Differentiability, Application of Derivatives, Integrals, Application of Integrals, Differential Equations and Vector Algebra.
Similarly, with the next question, we will complete Three Dimensional Geometry also with detailed, step-by-step solutions to help you develop a clear understanding of 3-D Geometry concepts and prepare confidently for your CBSE board examinations and CUET.
Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, where each solution is explained in a simple and student-friendly manner. Now, let’s proceed to the last question of Miscellaneous Exercise Chapter 11.
Question 5: Three Dimensional Geometry Miscellaneous
5. Find the vector equation of the line passing through the point (1,2,-4) and perpendicular to the two lines
\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}and
\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}.
Solution
Think First… To write the vector equation of a line, we need a point on the line and a direction vector. Since the required line is perpendicular to both the given lines, its direction vector must be perpendicular to both their direction vectors. The cross product of two vectors is perpendicular to each of them. Therefore, the required direction vector is parallel to \vec b_1\times\vec b_2.
The required line passes through the point
\displaystyle (1,2,-4)
We know, the vector equation of a line passing through a point with position vector \vec a and having direction vector \vec b is
\displaystyle \vec r=\vec a+\lambda\vec b \qquad ...(1)
Since the required line is perpendicular to both the given lines, its direction vector \vec b must be perpendicular to the direction vectors of both the given lines.
Let the direction vectors of the given lines are
\displaystyle \vec b_1=3\hat i-16\hat j+7\hat k,\qquad \vec b_2=3\hat i+8\hat j-5\hat k
Now, the cross product \vec b_1\times\vec b_2 is also perpendicular to both \vec b_1 and \vec b_2. Hence, it is parallel to the required direction vector \vec b. Therefore, we can take
\displaystyle \vec b=k\vec b_1\times\vec b_2\qquad ...(2)
Now, \displaystyle \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-16&7\\3&8&-5\end{vmatrix}
\displaystyle =(80-56)\hat i-\bigl(-15-21\bigr)\hat j+\bigl(24+48\bigr)\hat k
\displaystyle =24\hat i+36\hat j+72\hat k=12(2\hat i+3\hat j+6\hat k)
Therefore, using equation (2), we get
\displaystyle \vec b=12k(2\hat i+3\hat j+6\hat k)
or \displaystyle \vec b=m(2\hat i+3\hat j+6\hat k)
where \displaystyle m=12k is a non-zero scalar
Substituting \vec a=\hat i+2\hat j-4\hat k and \vec b=m(2\hat i+3\hat j+6\hat k) in equation (1), we get
\displaystyle \vec r=\hat i+2\hat j-4\hat k+\lambda\,m(2\hat i+3\hat j+6\hat k)
Hence, the required vector equation of the line is
\boxed{\vec r=\hat i+2\hat j-4\hat k+\mu(2\hat i+3\hat j+6\hat k)}
where \displaystyle \lambda m=\mu is an arbitrary scalar (a real number)
Maths Better Tip! If a line passes through a point and is perpendicular to two given lines having direction vectors \vec b_1 and \vec b_2, then its direction vector is parallel to \vec b_1\times\vec b_2. Hence, the vector equation of the required line can be written as
\displaystyle \vec r=\vec a+\lambda(\vec b_1\times\vec b_2)
Common Mistakes to Avoid
- Confusing direction ratios with direction cosines: Direction ratios need not satisfy a^2+b^2+c^2=1. Only direction cosines satisfy l^2+m^2+n^2=1.
- Using the wrong direction ratios: If two points are given, always find the direction ratios by subtracting the corresponding coordinates in the same order.
- Using the wrong condition for parallel or perpendicular lines: Parallel lines have proportional direction ratios, whereas perpendicular lines have zero dot product.
- Choosing the wrong shortest distance formula: First decide whether the given lines are parallel or skew, then apply the appropriate formula.
- Ignoring the cross product while finding a perpendicular line: If a line is perpendicular to two given lines, its direction vector is parallel to the cross product of their direction vectors.
- Not checking the required form of the equation: The final answer may be required in vector, parametric or Cartesian form. Write the form asked in the question.
- Forgetting to take the magnitude in the shortest distance formula: Always use the absolute value (or magnitude) while calculating the shortest distance.
- Skipping the final verification: Before writing the final answer, check whether the question asks for an angle, equation of a line, shortest distance or the value of an unknown.
Continue Learning
In the Three Dimensional Geometry chapter, you learned about direction cosines, direction ratios, the equations of a line in space, the angle between two lines, and the shortest distance between lines. These concepts are important in coordinate geometry and are frequently tested in CBSE board examinations as well as CUET.
Next, continue with Linear Programming. This chapter introduces a new approach to solving mathematical problems using graphical methods. Before starting the chapter, make sure you are confident with the following concepts from Three Dimensional Geometry, as they form the complete revision of this unit:
- The relationship between direction ratios and direction cosines, and the identity l^2+m^2+n^2=1.
- How to obtain the direction ratios of a line passing through two given points.
- The vector, parametric and Cartesian forms of the equation of a line.
- The conditions for two lines to be parallel or perpendicular.
- How to find the angle between two lines using their direction ratios or direction cosines.
- The formulas for the shortest distance between skew and parallel lines.
- How to identify whether two lines are intersecting, parallel or skew before choosing the appropriate formula.
- How to choose the correct form of the equation of a line and apply the appropriate concept in different types of problems.
Explore More
A good understanding of the concepts covered in Three Dimensional Geometry is essential for solving problems involving direction ratios, direction cosines, equations of lines, the angle between two lines, and the shortest distance between lines. Always begin by identifying the given information correctly, determine the required concept, and rewrite the equation in its standard (symmetric) form, whenever necessary, so that you can easily identify the point and the direction ratios.
While solving questions, first identify whether the problem involves the equation of a line, parallelism, perpendicularity, the angle between two lines, or the shortest distance. Then apply the appropriate formula carefully, simplify the calculations step by step, and express the final answer in the form expected in NCERT. With regular practice, you will be able to solve three-dimensional geometry problems accurately and confidently.
All the best and keep learning 👍
