Three Dimensional Geometry: Chapter 11 Links
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Three Dimensional Geometry 11.2 introduces the equation of a line in space, the angle between two lines, and the shortest distance between two lines. In this exercise, you will learn how to write the vector, parametric and Cartesian equations of a line, find the angle between intersecting or parallel lines using direction ratios or direction cosines, and calculate the shortest distance between skew and parallel lines.
\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|
The main idea is to understand how a line is represented in three dimensions using a point and a direction vector. You will also learn to convert between vector, parametric and Cartesian forms of the equation of a line, determine the angle between two lines, and distinguish between intersecting, parallel and skew lines. Finally, you will apply vector and cross product concepts to find the shortest distance between two lines in space.
By practising these questions, you will build a strong understanding of equations of lines and their applications in Three Dimensional Geometry. These concepts are frequently tested in CBSE board examinations as well as CUET, making this exercise an important part of your preparation.
Key Concepts
Before solving Exercise 11.2, it is important to understand the different forms of the equation of a line in space, direction ratios and direction cosines, and how these concepts are used to find the angle between two lines and the shortest distance between them.
1. Equation of a Line Through a Given Point and Parallel to a Vector
If a line passes through the point A(x_1,y_1,z_1) whose position vector is \vec a and is parallel to the vector \vec b=a\hat i+b\hat j+c\hat k, then
Vector form
\displaystyle \vec r=\vec a+\lambda\vec b,\qquad \lambda\in\mathbb R
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}
Here, a,b,c are the direction ratios (DRs) of the line.
2. Equation of a Line Passing Through Two Given Points
If a line passes through the points A(x_1,y_1,z_1) and B(x_2,y_2,z_2), then its direction vector is
\displaystyle \overrightarrow{AB}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k
Hence, the direction ratios are
\displaystyle a=x_2-x_1,\qquad b=y_2-y_1,\qquad c=z_2-z_1
Vector form
\displaystyle \vec r=\vec{OA}+\lambda\,\overrightarrow{AB}
or
\displaystyle \vec r=\vec a+\lambda(\vec b-\vec a)
Parametric form
\displaystyle x=x_1+\lambda(x_2-x_1),\qquad y=y_1+\lambda(y_2-y_1),\qquad z=z_1+\lambda(z_2-z_1)
Cartesian (Symmetric) form
\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}
This method is used whenever the coordinates of two points on the line are given directly.
3. Direction Ratios (DRs) and Direction Cosines (DCs)
Direction Ratios (DRs) are any three numbers proportional to the direction of a line.
If the direction ratios are a,b,c, then any proportional set such as 2a,2b,2c or -a,-b,-c also represents the same line.
Direction Cosines (DCs) are the cosines of the angles made by the line with the positive x, y and z axes.
If the direction cosines are l,m,n, then
\displaystyle l=\cos\alpha,\qquad m=\cos\beta,\qquad n=\cos\gamma
They satisfy the relation
\displaystyle l^2+m^2+n^2=1
The relation between direction ratios and direction cosines is
\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}}
Conversely, if l,m,n are the direction cosines, then the corresponding direction ratios are proportional to l:m:n.
Maths Better Tip! If two points are given, first find the direction ratios by subtracting the corresponding coordinates. Then write the vector, parametric or Cartesian equation of the line using these direction ratios. This saves time and avoids mistakes in examinations.
4. Angle Between Two Lines
Suppose two lines have direction ratios a_1,b_1,c_1 and a_2,b_2,c_2. Then the angle \theta between the lines is given by
\displaystyle \cos\theta=\pm\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}}
For the acute angle between the two lines, use
\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 of the two lines are l_1,m_1,n_1 and l_2,m_2,n_2, then
\displaystyle \cos\theta=l_1l_2+m_1m_2+n_1n_2
This formula is frequently used to determine whether two lines are parallel or perpendicular.
5. Conditions for Parallel and Perpendicular Lines
- Two lines are parallel if their direction ratios are proportional.
- \displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
- Two lines are perpendicular if the dot product of their direction vectors is zero.
- \displaystyle a_1a_2+b_1b_2+c_1c_2=0
- If two lines are parallel, the angle between them is 0^\circ or 180^\circ.
- If two lines are perpendicular, the angle between them is 90^\circ.
6. Coplanar and Skew Lines
Two lines are said to be coplanar if they lie in the same plane.
Coplanar lines may be
- Intersecting lines
- Parallel lines
Two lines are called skew lines if they are non-coplanar. They neither intersect nor are parallel.
Hence, skew lines satisfy all the following conditions:
- They lie in different planes.
- They do not intersect.
- They are not parallel.
- There exists a unique shortest distance between them.
The shortest distance between two skew lines is the length of the common perpendicular joining the two lines.
| Type of Lines | Coplanar? | Intersect? | Parallel? |
|---|---|---|---|
| Intersecting | ✔️ Yes | ✔️ Yes | ✖️ No |
| Parallel | ✔️ Yes | ✖️ No | ✔️ Yes |
| Skew | ✖️ No | ✖️ No | ✖️ No |
7. Shortest Distance Between Two Skew Lines
The shortest distance between two skew lines is the length of the common perpendicular joining the two lines.
If the equations of the lines are
\displaystyle \vec r=\vec a_1+\lambda\vec b_1,\qquad \vec r=\vec a_2+\mu\vec b_2
then the shortest distance between them is
\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|}
This formula is applicable only when the two lines are skew, i.e., they are neither parallel nor intersecting.
The line representing the shortest distance is perpendicular to both the given lines.
8. Shortest Distance Between Two Parallel Lines
If two lines are parallel and have equations
\displaystyle \vec r=\vec a_1+\lambda\vec b,\qquad \vec r=\vec a_2+\mu\vec b
then the shortest distance between them is
\displaystyle d=\frac{\left|\vec b\times(\vec a_2-\vec a_1)\right|}{|\vec b|}
Here also, the shortest distance is measured along a line perpendicular to both the parallel lines.
9. Important Points
- The equation of a line can be written in vector, parametric or Cartesian (symmetric) form.
- Any three numbers proportional to the direction of a line are its direction ratios (DRs).
- Direction cosines satisfy l^2+m^2+n^2=1.
- To find the angle between two lines, use their direction ratios or direction cosines.
- For the acute angle between two lines, always take the absolute value of the numerator in the cosine formula.
- Parallel lines have proportional direction ratios.
- Perpendicular lines satisfy a_1a_2+b_1b_2+c_1c_2=0.
- Intersecting and parallel lines are coplanar.
- Skew lines are non-coplanar; they neither intersect nor are parallel.
- There is a unique shortest distance between two skew lines.
- The shortest distance between two skew lines is always the common perpendicular to both the lines.
- For parallel lines, the shortest distance is also measured along a line perpendicular to both the given lines.
- If the shortest distance is non-zero, the lines are either parallel or skew.
- If the shortest distance between two lines is zero, the lines intersect and are therefore coplanar.
Quick Reference Table
| Concept | Details |
|---|---|
| Equation of a Line |
Vector: \displaystyle \vec r=\vec a+\lambda\vec b Cartesian: \displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} Two-point form: \displaystyle \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} |
| Direction Ratios & Direction Cosines |
DRs: Any three numbers proportional to the direction of a line. \displaystyle l=\frac{a}{\sqrt{a^2+b^2+c^2}},\quad m=\frac{b}{\sqrt{a^2+b^2+c^2}},\quad n=\frac{c}{\sqrt{a^2+b^2+c^2}} \displaystyle l^2+m^2+n^2=1 |
| Line Through Two Points |
If the line passes through P(x_1,y_1,z_1) and Q(x_2,y_2,z_2), then one set of direction ratios is \displaystyle (x_2-x_1,\;y_2-y_1,\;z_2-z_1) |
| Angle Between Two Lines | \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| |
| Parallel & Perpendicular Lines |
Parallel: \displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} Perpendicular: \displaystyle a_1a_2+b_1b_2+c_1c_2=0 |
| Shortest Distance |
Skew lines: \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|}Parallel lines: \displaystyle d= \frac{\left|\vec b\times(\vec a_2-\vec a_1)\right|} {|\vec b|} |
| Important Facts |
• Intersecting and parallel lines are coplanar. • Skew lines are non-coplanar. They neither intersect nor are parallel. • If the shortest distance between two lines is zero, the lines intersect and are therefore coplanar. • If the shortest distance is non-zero, the lines are either parallel or skew. • The shortest distance between two skew lines is measured along their common perpendicular. |
Maths Better Tip!
Whenever two lines are given, first check whether their direction ratios are proportional. If yes, the lines are parallel. Otherwise, check whether they intersect. If the shortest distance between the lines is zero, they intersect and are therefore coplanar. If they are neither parallel nor intersecting, they are skew (non-coplanar), and their shortest distance is measured along the common perpendicular.
Let us now solve all the NCERT questions of Three Dimensional Geometry 11.2 step by step.
Question 1: Mutually Perpendicular Lines
1. Show that the three lines with direction cosines \frac{12}{13},-\frac{3}{13},-\frac{4}{13}; \frac{4}{13},\frac{12}{13},\frac{3}{13}; and \frac{3}{13},-\frac{4}{13},\frac{12}{13} are mutually perpendicular.
Solution
Think First… Two lines are perpendicular if the dot product of their direction ratios (or direction cosines) is zero.
We have the direction cosines of the three lines as
L 1: \displaystyle \frac{12}{13},\;-\frac{3}{13},\;-\frac{4}{13}
L 2: \displaystyle \frac{4}{13},\;\frac{12}{13},\;\frac{3}{13}
and L 3: \displaystyle \frac{3}{13},\;-\frac{4}{13},\;\frac{12}{13}
First, consider the first and the second lines.
The dot product of their direction cosines is
\displaystyle \frac{12}{13}\cdot\frac{4}{13}+\left(-\frac{3}{13}\right)\cdot\frac{12}{13}+\left(-\frac{4}{13}\right)\cdot\frac{3}{13}
\displaystyle =\frac{48-36-12}{169}=0
Hence, the first and the second lines are perpendicular.
Now, consider the first and the third lines.
The dot product of their direction cosines is
\displaystyle \frac{12}{13}\cdot\frac{3}{13}+\left(-\frac{3}{13}\right)\cdot\left(-\frac{4}{13}\right)+\left(-\frac{4}{13}\right)\cdot\frac{12}{13}
\displaystyle =\frac{36+12-48}{169}=0
Hence, the first and the third lines are perpendicular.
Finally, consider the second and the third lines.
The dot product of their direction cosines is
\displaystyle \frac{4}{13}\cdot\frac{3}{13}+\frac{12}{13}\cdot\left(-\frac{4}{13}\right)+\frac{3}{13}\cdot\frac{12}{13}
\displaystyle =\frac{12-48+36}{169}=0
Hence, the second and the third lines are perpendicular.
Since each pair of lines is perpendicular, the three given lines are mutually perpendicular.
Question 2: Three Dimensional Geometry 11.2
2. Show that the line through the points (1,-1,2) and (3,4,-2) is perpendicular to the line through the points (0,3,2) and (3,5,6).
Solution
Think First… When two points on a line are given, first find its direction ratios by subtracting the corresponding coordinates. Then check whether the dot product of the direction ratios is zero.
For the first line, the direction ratios are
\displaystyle a_1=3-1=2,\qquad b_1=4-(-1)=5,\qquad c_1=-2-2=-4
i.e. \displaystyle 2,\;5,\;-4
For the second line, the direction ratios are
\displaystyle a_2=3-0=3,\qquad b_2=5-3=2,\qquad c_2=6-2=4
i.e. \displaystyle 3,\;2,\;4
Now,
\displaystyle a_1a_2+b_1b_2+c_1c_2=(2)(3)+(5)(2)+(-4)(4)
\displaystyle =6+10-16=0
Since the dot product of the direction ratios is zero, the two lines are perpendicular.
Hence, the given lines are perpendicular.
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 11.2 builds upon the concepts of direction ratios and direction cosines to study the equation of a line, the angle between two lines, and the shortest distance between them. Once you become comfortable with identifying the direction ratios of a line and applying the appropriate formula, solving questions on parallel, perpendicular and skew lines becomes much more straightforward. So, practise these concepts thoroughly before moving on to the next exercise. Now, let’s proceed to the next question.
Question 3: Three Dimensional Geometry 11.2
3. Show that the line through the points (4,7,8) and (2,3,4) is parallel to the line through the points (-1,-2,1) and (1,2,5).
Solution
Think First… When two points on a line are given, first find its direction ratios by subtracting the corresponding coordinates. Two lines are parallel if their direction ratios are proportional.
For the first line, the direction ratios are
\displaystyle 2-4=-2,\qquad 3-7=-4,\qquad 4-8=-4
i.e. \displaystyle -2,\;-4,\;-4
For the second line, the direction ratios are
\displaystyle 1-(-1)=2,\qquad 2-(-2)=4,\qquad 5-1=4
i.e. \displaystyle 2,\;4,\;4
Two lines are parallel if their corresponding direction ratios are proportional, that is,
\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
Substituting the given values, we get
\displaystyle \frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}=-1
Since the corresponding direction ratios are proportional, the given lines are parallel.
Question 4: Vector Equation of line in Three Dimension
4. Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector 3\hat i+2\hat j-2\hat k.
Solution
Think First… If a line passes through a point and is parallel to a given vector, then the components of the vector are the direction ratios of the line. Use the vector form of the equation of a line.
The given point is \displaystyle (1,2,3)
and the given vector is \displaystyle 3\hat i+2\hat j-2\hat k
We know, the vector equation of a line passing through a point with position vector \vec a and parallel to the vector \vec b is given by
\displaystyle \vec r=\vec a+\lambda\vec b
Here, we have
\displaystyle \vec a=\hat i+2\hat j+3\hat k,\qquad \vec b=3\hat i+2\hat j-2\hat k
Substituting these values, we get the required equation of line
\boxed{\displaystyle \vec r=(\hat i+2\hat j+3\hat k)+\lambda(3\hat i+2\hat j-2\hat k)}
Question 5: Cartesian Equation of line in Three Dimension
5. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2\hat i-\hat j+4\hat k and is in the direction \hat i+2\hat j-\hat k.
Solution
Think First… First write the vector equation of the line. Then convert it into Cartesian form using the direction ratios.
The given position vector is \displaystyle 2\hat i-\hat j+4\hat k
and the given direction vector is \displaystyle \hat i+2\hat j-\hat k
We know, the vector equation of a line passing through a point with position vector \vec a and parallel to the vector \vec b is given by
\displaystyle \vec r=\vec a+\lambda\vec b
Here, we have
\displaystyle \vec a=2\hat i-\hat j+4\hat k,\qquad \vec b=\hat i+2\hat j-\hat k
Substituting these values, we get the vector equation of the line
\boxed{\displaystyle \vec r=(2\hat i-\hat j+4\hat k)+\lambda(\hat i+2\hat j-\hat k)}
The Cartesian equation of a line passing through (x_1,y_1,z_1) with direction ratios a,b,c is
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
Here, \displaystyle (x_1,y_1,z_1)=(2,-1,4) and the direction ratios are \displaystyle 1,2,-1.
Therefore, the Cartesian equation of the line is
\boxed{\displaystyle \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}}
Question 6: Three Dimensional Geometry 11.2
6. Find the Cartesian equation of the line which passes through the point (-2,4,-5) and is parallel to the line given by
\displaystyle \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}
Solution
Think First… Parallel lines have the same (or proportional) direction ratios. So, first identify the direction ratios of the given line and then use the Cartesian form of the equation of a line passing through the given point.
The given point is \displaystyle (-2,4,-5).
The given line is
\displaystyle \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}
Comparing it with the standard Cartesian form
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\qquad ...(1)
we see that the direction ratios of the given line are
\displaystyle 3,\;5,\;6
Since the required line is parallel to the given line, it has the same direction ratios.
Hence, the direction ratios of the required line are also, \displaystyle 3,\;5,\;6
Substituting \displaystyle (x_1,y_1,z_1)=(-2,4,-5) and \displaystyle a=3,\;b=5,\;c=6, in equation (1), we get the required equation if line:
\boxed{\displaystyle \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}}
Maths Better Tip… You should be comfortable converting a line from its vector form to its Cartesian (symmetric) form and vice versa. From the vector form, the position vector gives the point on the line, while the coefficients of the parameter give the direction ratios. Similarly, by comparing a Cartesian equation with its standard form, you can directly identify the point and the direction ratios. This conversion is frequently tested in CBSE and CUET examinations.
Question 7: Three Dimensional Geometry 11.2
7. The Cartesian equation of a line is
\displaystyle \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}
Write its vector form.
Solution
Think First… Compare the given equation with the standard Cartesian form to identify the point and the direction ratios. Then use the standard vector equation of a line.
We know that the Cartesian equation of a line is
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
where (x_1,y_1,z_1) is a point on the line and a,\;b,\;c are its direction ratios.
The corresponding vector equation of the line is
\displaystyle \vec r=\vec a+\lambda\vec b
Comparing the given equation
\displaystyle \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}
with the standard Cartesian form, we get
\displaystyle \vec a=5\hat i-4\hat j+6\hat k,\qquad \vec b=3\hat i+7\hat j+2\hat k
Substituting these values in the vector equation, we get
\boxed{\displaystyle \vec r=(5\hat i-4\hat j+6\hat k)+\lambda(3\hat i+7\hat j+2\hat k)}
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, every exercise of Three Dimensional Geometry is explained here 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.
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 next question of Exercise 11.2.
Question 8: Angle Between Pair of Lines
8. Find the angle between the following pairs of lines:
(i) \displaystyle \vec r=2\hat i-5\hat j+\hat k+\lambda(3\hat i+2\hat j+6\hat k) and
\displaystyle \vec r=7\hat i-6\hat k+\mu(\hat i+2\hat j+2\hat k)
Solution
Think First… In the vector equation of a line, the vector multiplied by the parameter gives the direction of the line. Use its components as the direction ratios and then apply the angle formula.
The given lines are
\displaystyle \vec r=2\hat i-5\hat j+\hat k+\lambda(3\hat i+2\hat j+6\hat k)
and
\displaystyle \vec r=7\hat i-6\hat k+\mu(\hat i+2\hat j+2\hat k)
The vectors multiplied by the parameters represent the directions of the two lines.
Hence, the direction ratios are
Line 1: \displaystyle 3,\;2,\;6
Line 2: \displaystyle 1,\;2,\;2
We know that if a_1,b_1,c_1 and a_2,b_2,c_2 are the direction ratios of two lines, 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|
Substituting the values, we get
\displaystyle \cos\theta=\left|\frac{3(1)+2(2)+6(2)}{\sqrt{3^2+2^2+6^2}\sqrt{1^2+2^2+2^2}}\right|
\displaystyle =\frac{3+4+12}{\sqrt{49}\sqrt{9}}
i.e. \displaystyle \cos\theta=\frac{19}{21}
Hence, the required angle between the two lines is
\boxed{\displaystyle \theta=\cos^{-1}\!\left(\frac{19}{21}\right)}
(ii) \displaystyle \vec r=3\hat i+\hat j-2\hat k+\lambda(\hat i-\hat j-2\hat k) and
\displaystyle \vec r=2\hat i-\hat j-56\hat k+\mu(3\hat i-5\hat j-4\hat k)
Solution
Think First… In the vector equation of a line, the vector multiplied by the parameter gives the direction of the line. Use its components as the direction ratios and then apply the angle formula.
The given lines are
\displaystyle \vec r=3\hat i+\hat j-2\hat k+\lambda(\hat i-\hat j-2\hat k)
and
\displaystyle \vec r=2\hat i-\hat j-56\hat k+\mu(3\hat i-5\hat j-4\hat k)
The vectors multiplied by the parameters represent the directions of the two lines.
Hence, the direction ratios are
Line 1: \displaystyle 1,\;-1,\;-2
Line 2: \displaystyle 3,\;-5,\;-4
We know that if a_1,b_1,c_1 and a_2,b_2,c_2 are the direction ratios of two lines, 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|
Substituting the values, we get
\displaystyle \cos\theta=\left|\frac{1(3)+(-1)(-5)+(-2)(-4)}{\sqrt{1^2+(-1)^2+(-2)^2}\sqrt{3^2+(-5)^2+(-4)^2}}\right|
\displaystyle =\frac{3+5+8}{\sqrt{6}\sqrt{50}}
i.e. \displaystyle \cos\theta=\frac{16}{5\sqrt{12}}=\frac{8}{5\sqrt3}
Hence, the required angle between the two lines is
\boxed{\displaystyle \theta=\cos^{-1}\!\left(\frac{8}{5\sqrt3}\right)}
Question 9: Three Dimensional Geometry 11.2
9. Find the angle between the following pair of lines:
(i) \displaystyle \frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\quad\text{and}\quad\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}
Solution
Think First… In the Cartesian form of a line, the denominators are its direction ratios. Once the direction ratios are known, use the formula for the angle between two lines.
From the given equations, the direction ratios are
Line 1: \displaystyle 2,\;5,\;-3
Line 2: \displaystyle -1,\;8,\;4
We know that if a_1,b_1,c_1 and a_2,b_2,c_2 are the direction ratios of two lines, 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|
Substituting the given values, we get
\displaystyle \cos\theta=\left|\frac{2(-1)+5(8)+(-3)(4)}{\sqrt{2^2+5^2+(-3)^2}\sqrt{(-1)^2+8^2+4^2}}\right|
\displaystyle =\frac{|-2+40-12|}{\sqrt{38}\sqrt{81}}
i.e. \displaystyle \cos\theta=\frac{26}{9\sqrt{38}}
Hence, the required angle between the given lines is
\boxed{\displaystyle \theta=\cos^{-1}\!\left(\frac{26}{9\sqrt{38}}\right)}
(ii) \displaystyle \frac{x}{2}=\frac{y}{2}=\frac{z}{1}\quad\text{and}\quad\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}
Solution
Think First… In the Cartesian form of a line, the denominators represent its direction ratios. Use these direction ratios in the formula for the angle between two lines.
From the given equations, the direction ratios are
Line 1: \displaystyle 2,\;2,\;1
Line 2: \displaystyle 4,\;1,\;8
We know that if a_1,b_1,c_1 and a_2,b_2,c_2 are the direction ratios of two lines, 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|
Substituting the given values, we get
\displaystyle \cos\theta=\left|\frac{2(4)+2(1)+1(8)}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}}\right|
\displaystyle =\frac{|8+2+8|}{\sqrt{9}\sqrt{81}}
i.e. \displaystyle \cos\theta=\frac{18}{3\times9}=\frac{2}{3}
Hence, the required angle between the given lines is
\boxed{\displaystyle \theta=\cos^{-1}\!\left(\frac23\right)}
Maths Better Tip! Sometimes, the questions do not give the equation of a line directly in the standard (symmetrical) form. Before applying any formula, always rewrite the equation in the standard form. Once it is in standard form, you can immediately identify the point on the line and its direction ratios.
The standard (symmetrical) form of a line is
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
- The line passes through the point (x_1,y_1,z_1).
- Its direction ratios are a,\;b,\;c.
- If the equation is not in this form, simplify or rearrange it first, and then compare it with the standard form.
- After identifying the point and the direction ratios, apply the required formula for the given question.
This simple habit will make questions on parallel lines, perpendicular lines, angles and shortest distance much easier to solve.
Question 10: Three Dimensional Geometry 11.2
10. Find the values of p so that the lines
\displaystyle \frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}
and
\displaystyle \frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}
are at right angles.
Solution
Think First… Before comparing two lines, first convert each equation into the standard (symmetrical) form. Then identify their direction ratios and use the condition for perpendicular lines.
The first line is
\displaystyle \frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}
Since 1-x=-(x-1), we have
\displaystyle \frac{-(x-1)}{3}=\frac{7(y-2)}{2p}=\frac{z-3}{2}
Also,
\displaystyle \frac{7(y-2)}{2p}=\frac{y-2}{2p/7}
Therefore, the standard form of the first line is
\displaystyle \frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}
Hence, its direction ratios are
\displaystyle -3,\;\frac{2p}{7},\;2
The second line is
\displaystyle \frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}
Since 7-7x=-7(x-1) and 6-z=-(z-6), we get
\displaystyle \frac{-7(x-1)}{3p}=\frac{y-5}{1}=\frac{-(z-6)}{5}
Now,
\displaystyle \frac{-7(x-1)}{3p}=\frac{x-1}{-3p/7}\qquad\text{and}\qquad\frac{-(z-6)}{5}=\frac{z-6}{-5}
Therefore, the standard form of the second line is
\displaystyle \frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5}
Hence, its direction ratios are
\displaystyle -\frac{3p}{7},\;1,\;-5
Since the lines are perpendicular, the dot product of their direction ratios must be zero. Therefore,
We get, \displaystyle (-3)\left(-\frac{3p}{7}\right)+\frac{2p}{7}(1)+2(-5)=0
\displaystyle \frac{9p}{7}+\frac{2p}{7}-10=0
\displaystyle \frac{11p}{7}=10
Hence,
\boxed{\displaystyle p=\frac{70}{11}}
Question 11: Three Dimensional Geometry 11.2
11. Show that the lines
\displaystyle \frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\quad\text{and}\quad\frac{x}{1}=\frac{y}{2}=\frac{z}{3}
are perpendicular to each other.
Solution
Think First… Compare each line with the standard Cartesian form to identify its direction ratios. If the dot product of the direction ratios is zero, the two lines are perpendicular.
The first line is already in the standard Cartesian form
\displaystyle \frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}
Therefore, its direction ratios are
Line 1: \displaystyle 7,\;-5,\;1
Similarly, for the second line
\displaystyle \frac{x}{1}=\frac{y}{2}=\frac{z}{3}
the direction ratios are
Line 2: \displaystyle 1,\;2,\;3
We know that two lines are perpendicular if the dot product of their direction ratios is zero.
\displaystyle a_1a_2+b_1b_2+c_1c_2=0
Substituting the direction ratios,
We get, \displaystyle 7(1)+(-5)(2)+1(3)
\displaystyle =7-10+3
\displaystyle =0
Hence, the given lines are perpendicular to each other.
Question 12: Shortest Distance Between Two Skew Lines
12. Find the shortest distance between the lines
\displaystyle \vec r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)
and \displaystyle \vec r=2\hat i-\hat j-\hat k+\mu(2\hat i+\hat j+2\hat k)
Solution
Think First… Identify the position vectors and the direction vectors of the two lines. Then apply the shortest distance formula for skew lines.
From the given vector equations, we have
\displaystyle \vec a_1=\hat i+2\hat j+\hat k,\qquad \vec b_1=\hat i-\hat j+\hat k
\displaystyle \vec a_2=2\hat i-\hat j-\hat k,\qquad \vec b_2=2\hat i+\hat j+2\hat k
We know that the shortest distance between two skew lines is
\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 \vec a_2-\vec a_1.
\displaystyle \vec a_2-\vec a_1=(2-1)\hat i+(-1-2)\hat j+(-1-1)\hat k
\displaystyle =\hat i-3\hat j-2\hat k
Now find the cross product \vec b_1\times\vec b_2.
\displaystyle \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix}
\displaystyle =-3\hat i+0\hat j+3\hat k
Therefore,
\displaystyle |\vec b_1\times\vec b_2|=\sqrt{(-3)^2+0^2+3^2}=3\sqrt2
Also,
\displaystyle (\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=(-3)(1)+0(-3)+3(-2)
\displaystyle =-3+0-6=-9
Hence,
\displaystyle d=\frac{|-9|}{3\sqrt2}=\frac{3}{\sqrt2}
\boxed{\displaystyle d=\frac{3}{\sqrt2}=\frac{3\sqrt2}{2}}
Question 13: Three Dimensional Geometry 11.2
13. Find the shortest distance between the lines
\displaystyle \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}
and \displaystyle \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}
Solution
Think First… Compare each line with the standard Cartesian form to identify the position vectors and the direction vectors. Then apply the shortest distance formula.
Comparing the given lines with the standard form, we get
\displaystyle \vec a_1=-\hat i-\hat j-\hat k,\qquad \vec b_1=7\hat i-6\hat j+\hat k
\displaystyle \vec a_2=3\hat i+5\hat j+7\hat k,\qquad \vec b_2=\hat i-2\hat j+\hat k
We know that the shortest distance between two skew lines is
\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 \vec a_2-\vec a_1.
\displaystyle \vec a_2-\vec a_1=(3+1)\hat i+(5+1)\hat j+(7+1)\hat k
\displaystyle =4\hat i+6\hat j+8\hat k
Now find the cross product \vec b_1\times\vec b_2.
\displaystyle \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\7&-6&1\\1&-2&1\end{vmatrix}
\displaystyle =-4\hat i-6\hat j-8\hat k
Therefore,
\displaystyle |\vec b_1\times\vec b_2|=\sqrt{(-4)^2+(-6)^2+(-8)^2}=\sqrt{116}=2\sqrt{29}
Also,
\displaystyle (\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=(-4)(4)+(-6)(6)+(-8)(8)
\displaystyle =-16-36-64=-116
Hence,
\displaystyle d=\frac{|-116|}{2\sqrt{29}}=\frac{58}{\sqrt{29}}=2\sqrt{29}
\boxed{\displaystyle d=2\sqrt{29}}
Question 14: Three Dimensional Geometry 11.2
14. Find the shortest distance between the lines whose vector equations are
\displaystyle \vec r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)
and \displaystyle \vec r=4\hat i+5\hat j+6\hat k+\mu(2\hat i+3\hat j+\hat k)
Solution
Think First… From the vector equations, identify the position vectors and the direction vectors of the two lines. Then apply the shortest distance formula.
From the given vector equations, we have
\displaystyle \vec a_1=\hat i+2\hat j+3\hat k,\qquad \vec b_1=\hat i-3\hat j+2\hat k
\displaystyle \vec a_2=4\hat i+5\hat j+6\hat k,\qquad \vec b_2=2\hat i+3\hat j+\hat k
We know that the shortest distance between two skew lines is
\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 \vec a_2-\vec a_1.
\displaystyle \vec a_2-\vec a_1=(4-1)\hat i+(5-2)\hat j+(6-3)\hat k
\displaystyle =3\hat i+3\hat j+3\hat k
Now find the cross product \vec b_1\times\vec b_2.
\displaystyle \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-3&2\\2&3&1\end{vmatrix}
\displaystyle =-9\hat i+3\hat j+9\hat k
Therefore,
\displaystyle |\vec b_1\times\vec b_2|=\sqrt{(-9)^2+3^2+9^2}=\sqrt{171}=3\sqrt{19}
Also,
\displaystyle (\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=(-9)(3)+3(3)+9(3)
\displaystyle =-27+9+27=9
Hence,
\displaystyle d=\frac{|9|}{3\sqrt{19}}=\frac{3}{\sqrt{19}}
\boxed{\displaystyle d=\frac{3}{\sqrt{19}}}
Question 15: Three Dimensional Geometry 11.2
15. Find the shortest distance between the lines whose vector equations are
\displaystyle \vec r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k
and \displaystyle \vec r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k
Solution
Think First… First rewrite both lines in the standard vector form \vec r=\vec a+\lambda\vec b. Then identify the position vectors and direction vectors before applying the shortest distance formula.
Rewrite the first line by separating the constant and variable terms.
\displaystyle \vec r=(\hat i-2\hat j+3\hat k)+t(-\hat i+\hat j-2\hat k)
Similarly, the second line becomes
\displaystyle \vec r=(\hat i-\hat j-\hat k)+s(\hat i+2\hat j-2\hat k)
Hence,
\displaystyle \vec a_1=\hat i-2\hat j+3\hat k,\qquad \vec b_1=-\hat i+\hat j-2\hat k
\displaystyle \vec a_2=\hat i-\hat j-\hat k,\qquad \vec b_2=\hat i+2\hat j-2\hat k
We know that the shortest distance between two skew lines is
\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 \vec a_2-\vec a_1.
\displaystyle \vec a_2-\vec a_1=(1-1)\hat i+(-1-(-2))\hat j+(-1-3)\hat k
\displaystyle =\hat j-4\hat k
Now find the cross product \vec b_1\times\vec b_2.
\displaystyle \vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\-1&1&-2\\1&2&-2\end{vmatrix}
\displaystyle =2\hat i-4\hat j-3\hat k
Therefore,
\displaystyle |\vec b_1\times\vec b_2|=\sqrt{2^2+(-4)^2+(-3)^2}=\sqrt{29}
Also,
\displaystyle (\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=2(0)+(-4)(1)+(-3)(-4)
\displaystyle =0-4+12=8
Hence,
\boxed{\displaystyle d=\frac{8}{\sqrt{29}}}
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.
- Making sign errors while finding direction ratios from two points: Always subtract the corresponding coordinates in the same order, i.e., (x_2-x_1,\;y_2-y_1,\;z_2-z_1).
- Not converting the line into standard form first: Before identifying the point and the direction ratios, rewrite the equation in standard form whenever necessary.
- Using the wrong condition for parallel lines: Two lines are parallel only if their corresponding direction ratios are proportional.
- Using the wrong condition for perpendicular lines: Two lines are perpendicular only if the dot product of their direction ratios (or direction cosines) is zero.
- Using the wrong point in the vector equation: Convert the given point into its position vector before writing \vec r=\vec a+\lambda\vec b.
- Making mistakes while finding the shortest distance: Use the 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|} and take the magnitude in the numerator.
- Forgetting that the shortest distance is perpendicular to both lines: The shortest distance between two skew lines is always along their common perpendicular.
- Confusing intersecting and skew lines: If the shortest distance is zero, the lines intersect and are therefore coplanar. Skew lines are always non-coplanar and never intersect.
- Skipping the final verification: Check whether the question asks for the vector equation, Cartesian equation, angle, shortest distance, or a condition such as parallelism or perpendicularity before writing the final answer.
Continue Learning
In Three Dimensional Geometry 11.2, you learned how to write the vector, Cartesian and symmetric equations of a line, determine whether two lines are parallel or perpendicular, find the angle between two lines, and calculate the shortest distance between skew lines. These concepts form the foundation for solving more advanced problems in three-dimensional geometry.
Before starting the Miscellaneous Exercise, revise the important concepts from both Exercises 11.1 and 11.2. The Miscellaneous Exercise contains a mixture of questions based on direction cosines, direction ratios, equations of lines, angles between lines, and the shortest distance between lines. A quick revision of the following topics will make the problems much easier to solve:
- The relationship between direction cosines and direction ratios, 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, Cartesian and symmetric forms of the equation of a line, and how to convert one form into another.
- The conditions for two lines to be parallel or perpendicular using their direction ratios.
- How to find the angle between two lines using the direction ratios or direction cosines.
- The shortest distance formula between two skew lines and the fact that the shortest distance is always along the common perpendicular to both lines.
- How to identify whether two lines are intersecting, parallel or skew before choosing the appropriate formula.
- Always rewrite a line in its standard form, whenever required, before identifying the point and the direction ratios.
Explore More
A good understanding of the concepts covered in Exercise 11.2 is essential for solving problems involving the equations of lines, parallel and perpendicular lines, angles between lines, and the shortest distance between skew lines. Always begin by rewriting the equation in its standard (symmetrical) form, whenever required, so that you can easily identify the point through which the line passes and its direction ratios.
While solving questions, first identify the type of problem—whether it 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 👍
