In Class 12 Mathematics, Chapter 28, "The Straight Line in Space" deals with the concepts of the straight lines in three-dimensional space. This chapter builds upon the understanding of the straight lines in the two dimensions and extends it to three dimensions in which is crucial for the students aiming to excel in higher studies in the fields like engineering, architecture and physics.
The Straight Line in Space
The concept of a straight line in space involves understanding its direction ratios, direction cosines and various forms of equations of the straight line such as the vector form, parametric form and Cartesian form. This chapter also explores the shortest distance between the two skew lines and the angle between the two lines which are vital in solving the complex geometrical problems in the three-dimensional space.
Class 12 RD Sharma Solutions - Exercise 28.3
Question 1. Show that the lines \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} and \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} intersect and find their point of intersection.
Solution:
Given that the coordinates of any point on the first line are
\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}=\lambda
⇒ x = λ, y = 2λ + 2, z = 3λ - 3
The coordinates of a general point on the second line are given by:
\frac{-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}=\mu
⇒ x = 2μ + 2, y = 3μ + 6, z = 4μ + 3
If the lines intersect, for some values of λ and μ, we must have:
λ - 2μ = 2 ......(1)
2λ - 3μ = 4 ......(2)
3λ - 4μ = 6 .....(3)
Solving this system of equations, we get
λ = 2 and μ = 0
On substituting the values in eq(3), we have
LHS = 3(2) - 4(0)
= 6 = RHS
Thus, the given lines intersect at (2, 6, 3).
Question 2. Show that the lines \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} and \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2} do not intersect.
Solution:
Given that the coordinates of any point on the first line are
\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda
⇒ x = 3λ + 1, y = 2λ - 1, z = 5λ + 1
The coordinates of a general point on the second line are given by:
\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu
⇒ x = 4μ - 2, y = 3μ + 1, z = -2μ - 1
If the lines intersect, for some values of λ and μ, we must have:
3λ - 4μ = -3 ......(1)
2λ - 3μ = 2 ......(2)
5λ + 2μ = -2 .....(3)
Solving this system of equations, we get
λ = -17 and μ = -12
On substituting the values in eq(3), we have
LHS = 3(-17) + 2(-12)
= -75 ≠ RHS
Thus, the given lines do not intersect with each other.
Question 3. Show that the lines \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} and \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5} intersect and find their point of intersection.
Solution:
Given that the coordinates of any point on the first line are
\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda
⇒ x = 3λ - 1, y = 5λ - 3, z = 7λ - 5
The coordinates of a general point on the second line are given by:
\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu
⇒ x = 2μ + 2, y = 3μ + 6, z = 4μ + 3
If the lines intersect, for some values of λ and μ, we must have:
3λ - μ = 3 ......(1)
5λ - 3μ = 7 ......(2)
7λ - 5μ = 11 .....(3)
Solving this system of equations, we get
λ = 1/2 and μ = -3/2
On substituting the values in eq(3), we have
LHS = 3(2) - 4(0)
= -3/2 = RHS
Now put the value of λ in first equation and we get
x = 1/2, y = -1/2, z = -3/2
Thus, the given lines intersect at (1/2, -1/2, -3/2).
Question 4. Prove that the line through (0, -1, -1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(-4, 4, 4). Also, find their point of intersection.
Solution:
Given that the coordinates of any point on the line AB are
\frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}=\lambda
⇒ x = 4λ, y = 6λ - 1, z = 2λ - 1
Also, given that the coordinates of any point on the line CD are
\frac{x-3}{3+4}=\frac{y-9}{9-4}=\frac{z-4}{4-4}=\mu
⇒ x = 7μ + 3, y = 5μ + 9, z = 4
If the lines intersect, for some values of λ and μ, we must have:
4λ - 7μ = 3 ......(1)
6λ - 5μ = 10 ......(2)
λ = 5/2 .....(3)
⇒ λ = 5/2 and μ = 1.
On substituting the values in eq(3), we have
LHS = 4(5/2) - 7(1)
= 3 = RHS
Now put the value of λ in line AB, we get
x = 10, y = 14, z = 4
Thus, the given lines AB and CD intersect at point (10, 14, 4).
Question 5. Prove that the line \vec{r}=(\hat{i}+\hat{j}-\hat{k})+λ(3\hat{i}-\hat{j}) and \vec{r}=(4\hat{i}-\hat{k})+μ(2\hat{i}+3\hat{k}) intersect and find their point of intersection.
Solution:
According to the question, it is given that the position vector of two points on the lines are
\vec{r}=(\hat{i}+\hat{j}-\hat{k})+λ(3\hat{i}-\hat{j})
\vec{r}=(4\hat{i}-\hat{k})+μ(2\hat{i}+3\hat{k})
If the lines intersect, then for some value of λ and μ, we must have:
(1+3\lambda)\hat{i}+(1-\lambda)\hat{j}-\hat{k}=(4+2\mu)\hat{i}+0\hat{j}+(3\mu-1)\hat{k}
Now equate the coefficient of \hat{i}, \hat{j},\hat{k} we get
1 + 3λ = 4 + 2μ ......(1)
1 - λ = 0 .....(2)
-1 = -1 +3μ .....(3)
On solving the equation, we get
λ = 1 and μ = 0.
Now, substituting the values in eq(1), we get
1 + 3(1) = 4 + 2(0)
4 = 4
LHS = RHS
Thus, the coordinates of the point of intersection of the two lines are (4, 0, -1).
Question 6. Determine whether the following pairs of lines intersect or not:
(i) \vec{r}=(\hat{i}-\hat{j})+\lambda( 2\hat{i}+\hat{k}) and \vec{r}=(2\hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}-\hat{k}).
Solution:
Given that:
\vec{r}=(\hat{i}-\hat{j})+\lambda( 2\hat{i}+\hat{k})
\vec{r}=(2\hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}-\hat{k}).
If the lines intersect, then for some value of λ and μ, we must have:
(1+2\lambda)\hat{i}-\hat{j}+\lambda\hat{k}=(2+\mu)\hat{i}+(-1+\mu)\hat{j}-\mu\hat{k}
Now equate the coefficient of \hat{i}, \hat{j},\hat{k} we get
1 + 2λ = 2 + μ .....(1)
-1 = -1 + μ .....(2)
λ = -μ .....(3)
On solving the equations, we get
λ = 0 and μ = 0.
Now, substitute the values in eq(1), we get
1 + 2λ = 2 + μ
1 + 2(0) = 2 + 0
1 ≠ 2
LHS ≠ RHS
Thus, the given lines do not intersect.
(ii) \frac{x-1}{2}=\frac{y+1}{3}=z and \frac{x+1}{5}=\frac{y-2}{1};z=2
Solution:
Given that the coordinates of any point on the line AB are
\frac{x-1}{2}=\frac{y+1}{3}=z=\lambda
⇒ x = 2λ + 1, y = 3λ - 1, z = λ
The coordinates of a general point on the second line are given by
\frac{x+1}{5}=\frac{y-2}{1}=\mu;z=2
⇒ x = 5μ - 1, y = μ + 2, z = 2
If the lines intersect, for some values of λ and μ, we must have:
2λ - 5μ = -2 ......(1)
3λ - μ = 3 ......(2)
λ = 2 .....(3)
Solving this system of equations, we get
λ = 2 and μ = 3
On substituting the values in eq(3), we have
LHS = 2(2) - 5(3)
= -2 ≠ RHS
Thus, the given lines do not intersect each other.
(iii) \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0} and \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}
Solution:
Given that the coordinates of any point on the line AB are
\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda
⇒ x = λ, y = 2λ + 2, z = 3λ - 3
The coordinates of a general point on the second line are given by
\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}=\mu
⇒ x = 2μ + 4, y = 0, z = 3μ - 1
If the lines intersect, for some values of λ and μ, we must have:
λ - 2μ = 2 .......(1)
2λ - 3μ = 4 ......(2)
3λ - 4μ = 6 ......(3)
On solving this system of equations, we get
λ = 1 and μ = 0
On substituting the values in eq(3), we have
LHS = 3(1) - 2(0)
= 3 = RHS
Thus, the given lines intersect at (4, 0, -1).
(iv) \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5} and \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}
Solution:
Given that the coordinates of any point on the line AB are
\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda
⇒ x = 4λ + 5, y = 4λ + 7, z = -5λ - 3
The coordinates of a general point on the second line are given by:
\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu
⇒ x = 7μ + 8, y = μ + 4, z = 3μ + 5
If the lines intersect, for some values of λ and μ, we must have:
4λ - 7μ = 3 .......(1)
4λ - μ = -3 ......(2)
5λ + 3μ = -8 ......(3)
On solving this system of equations, we get
λ = -1 and μ = -1
On substituting the values in eq(3), we have
LHS = 5(-1) - 3(-1)
= -8 = RHS
Thus, the given lines intersect at (1, 3, 2).
Question 7. Show that the lines \vec{r}=(3\hat{i}+2\hat{j}-4\hat{k})+\lambda( \hat{i}+2\hat{j}+2\hat{k}) and \vec{r}=(5\hat{i}-2\hat{j})+\mu(3\hat{i}+2\hat{j}+6\hat{k}) are intersecting. Hence, find their point of intersection.
Solution:
Given that,
\vec{r}=(3\hat{i}+2\hat{j}-4\hat{k})+\lambda( \hat{i}+2\hat{j}+2\hat{k})
\vec{r}=(5\hat{i}-2\hat{j})+\mu(3\hat{i}+2\hat{j}+6\hat{k})
If the lines intersect, then for some value of λ and μ, we must have:
(3+\lambda)\hat{i}-\hat{j}+(2+2\lambda)\hat{j}+(2\lambda-4)\hat{k}=(2+\mu)\hat{i}+(-1+\mu)\hat{j}-\mu\hat{k}
Now equate the coefficient of \hat{i}, \hat{j},\hat{k} we get
3 + λ = 5 + 3μ ........(1)
2 + 2λ = -2 + 2μ ........(2)
2λ - 4 = 6μ ........(3)
Solving the equation, we have:
λ = -4 and μ = -2.
On substituting the values, we get
LHS = 2(-4) - 4
= -12
RHS = 6(-2)
= -12
Thus, the given lines intersect at point(-1, -6, -12).
Practice Questions:
1. Find the direction cosines of the line passing through the points (1, 2, 3) and (4, 6, 8).
2. Determine the direction ratios of the line with equation (x-2)/3 = (y+1)/4 = (z-5)/2.
3. Find the equation of the line passing through the point (1, -1, 2) with direction cosines proportional to 2, -3, 6.
4. Calculate the direction cosines of the line making equal angles with the coordinate axes.
5. Find the angle between the line 2x - y + z = 3, x + y + z = 1 and the positive direction of the x-axis.
6. Determine if the lines with direction ratios (1, 2, 3) and (2, 4, 6) are parallel.
7. Find the direction cosines of the line perpendicular to both the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5.
8. Calculate the angle between the line x = 1 + 2t, y = 2 - t, z = 3t and the xy-plane.
9. Find the equation of the line passing through the point (2, 3, -1) and parallel to the line x/2 = y/3 = z/4.
10. Determine the direction cosines of the line which makes angles 60°, 60°, and 45° with the positive directions of x, y, and z axes respectively.
Read more:
Conclusion
Exercise 28.3 in Chapter 28 "The Straight Line in Space" typically focuses on direction cosines and direction ratios of lines in three-dimensional space. Key concepts include, Definition and properties of direction cosines and direction ratios, Relationship between direction cosines and direction ratios, Finding equations of lines given direction cosines/ratios and a point, Determining angles between lines and coordinate axes ,Parallel and perpendicular lines in terms of direction cosines/ratios
Explore
Basic Arithmetic
Algebra
Geometry
Trigonometry & Vector Algebra
Calculus
Probability and Statistics
Practice