Question 1. Show that the lines
Solution:
Given that the coordinates of any point on the first line are
⇒ x = λ, y = 2λ + 2, z = 3λ – 3
The coordinates of a general point on the second line are given by:
⇒ 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
Solution:
Given that the coordinates of any point on the first line are
⇒ x = 3λ + 1, y = 2λ – 1, z = 5λ + 1
The coordinates of a general point on the second line are given by:
⇒ 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
Solution:
Given that the coordinates of any point on the first line are
⇒ x = 3λ – 1, y = 5λ – 3, z = 7λ – 5
The coordinates of a general point on the second line are given by:
⇒ 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
⇒ x = 4λ, y = 6λ – 1, z = 2λ – 1
Also, given that the coordinates of any point on the line CD are
⇒ 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
Solution:
According to the question, it is given that the position vector of two points on the lines are
If the lines intersect, then for some value of λ and μ, we must have:
Now equate the coefficient of
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)
Solution:
Given that:
If the lines intersect, then for some value of λ and μ, we must have:
Now equate the coefficient of
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)
Solution:
Given that the coordinates of any point on the line AB are
⇒ x = 2λ + 1, y = 3λ – 1, z = λ
The coordinates of a general point on the second line are given by
⇒ 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)
Solution:
Given that the coordinates of any point on the line AB are
⇒ x = λ, y = 2λ + 2, z = 3λ – 3
The coordinates of a general point on the second line are given by
⇒ 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)
Solution:
Given that the coordinates of any point on the line AB are
⇒ x = 4λ + 5, y = 4λ + 7, z = -5λ – 3
The coordinates of a general point on the second line are given by:
⇒ 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
Solution:
Given that,
If the lines intersect, then for some value of λ and μ, we must have:
Now equate the coefficient of
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).