RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space Exercise 28.3, is given here for students to excel in their board exams. Solutions that are provided here will help you in getting acquainted with a wide variety of questions and thus, develop problem-solving skills. The pdf of RD Sharma Solutions for Class 12 Maths can be easily downloaded from the links provided below.
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Q1.
Solution:
Let us consider the equation of line,
If lines (1) and (2) intersect, we get a common point
So for same value of λ and μ, must have,
Now, let us solve (3) and (4), we get
μ = 0
Now, let us substitute μ = 0 in equation (3), we get
By substituting the values λ and μ in equation (5), we get
Hence,
LHS = RHS
Q2.
Solution:
If lines (1) and (2) intersect, we get a common point
So for same value of λ and μ, must have,
Now, let us solve (3) and (4), we get
μ = -12
Now, let us substitute μ = -12 in equation (3), we get
By substituting the values λ and μ in equation (5), we get
Hence,
LHS ≠RHS
Q3.
Solution:
If lines (1) and (2) intersect, we get a common point
So for same value of λ and μ, must have,
Now, let us solve (3) and (4), we get
μ = -6/4
Now, let us substitute μ = -6/4 in equation (3), we get
By substituting the values λ and μ in equation (5), we get
LHS = RHS
Since, the value of λ and μ obtained by solving equations (3) and (4) satisfies equation (5).
Hence, the given lines intersect each other.
So,
Q4.
Solution:
Given points A(0, -1, -1) and B(4, 5, 1)
So, general point on line AB is
(4λ, 4λ, 2λ – 1)
Given points C(3, 9, 4) and D(-4, 4, 4)
The equation of line passing through the points is given by
Let
So, general point on line CD is
(-7μ + 3, -5μ + 9, 0μ + 4)
(-7μ + 3, -5μ + 9, 4)
If lines AB and CD intersect, there exists a common point.
So let us find the value of λ and μ.
So from equation (3),
2λ = 4 + 1
2λ = 5
λ = 5/2
By substituting the value of λ = 5/2 in equation (2), we get
6(5/2) + 5μ = 10
5μ = 10 – 15
= -5
μ = -1
Now, by substituting the values of λ and μ in equation (1), we get
4λ + 7μ = 3
4(5/2) + 7(-1) = 3
10 – 7 = 3
3 = 3
LHS = RHS
Since, the value of λ and μ obtained by solving equations (3) and (4) satisfies equation (1).
Hence, the given lines, AB and CD intersect each other.
So,
The point of intersection of AB and CD = (-7μ+3, -5μ+9, 4)
= (-7(-)+3, -5(-1)+9, 4)
= (7+3, 5+9, 4)
= (10, 14, 4)
Hence, the point of intersection of AB and CD is (10, 14, 4).
Q5.
Solution:
Given:
The equations of lines are
If these lines intersect, there exists a common point.
So for some value of λ and μ, we must have
So, the equation of coefficients of
we get
1 + 3λ = 4 + 2μ => 3λ – 2μ = 3 …….. (1)
1 – λ = 0 => λ = 1 …………… (2)
-1 = -1 + 3μ => μ = 0 …………… (3)
By substituting the values of λ and μ in equation (1)
3λ – 2μ = 3
3(1) – 2(0) = 3
3 = 3
LHS = RHS
Since, the value of λ and μ satisfies equation (1).
Hence, the given lines intersect each other.
So,
The point of intersection by substituting the value of λ in equation (1), we get
Hence, the point of intersection is (4, 0, -1).
