RD Sharma Solutions Class 12 Straight Line In Space Exercise 28.3

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.

Download PDF of Rd Sharma Solution for Class 12 Maths Chapter 28 Exercise 3

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rd sharma class 12 solution for maths chapter 28 ex 3 5
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Access Answers for Rd Sharma Solution Class 12 Maths Chapter 28 Exercise 3

Q1.

Solution:

Let us consider the equation of line,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 17

If lines (1) and (2) intersect, we get a common point

So for same value of λ and μ, must have,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 18

Now, let us solve (3) and (4), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 19

μ = 0

Now, let us substitute μ = 0 in equation (3), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 20

By substituting the values λ and μ in equation (5), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 21

Hence,

LHS = RHS

Q2.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 22

If lines (1) and (2) intersect, we get a common point

So for same value of λ and μ, must have,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 23

Now, let us solve (3) and (4), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 24

μ = -12

Now, let us substitute μ = -12 in equation (3), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 25

By substituting the values λ and μ in equation (5), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 26

Hence,

LHS ≠ RHS

Q3.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 27

If lines (1) and (2) intersect, we get a common point

So for same value of λ and μ, must have,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 28

Now, let us solve (3) and (4), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 29

μ = -6/4

Now, let us substitute μ = -6/4 in equation (3), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 30

By substituting the values λ and μ in equation (5), we get

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 31

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,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 32

Q4.

Solution:

Given points A(0, -1, -1) and B(4, 5, 1)

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 33

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

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 34

Let
RD Sharma Solutions for Class 12 Maths Chapter 28 – image 35

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.

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 36

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

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 37

If these lines intersect, there exists a common point.

So for some value of λ and μ, we must have

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 38

So, the equation of coefficients of
RD Sharma Solutions for Class 12 Maths Chapter 28 – image 39we 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

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 40

Hence, the point of intersection is (4, 0, -1).

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