RD Sharma Solutions Class 12 Straight Line In Space Exercise 28.4

RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space Exercise 28.4, is provided here. Students who wish to strengthen their knowledge and have a strong grip over the subject can refer to RD Sharma Solutions. Solutions are designed in a simple language by our experts to meet the needs of all students. Here, the solutions to this exercise are provided in pdf format, which can be downloaded easily from the links given below.

Download PDF of Rd Sharma Solution for Class 12 Maths Chapter 28 Exercise 4

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Access Answers for Rd Sharma Solution Class 12 Maths Chapter 28 Exercise 4

Q1.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 41

So, the Co-ordinates of Q are

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 42

Distance between P and Q is given as

 RD Sharma Solutions for Class 12 Maths Chapter 28 – image 43

Hence, the required distance is √(4901/841) units.

Q2.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 44

So, the Co-ordinates of Q are

(2λ+1, -3λ-1, 8λ-10) = [2(1)+1, -3(1)-1, 8(1)-10]

= [3, 4, -2]

Distance between P and Q is given as

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 45

Hence,

The foot of perpendicular = (3, -4, -2)

Length of perpendicular = 2√6 units.

Q3.

Solution:

Let us consider,

The foot of the perpendicular drawn from A(1, 0, 3) to the line joining the points B(4,7,1) and C(3,5,3) be D.

The equation of line passing through points B(4,7,1) and C(3,5,3) is

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 46

So, the direction ratio of AD is (-λ+4-1), (-2λ+7-0), (2λ+1-3)

= (-λ+3), (-2λ+7), (2λ-2)

Line AD is perpendicular to BC so,

a1a2 + b1b2 + c1c2 = 0

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 47

Hence,

Coordinates of D are:

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 48

Q4.

Solution:

Given:

D is the foot of perpendicular from A(1, 0, 4) on BC.

So,

Equation of line passing through B, C is

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 49

Coordinates of D = (2λ, 8λ-11, -2λ+3)

Direction ratios of AD is (2λ-1), (8λ-11-0), (-2λ+3-4)

= (2λ-1), (8λ-11), (-2λ-1)

Line AD is perpendicular to BC so,

a1a2 + b1b2 + c1c2 = 0

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 50

Hence,

Coordinates of D = (2λ, 8λ-11, -2λ+3)

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 51

Q5.

Solution:

Let us consider,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 52Coordinates of Q = (-2λ+4, 6λ, -3λ+1)

Direction ratios of PQ = (-2λ+4-2), (6λ-3), (-3λ+1-4)

= (-2λ+2), (6λ-3), (-3λ-3)

Since PQ is perpendicular to the given line,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 53

Coordinates of Q = (-2λ+4, 6λ, -3λ+1)

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 54

Distance between P and Q is given as

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 55

Hence,

Perpendicular distance from (2,3,4) to given line is (3√101)/49 units.

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