RD Sharma Solutions Class 12 Straight Line In Space Exercise 28.1

RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space Exercise 28.1, is provided here. BYJU’S tutor team has prepared solutions which help students attain good marks in Maths. From the exam point of view, the solutions are solved in a simple and structured manner where students can secure an excellent score in the board exams. RD Sharma Solutions pdf can be downloaded from the links given below.

Download PDF of Rd Sharma Solution for Class 12 Maths Chapter 28 Exercise 1

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Access Answers for Rd Sharma Solution Class 12 Maths Chapter 28 Exercise 1

Q1.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 1

Q2.

Solution:

It is given that,

The direction ratios of the line are: (3+1, 4-0, 6-2) = (4, 4, 4)

Since the given line passes through (-1, 0, 2)

We know that the vector equation of a line is given as

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 2

Q3.

Solution:

Let us consider,

Vector equation of line passing through a fixed point vector a and parallel to vector b is given as

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 3

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 4

Q4.

Solution:

Let us consider,

Vector equation of line passing through a fixed point vector a and parallel to vector b is given as

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 5

Hence,

The Cartesian form of equation of the line is

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 6

Q5.

Solution:

It is given that, ABCD is a parallelogram.

Let us consider, AC and BD bisects each other at point O.

So,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 7

Let us consider position vector of point O and B are represented by
RD Sharma Solutions for Class 12 Maths Chapter 28 – image 8

So,

Equation of the line BD is the line passing through O and B is given by

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 9

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 10

Now, let us compare the coefficients of vector i, j, R

x = 2 – λ, y = -3 – 13λ, z = 4 – 17λ

By equating to λ we get,

RD Sharma Solutions for Class 12 Maths Chapter 28 – image 11

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