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RD Sharma books are one of the best materials available for CBSE students for maths. CBSE schools use them all over the country, and most students prefer them as they are very informative. As maths is a vast subject that deals with many concepts, it is essential for students to refer to detailed material. This is where RD Sharma's class 12th exercise 27.1 comes in to help.

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RD Sharma solutions of Straight line in space exercise 27.1 is an informative material that helps students get a better insight on the concepts. RD Sharma class 12th exercise 27.1 consists of 18 questions that are of Level-1 difficulty and cover most of the fundamentals of the chapter. Students can complete this exercise fairly quickly as their difficulty is low.

Straight Line in Space exercise 27.1 question 1

Answer :-Hint :-

Given :-

Line is through point (5 , 2 , -4) and parallel to vector

Solution :-

Position vector of (5 , 2 , -4) is

Direction Ratio = (3 , 2 , -8 )

Vector equation of line which passes through (5 , 2 , -4 ) and parallel to vector is , where ? is constant parameter

Cartesian Equation :-

Straight Line in Space exercise 27.1 question 2

Answer :-Hint :-

Given :-

Points (-1 , 0 , 2) and (3 , 4 , 6)

Solution :-

Position Vector of the point ( -1 , 0 , 2)

Position Vector of the point (3 , 4 , 6)

Now equation of line passing through the point where position vector are a and b

straight line in space exercise 27.1 question 3

Answer :-Hint :-

Given :-

Line passes through (5 , -2 , 4) and parallel to

Solution :-

Position Vector of the point (5 , -2 , 4) is

and

parallel to

Direction ratio = (2 , -1 , 3)

Vector equation of the line passes through the point (5 , -2 , 4) and parallel to

, where ? is constant

And Cartisian form

Straight Line in Space exercise 27.1 question 4

Answer :-Hint :-

Given :-

Position vector of the point and in the direction of

Solution :-

Position vector of point

, point = ( 2 , -3 , 4)

And position vector of parallel line(in the direction)

, Direction ratio = (3 , 4 , -5 )

Vector Equation :-

, where ? is constant parameter

Cartesian Equation :-

Straight Line in Space exercise 27.1 question 5

Answer:Hint: Position vector of the mid point

Given : ABCD is a Parallelogram

Solution:

We know that the position vector of the mid-point of and is is

Let the position vector of D be

Position vector of mid-point of A and C = Position vector of mid-point of B and D

Comparing the coefficients of & we get

Position vector of point . The vector equation of line BD passing through the points with position vectors

Here,

Vector equation of the required line is

………….. (1)

Here, is a parameter.

Reducing (1) to Cartesian form, we get

Comparing the coefficients of & we get

Hence, the Cartesian form of (1) is

Straight Line in Space exercise 27.1 question 6

Answer :-Hint :-

Given :-

Two points A (1 , 2 , -1) and B ( 2 , 1 , 1)

Solution :-

Position vector of A (1 , 2 , -1)

Position vector of B (2 , 1 , 1)

Vector Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)

, where ? is parameter

Cartisian Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)

Straight Line in Space exercise 27.1 question 7

Answer :-Hint :-

Given :-

The line passes through (1 , 2 , 3) and parallel to the vector

Solution :-

Position vector of (1 , 2 , 3)

Direction Ratio (1 , -2 , 3)

Vector equation of the line passes through (1 , 2 , 3) and

parallel to

, where ? is parameter

Cartesian Equation

Straight Line in Space exercise 27.1 question 8

Answer :-Hint :-

Given :-

The line passes through (2 , -1 , 1) and parallel to equation

Solution :-

P.V of (2 , -1 , 1)

D.R of is (2 , 7 , -3)

Vector Equation

, where ? is parameter

Straight Line in Space exercise 27.1 question 9

Answer :-Hint :-

Given :-

Solution :-

We know that

….where ? is constant parameter.

Straight Line in Space exercise 27.1 question 10

Answer :-Hint :-

Given :-

The line passing through (1 , -1 , 2) and parallel to the line whose equation

Solution :-

Cartesian Equation of a line passing through a point (1,-1,2) and parallel to line

is the point and a,b,c are the direction ratios of the parallel line.)

is the required cartesian equation of the line.

Let,

We know that

is the required vector equation.

Straight Line in Space exercise 27.1 question 11

Answer :-Hint :-

Given :-

Solution :-

Direction ratios are (-2 , 6 , -3)

So, its directional cosines will be

We know that

Where ? is constant parameter.

Straight Line in Space exercise 27.1 question 12

Answer :-(a , 1 , c)

Hint :-

Take and and find y

Given :-

and

Solution :-

and

?

? ----- (1) ? ----- (2)

By (1) and (2)

D.R = ( a , 1 , c)

Let

We know that

Straight Line in Space exercise 27.1 question 13

Answer :-Hint :-

Given :-

The line passes through the point and parallel to the line joining with position vectors and

Solution :-

The points of position vector and are P(1 , -1 , 4) and Q(2 , 1 , 2)

D.R of PQ =Q-P= (1 , 2 , -2) ,

Points of P.V are (1 , -2 , -3)

The Cartesian equation of the line having the point (1 , -2 , -3) and

D.R (1 , 2 , -2) are

And vector equation are

Where is constant parameter.

Straight Line in Space exercise 27.1 question 14

Answer :-Hint :-

Distance Formula :

Given :-

Point are P (1 , 3 , 3)

And line are

Solution :-

-------- (1)

Let the point be

, ,

The distance of and P (1 , 3 , 3) = 5

Squaring both sides

∴ when ; , ,

& when ; , ,

Straight Line in Space exercise 27.1 question 15

Answer :- The points are collinearHint :-

Given :-

P.V are

Solution :-

The points of P.V are (-2 , 3 , 0) , (1 , 2 , 3) , (7 , 0 , 9)

The equation of the line passing through the point (-2,3,0) and (1 , 2 , 3) are

----- (1)

Putting (7 , 0 , 9) in (1) , we get

Since ( 7 , 0 , 9) satisfying equation (1)

Hence it is collinear.

Straight Line in Space exercise 27.1 question 16

Answer :-Hint :-

Given :-

The line passes through the point ( 1 , 2 , 3) and parallel to the line

Solution :-

The P.V of the point (1 , 2 , 3) will be

The line can be written as ---------- (1)

So the direction ratios can be

Cartesian equation of the line passing through ( 1 , 2 , 3) and parallel to (1)

Vector Equation

Straight Line in Space exercise 27.1 question 17

Answer :-…cartesian equation of line

…vector equation of line

…the fixed point through which the line passes ;

….direction ratios of the line

Hint :-

Given :-

Solution :-

The given line can be written as

Thus the line passes through the point

vector equation of the point will be

The direction ratios are proportional to

and its vector equation will be

Vector Equation of line

Straight Line in Space exercise 27.1 question 18

Answer :-Hint :-

Given :-

Solution :-

-------------- (1)

Direction ratios of (1) will be ( 7 , -5 , 1 )

hence,

P.V of

**Here are the benefits of using RD Sharma class 12th exercise 27.1 material:**

RD Sharma class 12 solutions chapter 27 exercise 27.1 material follows the CBSE syllabus and is updated to the latest version of the book. Students can use it to prepare for their exams and also for their homework. Moreover, they can stay ahead of the classes by preparing this material in advance to get more time for revision. This way, students will have fewer doubts and can understand more clearly.

RD Sharma class 12th exercise 27.1 solutions are prepared by experts and are exam-oriented. Every answer has step-by-step solutions which make it easier for students to understand. Additionally, each answer goes through a series of quality checks to ensure that the most accurate information is available to students.

RD Sharma class 12 chapter 27 exercise 27.1 material is also beneficial for revision purposes. Students can come back to it any time if they do not understand a particular question. Moreover, as the questions and answers are available in one place, it saves a lot of time for preparation.

RD Sharma class 12 chapter 27 exercise 27.1 is available on Career360’s website. Students can access them through any device with an internet connection. This is a convenient and paperless alternative to textbooks which is made available free of cost.

As maths is a vast subject, students must get as much knowledge as possible. RD Sharma class 12th exercise 27.1 is made to fulfill this requirement. Students can systematically divide up their work and study to reduce the burden and re-collect as many concepts as possible. This ensures that they score good marks in exams and excel in their math game.

**RD Sharma Chapter wise Solutions**

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

1. Who can use this material?

Students who want to gain more knowledge on the subject and discover alternative methods of solving the problems can use Class 12 RD Sharma chapter 27 exercise 27.1 solution material.

2. Does this material follow CBSE syllabus?

Yes, Class 12 RD Sharma chapter 27 exercise 27.1 solution material complies with the CBSE syllabus and is updated to the latest version.

3. Can I finish my homework through this material?

This material covers all concepts which means that students can use it as a guide to finish their homework.

4. Where can I find this material?

Students can access this material through career360's website by searching the book name and exercise number.

5. Are there any hidden charges?

No, RD Sharma class 12 solution of Straight line in space exercise 27.1 material is absolutely free without any additional costs.

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