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RD Sharma books are one of the best materials available in the country for maths. They have a reputation for being comprehensive and detailed, so they are the most beneficial for exams. Many schools and teachers all over the country use this material for their exam paper preparation.

**Also Read -** RD Sharma Solutions For Class 9 to 12 Maths

The RD Sharma class 12th exercise 27.4 consists of 14 questions that are based on fundamentals. These questions are of Level-1 difficulty and can be solved easily with a bit of basic knowledge. Moreover, RD Sharma class 12th exercise 27.4 contains questions and answers in one place and can be a more accessible alternative for revision.

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Straight Line in Space Exercise 27.4 Question 1

(

Thus to find distance PQ we have to first find co-ordinator of Q .

Therefore ,Co-ordinates of Q

Hence

Direction of PQ is

And by comparing with given line equation, direction ratios of the given line are

(Hint – denominator terms of line equation)

PQ is perpendicular to given line,

:: By “Condition of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

::Co-ordinate of Q i.e foot of perpendicular , by putting the value of γ

In Q co-ordinate equation ,we got

Now

Distance between PQ,

Straight Line in Space Exercise 27.4 Question 2

(

:: Co-ordinates of Q

Hence,

Direction ratio of PQ is

and by comparing the given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2 , -3 , 8)

Since PQ is perpendicular at the given line.

Therefore, By “Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

::Co-ordinate of Q i.e foot of perpendicular , by putting the value of γ in Q co-ordinate equation ,we get

NOW

Distance between PQ

Straight Line in Space Exercise 27.4 Question 3

Answer - The answer of this question is(Hint – By using )

Given – Perpendicular from A ( 1,0,3)drawn at line joining points B (4,7,1) and (3,5,3).

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

Now,

Now,

:: Co-ordinates of

__Hence__

__Direction ratios pf AD__

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (-1,-2,2)

Since PQ is perpendicular at the given line.

Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

Co-ordinates of D i.e. foot of perpendicular

By putting value of *γ * in D co-ordinate equation. we get,

Straight Line in Space Exercise 27.4 Question 4

(

3,1).

points B(0,-11,3) and C(2,-3,1).

Now, lets find the equation of the line which is formed by joining points B(0,-11,3) and C(2,-3,1).

::Co-ordinates of*.*

__Hence,__

Direction ratios of AD

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,8,-2)

AD is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of D i.e.foot of the perpendicular.

By putting the value of * in D co-ordinate equation, we get*

Straight Line in Space Exercise 27.4 Question 5

Answer - The answer of this question is(Hint – By using the distance between two point formula).

Given – Point (2,3,4) and equation of the line

Solution – Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

Now,

:: Co-ordinates of

__Hence __

Direction of PQ is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (-2,6,-3)

PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of Q.,

By putting the value of γ in Q co-ordinate equation, we get

Now,

Distance between PQ

Straight Line in Space Exercise 27.4 Question 6

Answer – The answer of the given question is(Hint – By using two point formula).

Given – Point (2,4,-1) and equation of the line

Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

:: Co-ordinates of

__Hence,__

Direction of PQ is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (1,4,9)

PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of Q.,

By putting the value of *γ * in Q co-ordinate equation, we get

Now,

So, equation of perpendicular PQ is

Straight Line in Space Exercise 27.4 Question 7

As we know positive vector is given by

:: position vector of point P is

Ad from a given line we get

On comparing both side we get,

Thus, Co-ordinates of Q i.e. General point of the given line.

__Hence__

Direction ratio of PQ is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,9,5).

PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of Q.,

By putting the value of *γ * in Q co-ordinate equation, we get

:: Co-ordinates of Q i.e. foot of perpendicular

By putting the value of *γ in Q*

Now,

Distance between PQ

Straight Line in Space Exercise 27.4 Question 8

Q is on line.

Hence,

:: PQ is perpendicular on line

:: Their dot product is zero.

Compare given line equation with

Hence, Position vector of Q ,By putting the value of

Putting the value of *γ** in PQ.*

Now, Magnitude of PQ , we know that

Hence

Straight Line in Space Exercise 27.4 Question 9

Answer- The answer of the given question is(Hint- By comparing both the equation).

Given – Point P(-1,3,2) and equation of line

Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

As we know position vector is given by

:: Position vector is given by

And, from a given line, we get

On comparing both sides we get,

Thus, co-ordinate of Q i.e. general point on the given line.

Hence

Direction of PQ is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,1,3).

PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of Q,

By putting the value of *γ * in Q co-ordinate equation, we get

Position vector of Q

Now, Equation of line passing through two points with position vector

Straight Line in Space Exercise 27.4 Question 10

Answer – The answer of the question is(Hint – By putting the value of ).

Given – Point (0,2,7) and equation of the line

Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

Thus to find Distance PQ we have to first find co-ordinates of Q.

__Hence,__

Direction of PQ is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (-1,3,-2)

:: PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of Q ,

By putting the value of * in Q*

Straight Line in Space Exercise 27.4 Question 11

Answer - The answer of the question is(Hint – By comparing with line equation).

Given – Point and equation of the line

Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.

Thus to find Distance PQ we have to first find co-ordinates of Q.

__Hence,__

Direction ratio of PQ is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,-2,-1)

:: PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of Q ,

By putting the value of in Q

Straight Line in Space Exercise 27.4 Question 12

Answer - The answer of the given question is(Hint – By using two point formula).

Given – Line passing through the point

Solution –

Hence equation of line AB

__Hence,__

Direction ratio of CD is

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

CD is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

:: Co-ordinates of D,

By putting the value of * in D , we get*

Hence, Equation of line

Straight Line in Space Exercise 27.4 Question 13

(

Hence,

Line parallel to

Now,

Now, let’s find cross product of the two vectors.

And magnitude of

Thus, distance of Point from line is

**The benefits of RD Sharma class 12th exercise 27.4 material are:**

Students can refer to RD Sharma class 12 chapter 27 exercise 27.4 to clear their doubts and use it as a guide for their RD Sharma math textbook. The answers are prepared by subject experts who are easy to understand. RD Sharma class 12 solutions chapter 27 exercise 27.4 is a paperless and convenient alternative to textbooks as it covers all topics and can be accessed through any device with a browser.

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As this material is easily accessible, students can use it for revision purposes and get back to any problem they do not understand. As there are step-by-step solutions provided in this material, it is easier for students to get a good insight into the subject.

**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. How can I benefit from this material?

As it contains questions and answers in the same place, students can use this material as a guide to prepare well for their exams and score good marks.

2. Where can I find this material?

Students can find Class 12 RD Sharma chapter 27.4 exercise 27.4 on career360's website by searching the book name and exercise number.

3. Who can use the solutions?

CBSE students who want to gain more insight into the subject and score good marks in exams can use this material.

4. Can I solve NCERT questions after referring to this material?

Yes, both RD Sharma and NCERT books follow similar concepts, which is why students will also be able to solve NCERT questions.

5. Is the entire material available for free?

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