Careers360 Logo
RD Sharma Class 12 Exercise 27.1 Straight Line in Space Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 27.1 Straight Line in Space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 08:09 PM IST

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.

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.

RD Sharma Class 12 Solutions Chapter27 Straight line in space - Other Exercise

Straight Line in Space Excercise: 27.1

Straight Line in Space exercise 27.1 question 1

Answer :-
r=5ı^+2ȷ^4k^+λ(3ı^+2ȷ^8k^):x53=y22=z+48
Hint :-
r=a+λb
Given :-
Line is through point (5 , 2 , -4) and parallel to vector 3i^+2j^8k^
Solution :-
Position vector of (5 , 2 , -4) is
a=5ı^+2ȷ^4k^ and b=3ı^+2ȷ^8k^
Direction Ratio = (3 , 2 , -8 )
Vector equation of line which passes through (5 , 2 , -4 ) and parallel to vector 3i^+2j^8k^ is a=5ı^+2ȷ^4k^ and b=3ı^+2ȷ^8k^ , where ? is constant parameter
=(5ı^+2ȷ^4k^)+λ(3ı^+2ȷ^8k^)
Cartesian Equation :-
xx1a=yy1b=zz1cx53=y22=z+48

Straight Line in Space exercise 27.1 question 2

Answer :-
r=(ı^+2k^)+λ(4ı^+4ȷ^+4k^)
Hint :-
r=a+λ(ba)
Given :-
Points (-1 , 0 , 2) and (3 , 4 , 6)
Solution :-
Position Vector of the point ( -1 , 0 , 2)
a=i^+2k^
Position Vector of the point (3 , 4 , 6)
b=3i^+4j^+6k^
Now equation of line passing through the point where position vector are a and b
r=a+λ(ba)r=(ı^+2k^)+λ(3ı^+4ȷ^+6k^+ı^2k^)=(ı^+2k^)+λ(4ı^+4ȷ^+4k^)

straight line in space exercise 27.1 question 3

Answer :-
r=5ı^2^ȷ^+4k^+λ(2ı^ȷ^+3k^)x52=y+21=z43
Hint :-
r=a+λb
Given :-
Line passes through (5 , -2 , 4) and parallel to 2ı^ȷ^+3k^
Solution :-
Position Vector of the point (5 , -2 , 4) is
a=5ı^2^ȷ^+4k^ and
parallel to b=2ı^ȷ^+3k^
Direction ratio = (2 , -1 , 3)
Vector equation of the line passes through the point (5 , -2 , 4) and parallel to 5ı^2^ȷ^+4k^
r=a+λb
=5ı^2^ȷ^+4k^+λ(2ı^ȷ^+3k^), where ? is constant
And Cartisian form
xx1a=yy1b=zz1cx52=y+21=z43

Straight Line in Space exercise 27.1 question 4

Answer :-
r=(2ı^3ȷ^+4k^)+λ(3ı^+4ȷ^5k^)x23=y+34=z45
Hint :-
r=a+λb
Given :-
Position vector of the point =2ı^3ȷ^+4k^ and in the direction of 3ı^+4ȷ^5k^
Solution :-
Position vector of point
a=2ı^3ȷ^+4k^ , point = ( 2 , -3 , 4)
And position vector of parallel line(in the direction)
b=3ı^+4ȷ^5k^ , Direction ratio = (3 , 4 , -5 )
Vector Equation :-
r=a+λb , where ? is constant parameter
=(2ı^3ȷ^+4k^)+λ(3ı^+4ȷ^5k^)
Cartesian Equation :-
xx1a=yy1b=zz1cx23=y+34=z45

Straight Line in Space exercise 27.1 question 5

Answer:
r=(2ı^3ȷ^+4k^)+λ(ı^13ȷ^+17k^)x21=y+313=z417
Hint: Position vector of the mid point
Given : ABCD is a Parallelogram
Solution:
We know that the position vector of the mid-point of a and is bis a+b2
Let the position vector of D be xi^+yj^+zk^
Position vector of mid-point of A and C = Position vector of mid-point of B and D
(4i^+5j^10k^)+(i^+2j^+k^)2=(2i^3j^+4k^)+(xi^+yj^+zk^)232i^+72j^92k^=(x+22)i^+(3+y2)j^+(4+z2)k^
Comparing the coefficients of i^,j^ & k^ we get
(x+22)=32x=13+y2=72y=104+z2=92z=13
Position vector of point D=i^+10j^13k^. The vector equation of line BD passing through the points with position vectors
a(B)andb(D)
r=a+λ(ba)
Here,
a=2i^3j^+4k^b=i^+10j^13k^
Vector equation of the required line is
r=(2i^3j^+4k^)+λ{(i^+10j^13k^)(2i^3j^+4k^)}r=(2i^3j^+4k^)+λ(i^+13j^17k^) ………….. (1)
Here, λis a parameter.
Reducing (1) to Cartesian form, we get
xi^+yj^+zk^=(2i^3j^+4k^)+λ(i^+13j^17k^)[ Putting r=xi^+yj^+zk^ in (1)]xi^+yj^+zk^=(2λ)i^+(3+13λ)j^+(417λ)k^
Comparing the coefficients of i^,j^ & k^ we get
x=2λ,y=3+13λ,z=417λx21=λ,y+313=λ,z417=λx21=y+313=z417=λx21=y+313=z417=λ
Hence, the Cartesian form of (1) is
x21=y+313=z417

Straight Line in Space exercise 27.1 question 6

Answer :-
r=(ı^+2ȷ^k^)+λ(ı^ȷ^+2k^)x11=y21=z+12
Hint :-
r=a+λ(ba)xx1x2x1=yy1y2y1=zz1z2z1
Given :-
Two points A (1 , 2 , -1) and B ( 2 , 1 , 1)
Solution :-
Position vector of A (1 , 2 , -1)
a=ı^ȷ^+2k^
Position vector of B (2 , 1 , 1)
b=2ı^+ȷ^+k^ba=ı^ȷ^+2k^
Vector Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)
r=a+λ(ba) , where ? is parameter
=(ı^2ȷ^+k^)+λ(ı^ȷ^+2k^)
Cartisian Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)
xx1x2x1=yy1y2y1=zz1z2z1x11=y21=z+12

Straight Line in Space exercise 27.1 question 7

Answer :-
r=(ı^+2ȷ^+3k^)+λ(ı^2ȷ^+3k^)x11=y22=x33
Hint :-
r=a+λb
Given :-
The line passes through (1 , 2 , 3) and parallel to the vector ı^+2ȷ^+3k^
Solution :-
Position vector of (1 , 2 , 3)a=(ı^+2ȷ^+3k^)
b=ı^2ȷ^+3k^
Direction Ratio (1 , -2 , 3)
Vector equation of the line passes through (1 , 2 , 3) and
parallel to ı^2ȷ^+3k^
r=a+λb , where ? is parameter
=(ı^+2ȷ^+3k^)+λ(ı^2ȷ^+3k^)
Cartesian Equation
xx1a=yy1b=zz1cx11=y22=x33

Straight Line in Space exercise 27.1 question 8

Answer :-
(2ı^ȷ^+k^)+λ(2ı^+7ȷ^3k^)
Hint :-
a+λb
Given :-
The line passes through (2 , -1 , 1) and parallel to equation x32=y+17=z23
Solution :-
P.V of (2 , -1 , 1) a=2ı^ȷ^+k^
D.R of x32=y+17=z23 is (2 , 7 , -3)
b=2i^+7j^3k^
Vector Equation
r=a+λb , where ? is parameter
= (2ı^ȷ^+k^)+λ(2ı^+7ȷ^3k^)

Straight Line in Space exercise 27.1 question 9

Answer :-
(5ı^4ȷ^+6k^)+λ(3ı^+7ȷ^+2k^)
Hint :-
r=xı^yȷ^+zk^
Given :-
x53=y+47=z62
Solution :-
$$x53=y+47=z62=λ( say )x=3λ+5,y=7λ4,z=2λ+6
We know that
r=xı^yȷ^+zk^=(3λ+5)ı^+(7λ4)ȷ^+(2λ+6)k^=(5ı^4ȷ^+6k^)+λ(3ı^+7ȷ^+2k^)….where ? is constant parameter.

Straight Line in Space exercise 27.1 question 10

Answer :-
x11=y+12=z22r=(ı^ȷ^+2k^)+λ(ı^+2ȷ^2k^)
Hint :-
r=a+λb
Given :-
The line passing through (1 , -1 , 2) and parallel to the line whose equation x31=y12=z+12
Solution :-
Cartesian Equation of a line passing through a point (1,-1,2) and parallel to line
x31=y12=z+12is
xx1a=yy1b=zz1c
(x1,y1,z1) is the point and a,b,c are the direction ratios of the parallel line.)
x11=y+12=z22is the required cartesian equation of the line.
Let, x11=y+12=z22=λ (say) 
We know that
r=xı^yȷ^+zk^=(λ+1)ı^+(2λ1)ȷ^+(2λ+2)k^=(ı^ȷ^+2k^)+λ(ı^+2ȷ^2k^)is the required vector equation.

Straight Line in Space exercise 27.1 question 11

Answer :-
27,67,37;r=(4ı^+k^)+λ(2ı^+6ȷ^3k^)
Hint :-
r=xı^yȷ^+zk^
Given :-
4x2=y6=1z3
Solution :-
4x2=y6=1z3x42=y6=z13
Direction ratios are (-2 , 6 , -3)
So, its directional cosines will be
2(2)2+62+(3)2,6(2)2+62+(3)2,3(2)2+62+(3)2
=27,67,37
Letx42=y6=z13=λ(say)
x=2λ+4,y=6λ,z=3λ+1
We know that
r=xı^+yȷ^+zk^
=(2λ+4)ı^+(6λ)ȷ^+(3λ+1)k^
=(4ı^+k^)+λ(2ı^+6ȷ^3k^)
Where ? is constant parameter.

Straight Line in Space exercise 27.1 question 12

Answer :-
xba=y01=zdcr=(bı^+dk^)+λ(aı^+ȷ^+ck^)
(a , 1 , c)
Hint :-
Take x=ay+b and z=cy+d and find y
Given :-
x=ay+b and z=cy+d
Solution :-
x=ay+b and z=cy+d
? xb=ay cy=zd
? y=xba ----- (1) ? y=zdc ----- (2)
By (1) and (2)
xba=y=zdcxba=y01=zdc
D.R = ( a , 1 , c)
Let xba=y=zdc=λ(say)
x=aλ+b,y=λ,z=cλ+d
We know that
r=xı^+yȷ^+zk^
=(aλ+b)ı^+λȷ^+(cλ+d)k^
=(bı^+dk^)+λ(aı^+ȷ^+ck^)

Straight Line in Space exercise 27.1 question 13

Answer :-
x11=y+22=z+32r=(ı^2ȷ^3k^)+λ(ı^+2ȷ^2k^)
Hint :-
r=xı^yȷ^+zk^
Given :-
The line passes through the point ı^2ȷ^3k^ and parallel to the line joining with position vectors ı^ȷ^+4k^ and 2ı^+ȷ^+2k^
Solution :-
The points of position vector ı^ȷ^+4k^ and 2ı^+ȷ^+2k^ are P(1 , -1 , 4) and Q(2 , 1 , 2)
D.R of PQ =Q-P= (1 , 2 , -2) , b=ı^+2ȷ^2k^
Points of P.V b=ı^2ȷ^3k^ are (1 , -2 , -3)
The Cartesian equation of the line having the point (1 , -2 , -3) and
D.R (1 , 2 , -2) are
x11=y+22=z+32
And vector equation are
r=a+λb=(ı^2ȷ^3k^)+λ(ı^+2ȷ^2k^)
Where λ is constant parameter.

Straight Line in Space exercise 27.1 question 14

Answer :-
(x,y,z)=(2,1,3)or(4,3,7)
Hint :-
Distance Formula : (x1x2)2+(y1y2)2+(z1z2)2
Given :-
Point are P (1 , 3 , 3)
And line are x+23=y+12=z32
Solution :-
x+23=y+12=z32 -------- (1)
Let the point be x1,y1,z1
x+23=y+12=z32=λ(say)
x1=3λ2 , y1=2λ1 , z1=2λ+3
The distance of (x1,y1,z1) and P (1 , 3 , 3) = 5
(3λ21)2+(2λ13)2+(2λ+33)2=5
Squaring both sides
(3λ3)2+(2λ4)2+(2λ)2=259λ218λ+9+4λ216λ+16+4λ2=2517λ234λ=017λ(λ2)=0λ=0,λ=2
∴ when λ=0;x=2 , y=1 , z=3
& when λ=2;x=4 , y=3 , z=7
(x,y,z)=(2,1,3)or(4,3,7)

Straight Line in Space exercise 27.1 question 15

Answer :- The points are collinear
Hint :-
xx1x2x1=yy1y2y1=zz1z2z1
Given :-
P.V are 2ı^+3ȷ^,ı^+2ȷ^+3k^,7ı^+9k^
Solution :-
The points of P.V 2ı^+3ȷ^,ı^+2ȷ^+3k^,7ı^+9k^ 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
x+21+2=y323=z030x+23=y31=z03 ----- (1)
Putting (7 , 0 , 9) in (1) , we get
7+23=031=93
3,3,3
Since ( 7 , 0 , 9) satisfying equation (1)
Hence it is collinear.

Straight Line in Space exercise 27.1 question 16

Answer :-
x12=y214=z33(ı^+2ȷ^+3k^)+λ(2^ı+14ȷ^+3k^)
Hint :-
r=a+λb
Given :-
The line passes through the point ( 1 , 2 , 3) and parallel to the line
x21=y+37=2z63
Solution :-
The P.V of the point (1 , 2 , 3) will be r=i^+2j^+3k^
The line x21=y+37=2z63 can be written as x+21=y+37=z332 ---------- (1)
So the direction ratios can be 1,7,32or2,14,3 PV=2^i+14j^+3k^
Cartesian equation of the line passing through ( 1 , 2 , 3) and parallel to (1)
x12=y214=z33
Vector Equation
r=a+λb
=(ı^+2ȷ^+3k^)+λ(2^ı+14ȷ^+3k^)

Straight Line in Space exercise 27.1 question 17

Answer :-
x+132=y131=z16…cartesian equation of line
(13ı^+13ȷ^+k^)+λ(2ı^+ȷ^6k^)…vector equation of line
(13,13,1)…the fixed point through which the line passes ;
(2,1,6)….direction ratios of the line
Hint :-
r=a+λb
Given :-
3x+1=6y2=1z
Solution :-
The given line can be written as x+1313=y2616=z11
x+132=y131=z16
Thus the line passes through the point (13,13,1)
vector equation of the point will be (a=13ı^+13ȷ^+k^)
The direction ratios are proportional to (2,1,6)
and its vector equation will be b=2ı^+ȷ^6k^
Vector Equation of line
r=a+λb
=(13ı^+13ȷ^+k^)+λ(2ı^+ȷ^6k^)

Straight Line in Space exercise 27.1 question 18

Answer :-
(ı^+2ȷ^k^)+λ(7ı^5ȷ^+k^)
Hint :-
r=a+λb
Given :-
A(1,2,1)&5x25=147y=35z
Solution :-
5x25=147y=35zx515=y217=z135x57=y25=z1 -------------- (1)
Direction ratios of (1) will be ( 7 , -5 , 1 )
hence, b=7ı^5ȷ^+k^
P.V of A(1,2,1)=>a=ı^+2ȷ^k^
r=a+λb
=(ı^+2ȷ^k^)+λ(7ı^5ȷ^+k^)

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

Frequently Asked Questions (FAQs)

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.

Articles

Upcoming School Exams

Application Date:24 March,2025 - 23 April,2025

Admit Card Date:25 March,2025 - 21 April,2025

Admit Card Date:25 March,2025 - 17 April,2025

View All School Exams
Back to top