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RD Sharma Class 12 Exercise VSA Straight line in space Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise VSA Straight line in space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 09:48 AM IST

Planning a schedule that every 12th grader must follow to keep in touch with their ongoing syllabus and preparing for the board exams is essential. This task can be accomplished easily only with the help of the right reference books. Many students struggle with mathematics to arrive at the right solution in various concepts. If any student faces challenges in the Straight Line in Space topic, they can refer to the RD Sharma Class 12th Chapter 27 VSA book.

RD Sharma Class 12 Solutions Chapter27VSA Straight line in space - Other Exercise

Straight Line in Space Excercise: 27 VSA

Straight Line in Space Exercise Very Short Answer Question 1

Answer:
Required answer is λi^
Hint:
Use properties of vector
Given:
Cartesian and vector equation of x-axis
Solution:
Since x-axis passes through the point (0,0,0) having position vector
a=ai^+0j^+0k^ and parallel to the vector b=1i^+0j^+0k^ having direction ratios proportional to 1, 0, 0.
The Cartesian equation of x-axis is
x01=y00=z00
x1=y0=z0
Also its vector equation is
r=a+λb
=0i^+0j^+0k^+λ(i^+0j^+0k^)
=λi^

Straight Line in Space Exercise Very Short Answer Question 2

Answer:
Required answer is λj^
Hint:
Use properties of vector
Given:
Cartesian and vector equation of y-axis
Solution:
Since y-axis passes through the point (0,0,0) having position vector
a=0i^+0j^+0k^ and is parallel to the vector b=0i^+1j^+0k^ having direction ratios proportional to 0, 1, 0
The Cartesian equation of y-axis is
x00=y01=z00
x0=y1=z0
Also the vector equation is
r=a+λb
=0i^+0j^+0k^+λ(0i^+j^+0k^)
=λj^

Straight Line in Space Exercise Very Short Answer Question 3

Answer:
Required answer is λk^
Hint:
Use properties of vector
Given:
Cartesian and vector equation of z-axis
Solution:
Since z-axis passes through the point (0,0,0) having position vector
a=0i^+0j^+0k^ and is parallel to the vector b=0i^+0j^+1k^ having direction ratios proportional to 0, 0, 1
The Cartesian of z-axis is
x00=y00=z01
x0=y0=z1
Also its vector equation
r=a+λb
=0i^+0j^+0k^+λ(0i^+0j^+k^)
=λk^

Striaght Line in Space Exercise Very Short Answer question 4

Answer:
Required answer is r=α+λβ
Hint:
Use properties of vector
Given:
α Vector is parallel to β vector
Solution:
The vector equation of the line passing through the point having position vector α and parallel to the vector β is r=α+λβ

Straight Line in Space Exercise Very Short Answer question 5

Answer:
Required answer is 1,7,2
Hint:
Use properties of vector
Given:
2x12=4y7=z+12
Solution:
We have
2x12=4y7=z+12
Now the equation of the line AB can be rewritten as,
x121=y47=z+12
The direction ratios of the line parallel to AB proportional to 1,7,2

Straight Line in Space Exercise Very Short Answer question 6

Answer:
Required answer is (114,214,314)
Hint:
Use properties of vector
Given:
6x2=3y+1=2z4
Solution:
We have
6x2=3y+1=2z4
The equation of the given line can be rewritten as,
x1316=y+1313=z212
=x131=y+132=z23
The direction ratios of the line parallel to AB are proportional to 1,2,3
The direction cosines of the line parallel to AB are proportional to
112+22+32,212+22+32,312+22+32
=114,214,314

Straight Line in Space Exercise Very Short Answer Question 7

Answer:
Required answer is 429,329,229
Hint:
Use properties of vector
Given:
x22=2y53,z=2
Solution:
We have,
x22=2y53,z=2
The equation of the given line can be rewritten as,
x22=y5232=z21
x24=y523=z22
The direction ratios of the line parallel to AB are proportional to 4, -3, 1
The direction cosines of the line parallel to AB proportional to
442+(3)2+22,342+(3)2+22,242+(3)2+22
=429,329,229

Straight Line in Space Exercise Very Short Answer Question 8

Answer:
The given line is perpendicular to z-axis
Hint:
Use properties of vector
Given:
x23=y+14=z10
Solution:
We have
x23=y+14=z10
The given line is parallel to the vector b=3i^+4j^+0k^
Let, xi^+yj^+zk^ be perpendicular to the given line
Now, 3x+4y+0z=0
Direction ratios are (a1,b1,c1)=(3,4,0)
Check perpendicular condition with x axis
Direction ratio of x axis is (a2,b2,c2)=(1,0,0)
a1a2+b1b2+c1c2
=3(1)+4(0)+0(0)
=3
0 Hence x axis is not perpendicular
Check perpendicular condition with y axis
Direction ratio of y axis is (a2,b2,c2)=(0,1,0)
a1a2+b1b2+c1c2
=3(0)+4(1)+0(0)
=40
Hence y axis is not perpendicular
Check perpendicular condition with z axis
Direction ratio of z axis is (a2,b2,c2)=(0,0,1)
a1a2+b1b2+c1c2
=3(0)+4(0)+0(1)
=0
Hence, the given line is perpendicular to the z-axis

Straight Line in Space Exercise Very Short Answer Question 9

Answer:
Required answer is
Hint:
Use properties of vector
Given:
x57=y+25=z21 and x11=y2=z13
Solution:
We have,
x57=y+25=z21 and x11=y2=z13
The given lines are parallel to the vectors
b1=7i^5j^+k^
b2=i^+2j^+3k^
Let θ be the angle between the given lines
Now,
cosθ=b1b2|b1||b2|cosθ=b1b2|b1||b2|
=(7i^5j^+k^)(i^+2j^+3k^)72+(5)2+1212+22+32
=710+349+25+11+4+9
=0
θ=π2

Straight Line of Space Excercise Very Short Anwer Question 10

Answer:
Required answer is 37,27,67
Hint:
Use equation of a line in space.
Given:
Cartesian equations are 2x=3y=z
Solution:
We have,
2x=3y=z
The equation of the given line can be rewritten as,
x12=y13=z1
x3=y2=z6
The direction ratios of the line parallel to AB are proportional to 3,2,6
The direction cosines of the line parallel to AB are proportional to
332+22+(6)2,232+22+(6)2,632+22+(6)2
=37,27,67

Straight Line of Space Excercise Very Short Anwer Question 11

Answer:
Required answer is θ=π2
Hint:
Use formula cosθ=b1b2|b1||b2|
Given:
2x=3y=z and 6x=y=4z
Solution:
We have,
2x=3y=z and 6x=y=4z
The given lines can be rewritten as,
x3=y2=z6 and x2=y12=z3
These lines are parallel to vectors b1=3i^+2j^6k^ and b2=2i^12j^3k^
Let θ be the angle between these lines.
Now,
cosθ=b1b2|b1||b2|
=624+189+4+364+144+9=0
θ=π2

Straight Line of Space Excercise Very Short Anwer Question 12

Answer:
Required answer is λ=107
Hint:
Use equation of line in space
Given:
x33=y+22λ=z+42 And x+13λ=y21=z+65
Solution:
We have,
x33=y+22λ=z+42 And x+13λ=y21=z+65
The given lines are parallel to vectors
b1=3i^+2λj^+2k^ and b2=3λi^+j^5k^
Now, for b1b2 we must have,
b1b2=0
(3i^+2λj^+2k^)(3λi^+j^5k^)=0
7λ10=0
λ=107

Straight Line in Space Exercise Very Short Answer Question 13

Answer:
Required answer is d=|(a2a1)b|b||
Hint:
Use equation of line in space
Given:
r=a1+λb and r=a2+μb
Solution:
The shortest distance between the parallel lines
r=a1+λb
r=a2+μb is given by,
d=|(a2a1)b|b||

Straight Line in Space Exercise Very Short Answer Question 14

Answer:
Required answer is (a2a1)(b1×b2)=0
Hint:
Use formula d=|(a2a1)(b1×b2)|b1×b2||
Given:
r=a1+λb1 And r=a2+μb2
Solution:
The shortest distance between the lines r=a1+λb1 And r=a2+μb2 is given by
d=|(a2a1)(b1×b2)|b1×b2||
For the lines to be intersecting, d=0
|(a2a1)(b1×b2)|b1×b2||=0
(a2a1)(b1×b2)=0

Straight Line in Space Exercise Very Short Answer Question 15

Answer:
Required answer is 355,455,655
Hint:
Use equation of line in space
Given:
2x13=y+22=z33
Solution:
We have,
2x13=y+22=z33
The equation of the given line can be rewritten as,
x1232=y+22=z33
x123=y+24=z36
The direction ratios of the line parallel to AB are proportional to 3,4,6
The direction cosines of the line parallel to AB are proportional to,
3(3)2+42+62,4(3)2+42+62,6(3)2+42+62
=355,455,655

Straight Line in Space Exercise Very Short Answer question 16

Answer:
Required answer is 1,2,4
Hint:
Use equation of line in space
Given:
3x1=y+22=z54
Solution:
We have,
3x1=y+22=z54
The equation of the line AB can be rewritten as,
x31=y+22=z54
Thus, the direction ratios of the line parallel to AB are proportional to 1,2,4

Straight Line in Space Exercise Very Short Answer question 17

Answer:
Required answer is r=5i^4j^+6k^+λ(3i^+7j^+2k^)
Hint:
Use properties of vector
Given:
x53=y+47=z62
Solution:
We have,
x53=y+47=z62
The given line passes through the point (5,4,6) and has direction ratios proportional to 3,7,2
Vector equation of the given line passing through the point having position vector
a=5i^4j^+6k^ and parallel to a vector b=3i^+7j^+2k^ is
r=a+λb
r=5i^4j^+6k^+λ(3i^+7j^+2k^)

Straight Line in Space Exercise Very Short Answer question 18

Answer:
Required answer is 16,16,26
Hint:
Use the equation of a line in space.
Given:
4x3=y+33=z+26
Solution:
We have,
4x3=y+33=z+26
The equation of the given line can be rewritten as,
x43=y+33=z+26
The direction ratios of the line parallel to the given line are proportional to 3,3,6
The direction cosines of the line parallel to the given line are proportional to
3(3)2+32+62,3(3)2+32+62,6(3)2+32+62
=16,16,26

Straight Line in Space Exercise Very Short Answer question 19

Answer: (2,4,5)
Required answer is
Hint:
Use the equation of a line in space.
Given:
Point x+23=y45=z+56
Line x+33=y45=z+86
Solution:
The equation of the given line is
x+33=y45=z+86
Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3,5,6
Hence the Cartesian equation of the line passing through the point (2,4,5) and parallel to a vector having direction ratios proportional to 3,5,6 is
x+23=y45=z+56

Straight Line in Space Exercise Very Short Answer Question 20

Answer:
Required answer is cos11921
Hint:
Use the equation of a line in space.
Given:
r=(2i^5j^+k^)+λ(3i^+2j^+6k^)
r=(7i^6k^)+μ(i^+2j^+2k)
Solution:
Let θ be the angle between the given lines
The given lines are parallel to the vectors b1=3i^+2j^+6k^ and b2=i^+2j^+2k^ respectively
So, the angle θ between the given lines are given by
cosθ=b1b2|b1||b2|
=(3i^+2j^+6k^)(i^+2j^+2k^)32+22+6212+22+22
=19499
=1921
θ=cos11921

Straight Line in Space Exercise Very Short Answer Question 21

Answer:
Required angle is π2
Hint:
Use equation of a line in space.
Given:
2x=3y=z and 6x=y=4z
Solution:
The equation of the given lines can be rewritten as,
x3=y2=z6 and x2=y12=z3
We know that angle between the lines
xx1a1=yy1b1=zz1c1 and xx2a2=yy2b2=zz2c2 is given by,
cosθ=a1a2+b1b2+c1c2
Let,
θ=a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22
Let θ be the angle between the given lines
cosθ=3×2+2×(12)+(6)×(3)32+22+(6)222+(12)2+(3)2
=624+1849157
=0
θ=π2
Thus, the angle between the given lines is π2

Straight Line in Space Exercise Very Short Answer Question 22

Answer:
Required answer is r=(3i^+4j^+5k^)+λ(2i^+2j^3k^)
Hint:
Use equation of a line in space.
Given:
Point (3,4,5) and parallel vector 2i^+2j^3k^
Solution:
The line is passing through the point (3,4,5) through the point having position vector is a=3i^+4j^+5k^ and parallel to the vector b=2i^+2j^3k^
r=a+λb
r=(3i^+4j^+5k^)+λ(2i^+2j^3k^)

Also, see,

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.1

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.2

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.3

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.4

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.5

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise MCQ

  • RD Sharma Solutions Class 12 mathematics chapter 27 exercise FBQ

The number of exercises in class 12, chapter 27, Straight Line in Space, is five. The concepts in the ex 27.1 to ex 27.5 are included in the Very Short Answers part, where there are 22 questions in total. The topics from which the questions are asked include finding the angle between the lines, symmetric and unsymmetric, vector equations, cartesian equations, equation of a plane, equation of a line, direction cosines, conditions for lines to intersect, direction cosine, etc. The RD Sharma Class 12 Chapter 27 VSA solution guide directs the students on how these sums can be solved easily.

Students cannot contact their teachers or tutors every time a doubt strikes them; simple and complex doubts will also vanish when a glance is made into the RD Sharma Class 12th Chapter 27 VSA reference book. The various practice questions in this book will motivate the students to work out more sums that eventually sharpen their knowledge. Soon, they would cross their benchmark scores effortlessly due to the high confidence level they gained from the practice.

The Class 12 RD Sharma Chapter 27 VSA Solution book is updated according to the recent NCERT pattern. And it also provides additional benefits for the students who refer to these books. The well-known source from where most of the teachers pick questions to conduct tests is the RD Sharma books. Therefore, when a student uses the RD Sharma Class 12 Solutions Straight Line in Space Chapter 27 VSA, he is getting ready for his tests and exams.

As the Career 360 website provides all the RD Sharma solution books, including the RD Sharma Class 12th Chapter 27 VSA, for free of cost, many students flock to this website to get a copy. In addition, due to the presence of the Download option, it is easier to save the PDF material at your device to access it offline. As a result, a great number of students have benefitted from using these books. So, download your copy of RD Sharma Class 12 Solutions Chapter 27 VSA and start your preparation for the board exams.

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 23: 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 23: Scalar and dot products

Chapter 17: Maxima and Minima

Chapter 123: 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 223: The plane

Chapter 29: Linear programming

Chapter 30: Probability

Chapter 31: Mean and variance of a random variable

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