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All questions of Vectors for SSS 2 Exam
If 
what is the angle between 
and 
- a)900
- b)300
- c)600
- d)450
Correct answer is option 'D'. Can you explain this answer?
If what is the angle between and
what is the angle between and a)
900
b)
300
c)
600
d)
450
|
Gaurav Kumar answered |
We know that for two vectors P and Q
P.Q = |P||Q| cos a
And PxQ = |P||Q| sin a
Where a is angle between them
When P.Q = PxQ
We get sin a = cos a
Thus a = 450
P.Q = |P||Q| cos a
And PxQ = |P||Q| sin a
Where a is angle between them
When P.Q = PxQ
We get sin a = cos a
Thus a = 450
In the triangle ABC, which statement is not true?
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In the triangle ABC, which statement is not true?
a)
b)
c)
d)
|
sanju sura answered |
Ab + bc > ac ...it is property of a triangle .... that sum of two sides is greater than the third side ... so I think it is no possible that ab +bc -ca =0...but I'm not understanding the logic behind the question.......😊
The points with position vectors 
are collinear vectors, Value of a =- a)-20
- b)20
- c)-40
- d)40
Correct answer is option 'C'. Can you explain this answer?
The points with position vectors are collinear vectors, Value of a =
are collinear vectors, Value of a =a)
-20
b)
20
c)
-40
d)
40
|
Om Desai answered |
Position vector A = 60i+3j
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.
ABCD is a parallelogram. If coordinates of A,B,C are (2,3), (1,4) and (0, -2). Coordinates of D =- a)(-1,3)
- b)(1,-3)
- c)(1,3)
- d)(-1,-3)
Correct answer is option 'B'. Can you explain this answer?
ABCD is a parallelogram. If coordinates of A,B,C are (2,3), (1,4) and (0, -2). Coordinates of D =
a)
(-1,3)
b)
(1,-3)
c)
(1,3)
d)
(-1,-3)
|
Naina Bansal answered |
Here's a generic way to solve problems like this.
The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:
a)
b)
c)
d)
|
Siddhant Kumar answered |
The vector goes from B to A means that its initial coordinates is at B and final at A.
so the vector BA will be [ (2-1)i + (-3+2) j + (1+5)k ] = i-j+6k.
so the vector BA will be [ (2-1)i + (-3+2) j + (1+5)k ] = i-j+6k.
The distance between the point (2, 3, 1) and (–1, 2, – 3) is:
- a)√26 units
- b)5 units
- c)√20 units
- d)4 units
Correct answer is option 'A'. Can you explain this answer?
The distance between the point (2, 3, 1) and (–1, 2, – 3) is:
a)
√26 units
b)
5 units
c)
√20 units
d)
4 units
|
|
Giriraj Gupta answered |
Use the distance formula
distance=root of [ (x1-x2)^2+(y1-y2)^2+(z1-z2)^2]
distance=root of [ (x1-x2)^2+(y1-y2)^2+(z1-z2)^2]
The position vectors of the end points of diameter of a circle are 
and 
, then the position vector of the centre of the circle is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The position vectors of the end points of diameter of a circle are and , then the position vector of the centre of the circle is:
and , then the position vector of the centre of the circle is:a)
b)
c)
d)
|
|
Om Desai answered |
{(1+5)î +(1-3)j + (1-1)k} / 2
= {6i - 2j + 0k}/2
= 3i - j
= {6i - 2j + 0k}/2
= 3i - j
, then 
........
The vector that must be added to the vector 
so that the resultant vector is a unit vector along the y-axis is
- a)
- b)
- c)
- d)Null vector
Correct answer is option 'B'. Can you explain this answer?
The vector that must be added to the vector so that the resultant vector is a unit vector along the y-axis is
so that the resultant vector is a unit vector along the y-axis isa)
b)
c)
d)
Null vector
|
Ambition Institute answered |
Unit vector along y axis = 
so the required vector = 
so the required vector = Any vector in an arbitrary direction can be replaced by two or three vectors - a)Perpendicular to each other and have the original vector as their resultant
- b)Parallel to each other and have the original vector as their resultant
- c)Arbitrary vectors which have original vectors as their resultant
- d)It is impossible to resolve a vector
Correct answer is option 'C'. Can you explain this answer?
Any vector in an arbitrary direction can be replaced by two or three vectors
a)
Perpendicular to each other and have the original vector as their resultant
b)
Parallel to each other and have the original vector as their resultant
c)
Arbitrary vectors which have original vectors as their resultant
d)
It is impossible to resolve a vector
|
Varun Kapoor answered |
If we slide a vector parallel to its position it will be the same as before. Any vector in an arbitrary direction can always be replaced by two (or three) arbitrary vectors which have the original vector as their resultant.
. The distance traversed in first 10 sec is
, the angle between 
and y-axis is
Direction of the resultant of two vectors is given by- a)cos-1(Ay + By/Ax + Bx)
- b)cos-1(Ax + By/Ay + Bx)
- c)tan-1(Ay + By/Ax + Bx)
- d)tan-1(Ay - By/Ax - Bx)
Correct answer is option 'C'. Can you explain this answer?
Direction of the resultant of two vectors is given by
a)
cos-1(Ay + By/Ax + Bx)
b)
cos-1(Ax + By/Ay + Bx)
c)
tan-1(Ay + By/Ax + Bx)
d)
tan-1(Ay - By/Ax - Bx)
|
Lakshmi Sharma answered |
Assume the two vector A and B with an angle alpha whose resultant be vector C given by vector C = vector A + vector B. In addition drop a line from the point B and C meeting to form the right angled triangle, so AC2 = AD2 + CD2
⇒
⇒
⇒
⇒ From triangle BCD, cos alpha = BD / BC
⇒ BD = BC x cos∝
⇒
⇒
The resultant vector makes angle theta therefore, in triangle BCD,
sin∝ = CD/BC
⇒ CD = BCsinα,
and in triangle ADC,
tanθ = CD/AD = BC sinα/a + bcos∝
⇒ θ = tan -1 bsin∝/a + bcosα
⇒
⇒
⇒
⇒ From triangle BCD, cos alpha = BD / BC
⇒ BD = BC x cos∝
⇒
⇒
The resultant vector makes angle theta therefore, in triangle BCD,
sin∝ = CD/BC
⇒ CD = BCsinα,
and in triangle ADC,
tanθ = CD/AD = BC sinα/a + bcos∝
⇒ θ = tan -1 bsin∝/a + bcosα
The position vector of a particle is 
.The velocity of the particle is- a)Parallel to the position vector
- b)Perpendicular to the position vector
- c)Directed towards the origin
- d)Directed away from the origin
Correct answer is option 'B'. Can you explain this answer?
The position vector of a particle is .The velocity of the particle is
.The velocity of the particle isa)
Parallel to the position vector
b)
Perpendicular to the position vector
c)
Directed towards the origin
d)
Directed away from the origin
|
EduRev Class 9 answered |
As
therefore velocity of the particle is perpendicular to the position vector.
is a 
will be
then which of the following is NOT correct
Its length in XY plane is 
Chapter doubts & questions for Vectors - Physics for SSS 2 2025 is part of SSS 2 exam preparation. The chapters have been prepared according to the SSS 2 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for SSS 2 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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Physics for SSS 2
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