Solutions to the Sat Physics subject questions on vectors, with detailed explanations.

Which of the following is represented by a vector?
I) velocity II) speed III) displacement IV) distance V) force VI) acceleration
A) I only
B) III and IV only
C) I , III, V and VI
D) III, V and VI only
E) None
Solution  Explanations
In physics velocity, displacement, force and acceleration are defined as vector quantities.

Which of the following is not represented by a vector?
I) mass II) electric field III) magnetic field IV) current V) voltage VI) work
A)I , IV, V and VI
B) II and III
C) All
D) I , IV and V only
E) I and IV only
Solution  Explanations
In physics mass, current, voltage and work are defined as scalar quantities and therefore are NOT represented by vectors.

A andB are vectors with angle θ between them such that 0 < θ < 90°. If A , B  and A +B  are the magnitudes of vectorsA ,B andA +B respectively, which of the following is true?
A) A +B  > A  + B 
B) A +B  = A  + B 
C) A +B  = √(A ^{2} + B ^{2})
D) A +B  < √(A ^{2} + B ^{2})
E) A +B  < A  + B 
Solution  Explanations
Start with the following identity
(A +B )·(A +B ) = (A +B )·(A +B )
use scalar product to rewrite the above as follows
A +B ^{ 2} = A ^{ 2} + B ^{ 2} + 2A ·B
rewrite as
A +B ^{ 2} = A ^{ 2} + B ^{ 2} + 2 A  B  cos (θ)
Since 0 < θ <90°, we can write
0 < cos (θ) < 1
multiply all terms of inequality by 2 A  B  to obtain
0 < 2 A  B  cos (θ) < 2 A  B 
add A ^{ 2} + B ^{ 2} to all terms of above inequality to obtain
A ^{ 2} + B ^{ 2} < A ^{ 2} + B ^{ 2} + 2 A  B  cos (θ) < A ^{ 2} + B ^{ 2} + 2 A  B 
the above inequality can now be written as
A ^{ 2} + B ^{ 2} < A +B ^{ 2} < ( A  + B  )^{ 2}
all terms of the above inequality are positive, we can write the following inequality taking the square root as follows
√[ A ^{ 2} + B ^{ 2} ] < A +B  < ( A  + B  )

Which of the following is a unit vector in the same direction as vectorA = 3 i  4 j, where i and j are the unit vector along the x and y axis respectively?
A) (3/5) i + (4/5) j
B) (9/5) i  (16/5) j
C) (3/5) i  (4/5) j
D)  (3/5) i + (4/5) j
E) (3/25) i  (4/25) j
Solution  Explanations
The unit vector u in the same direction as vectorA = 3 i  4 j is given by
u =A / A  = ( 3 i  4 j ) / √(3^{ 2} + (4)^{ 2})
= (3/5) i  (4/5)j

IfA andB are vectors, thenA ·(A ×B ) =
A) A ^{ 3}
B) 0
C) A ^{ 2}
D) 1
E) A 
Solution  Explanations
The cross productA ×B gives a vector perpendicular to both vectorsA andB and therefore the scalar product between vectorsA andA ×B , which is perpendicular to A, is equal to zero.

VectorU has a magnitude of 3 and points Northward. VectorV has a magnitude of 7 and points Eastward. U +V  is the magnitude of vectorU +V . Which of the following is true?
A) U +V  > 10
B) U +V  = 10
C) U +V  = √58
D) U +V  = √10
E) U +V  = 4
Solution  Explanations
The x and y axes are directed Eastward and Northward respectively and therefore the components ofU +V are given as follows
U +V = 3j + 7 i
magnitude is now calculated
U +V  = √(7^{ 2} + 3^{ 2}) = √ 58

Given vectorsU = 2 i + 2j andV = 2i  2j. What angle does the vectorU V make with the positive x  axis?
A) 0 °
B) 45 °
C) 90 °
D) 90 °
E) 45 °
Solution  Explanations
U V = 2 i + 2j  (2i  2j) = 4j
U V is proportional to the unit vector j which makes 90° with the positive xaxis. HenceU V makes 90° with the positive xaxis

Two forcesF1 andF2 are used to pull an object. The angle between the two forces is θ. For what value of θ is the magnitude of the resultant force equal to √(F1 ^{2} + F2 ^{2})
A) 90 °
B) 135 °
C) 45 °
D) 180 °
E) 0 °
Solution  Explanations
LetR be the resultant and write
R =F1 +F2
use scalar product to write
R R = (F1 +F2 ) · (F1 +F2 )
expand the terms on the right
R ^{2} = F1 ^{2} + F2 ^{2} + F1  F2  cos (θ)
For θ = 90°, cos (θ) = 0
and
R ^{2} = F1 ^{2} + F2 ^{2}
taking the square root gives
R  = √ (F1 ^{2} + F2 ^{2})

Two forcesF1 = 3i + bj andF2 = 9i + 12j act on the same object.(i and j are the unit vectors along the positive xand y axes). For what value of b will the magnitude of the resultant force be minimum?
A) 0
B)  12
C) 9
D)  10
E) 4
Solution  Explanations
LetR be the resultant force
R =F1 +F2 = 3i + bj + 9i + 12j = 12i + (b + 12)j
calculate magnitude
R = √( 144 + (b + 12)^{2} )
the quantity under the radical is positive and therefore a value of b that minimizes 144 + (b + 12)^{2} will also minimizeR the magnitude
144 + (b + 12)^{2} is a quadratic expression and it has a minimum value at b = 12 (position of vertex)

Find m so that the vectorsA = 5 i  10 j andB = 2 m i + (1 / 2) j are parallel.
A) 5
B)  40
C) 8
D)  1 / 8
E)  5
Solution  Explanations
For vectorsA andB to be parallel, there must be a real K so thatA = KB
A = 5 i  10 j = K (B = 2 m i + (1 / 2) j )
components are equal, hence
5 = K (2 m)
10 = K (1 / 2)
Solve the second equation fo K: K = 20
substitute K by 20 in the equation 5 = K (2 m)
5 = 20 (2 m)
solve for m
m =  5 / 40 =  1 / 8
Answers to the Above questions
 C
 A
 E
 C
 B
 C
 D
 A
 B
 D