HC Verma Solutions: Chapter 2 - Physics & Mathematics | Physics Class 11 - NEET PDF Download

 Page 1


HC VERMA Solutions for Class 11 Physics Chapter 2 - Physics 
and Mathematics 
Question 1 
A vector  makes an angle of  and  makes an angle of  with the X-axis. The magnitudes of 
these vectors are 3 m and 4 m respectively. Find the resultant. 
Solution 1 
The angle between A and B from the x-axis are 20° and 110° respectively. 
Their magnitudes are 3 units and 4 units respectively. 
Thus the angle between A and B is = 110 - 20 = 90°  
Now, R
2
 = A
2
 + B
2
 + 2ABcos?  
 = 3
2
 + 4
2
 + 2.3.4 Cos(90) 
   = 5
2
 
 Or, R = 5 
Let ? is the angle between R and A, 
Then tan ? = , or ? = 53°. 
The resultant makes an angle of (53+20)° = 73° with the x axis. 
Question 2 
Let  and  be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at 
angles  and  respectively, find the resultant.  
Solution 2 
A and B are inclined at angles of 30 degrees and 60 degrees with respect to the x axis 
Angle between them = (60 -30) = 90 degrees 
Given that |A| = |B| = 10 units, we get 
R
2
 = A
2
 + B
2
 +2ABcos?  
 = 10
2
 + 10
2
 + 2.10.10 Cos(30) 
 R = 20cos15°
 
 
Or, R = 19.3 units 
And tan ? = , or ? = 15°. 
Therefore, this resultant makes an angle of (15+30) = 45 degrees with the x axis 
Question 3 
Add vectors  and  each having magnitude of 100 unit and inclined to the X-axis at 
angles  and  respectively. 
Solution 3 
Vectors A, B and C are oriented at 45°, 135° and 315° respectively. 
|A|=|B|=|C|=100 units 
Let A = Axi + A yj +Azk, B = Bxi + B yj +B zk, and C = Cxi + C yj +C zk, and we can write that, 
Ax= Cx=100cos(45°)=100/v2 , by considering their components 
Bx=-100/v2 
Now Ay = 100sin(45°)= 100/v2, 
Page 2


HC VERMA Solutions for Class 11 Physics Chapter 2 - Physics 
and Mathematics 
Question 1 
A vector  makes an angle of  and  makes an angle of  with the X-axis. The magnitudes of 
these vectors are 3 m and 4 m respectively. Find the resultant. 
Solution 1 
The angle between A and B from the x-axis are 20° and 110° respectively. 
Their magnitudes are 3 units and 4 units respectively. 
Thus the angle between A and B is = 110 - 20 = 90°  
Now, R
2
 = A
2
 + B
2
 + 2ABcos?  
 = 3
2
 + 4
2
 + 2.3.4 Cos(90) 
   = 5
2
 
 Or, R = 5 
Let ? is the angle between R and A, 
Then tan ? = , or ? = 53°. 
The resultant makes an angle of (53+20)° = 73° with the x axis. 
Question 2 
Let  and  be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at 
angles  and  respectively, find the resultant.  
Solution 2 
A and B are inclined at angles of 30 degrees and 60 degrees with respect to the x axis 
Angle between them = (60 -30) = 90 degrees 
Given that |A| = |B| = 10 units, we get 
R
2
 = A
2
 + B
2
 +2ABcos?  
 = 10
2
 + 10
2
 + 2.10.10 Cos(30) 
 R = 20cos15°
 
 
Or, R = 19.3 units 
And tan ? = , or ? = 15°. 
Therefore, this resultant makes an angle of (15+30) = 45 degrees with the x axis 
Question 3 
Add vectors  and  each having magnitude of 100 unit and inclined to the X-axis at 
angles  and  respectively. 
Solution 3 
Vectors A, B and C are oriented at 45°, 135° and 315° respectively. 
|A|=|B|=|C|=100 units 
Let A = Axi + A yj +Azk, B = Bxi + B yj +B zk, and C = Cxi + C yj +C zk, and we can write that, 
Ax= Cx=100cos(45°)=100/v2 , by considering their components 
Bx=-100/v2 
Now Ay = 100sin(45°)= 100/v2, 
By= 100sin(135
0
)= 100/v2 
Similarly, Cy= -100/v2 
Net x component = 100/v2+100/v2-100/v2=100/v2 
Net y component = 100/v2+100/v2-100/v2=100/v2 
R
2
=x
2
+y
2
=100
2
 
R=100 and tan ? = (100/v2)/( 100/v2)=1, and ? = 45°  
Question 4 
Let  and . (a) Find the magnitudes of (a) , (b) , (c) and (d) . 
Solution 4 
a = 4i+3j, b = 3i+4j 
|a|=|b|=
 
=5 
a+b = 7i+7j and a-b = i-j 
|a+b| 
= 
 
7v2 and |a-b| =  = v2 
 
Question 5 
Refer to the figure. (a) Find the magnitude, (b) x and y 
components and (c) the angle with the X-axis of the resultant of  and . 
  
  
Solution 5 
x component of OA = 2cos30° = v3 
x component of BC = 1.5 cos 120° = -0.75 
x component of DE = 1 cos 270° = 0 
y component of OA = 2 sin 30° = 1 
component of BC = 1.5 sin 120° = 1.3 
component of DE = 1 sin 270° = -1 
Rx= x component of resultant = 3 - 0.75 + 0 = 0.98 m 
Ry = resultant y component = 1 + 1.3 - 1 = 1.3 m 
So, R = (Rx
2
+R y
2
)
1/2
 = 1.6 m 
It makes and angle ? with positive x-axis then tan ? = R y/Rx = 1.32 ? = tan
-1
(1.32) 
Question 6 
Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them 
if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit. 
Solution 6 
|a| = 3 and |b| = 4 
Page 3


HC VERMA Solutions for Class 11 Physics Chapter 2 - Physics 
and Mathematics 
Question 1 
A vector  makes an angle of  and  makes an angle of  with the X-axis. The magnitudes of 
these vectors are 3 m and 4 m respectively. Find the resultant. 
Solution 1 
The angle between A and B from the x-axis are 20° and 110° respectively. 
Their magnitudes are 3 units and 4 units respectively. 
Thus the angle between A and B is = 110 - 20 = 90°  
Now, R
2
 = A
2
 + B
2
 + 2ABcos?  
 = 3
2
 + 4
2
 + 2.3.4 Cos(90) 
   = 5
2
 
 Or, R = 5 
Let ? is the angle between R and A, 
Then tan ? = , or ? = 53°. 
The resultant makes an angle of (53+20)° = 73° with the x axis. 
Question 2 
Let  and  be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at 
angles  and  respectively, find the resultant.  
Solution 2 
A and B are inclined at angles of 30 degrees and 60 degrees with respect to the x axis 
Angle between them = (60 -30) = 90 degrees 
Given that |A| = |B| = 10 units, we get 
R
2
 = A
2
 + B
2
 +2ABcos?  
 = 10
2
 + 10
2
 + 2.10.10 Cos(30) 
 R = 20cos15°
 
 
Or, R = 19.3 units 
And tan ? = , or ? = 15°. 
Therefore, this resultant makes an angle of (15+30) = 45 degrees with the x axis 
Question 3 
Add vectors  and  each having magnitude of 100 unit and inclined to the X-axis at 
angles  and  respectively. 
Solution 3 
Vectors A, B and C are oriented at 45°, 135° and 315° respectively. 
|A|=|B|=|C|=100 units 
Let A = Axi + A yj +Azk, B = Bxi + B yj +B zk, and C = Cxi + C yj +C zk, and we can write that, 
Ax= Cx=100cos(45°)=100/v2 , by considering their components 
Bx=-100/v2 
Now Ay = 100sin(45°)= 100/v2, 
By= 100sin(135
0
)= 100/v2 
Similarly, Cy= -100/v2 
Net x component = 100/v2+100/v2-100/v2=100/v2 
Net y component = 100/v2+100/v2-100/v2=100/v2 
R
2
=x
2
+y
2
=100
2
 
R=100 and tan ? = (100/v2)/( 100/v2)=1, and ? = 45°  
Question 4 
Let  and . (a) Find the magnitudes of (a) , (b) , (c) and (d) . 
Solution 4 
a = 4i+3j, b = 3i+4j 
|a|=|b|=
 
=5 
a+b = 7i+7j and a-b = i-j 
|a+b| 
= 
 
7v2 and |a-b| =  = v2 
 
Question 5 
Refer to the figure. (a) Find the magnitude, (b) x and y 
components and (c) the angle with the X-axis of the resultant of  and . 
  
  
Solution 5 
x component of OA = 2cos30° = v3 
x component of BC = 1.5 cos 120° = -0.75 
x component of DE = 1 cos 270° = 0 
y component of OA = 2 sin 30° = 1 
component of BC = 1.5 sin 120° = 1.3 
component of DE = 1 sin 270° = -1 
Rx= x component of resultant = 3 - 0.75 + 0 = 0.98 m 
Ry = resultant y component = 1 + 1.3 - 1 = 1.3 m 
So, R = (Rx
2
+R y
2
)
1/2
 = 1.6 m 
It makes and angle ? with positive x-axis then tan ? = R y/Rx = 1.32 ? = tan
-1
(1.32) 
Question 6 
Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them 
if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit. 
Solution 6 
|a| = 3 and |b| = 4 
Let ? be the angle between them. 
Then, using the relation R
2
 = A
2
 + B
2
 +2ABcos?, 
a) 
We get for R = 1, 
1 = 9+16+24Cos ?  
Or, ? = 180°
 
 
b) 
For, R = 5, we have 
25 = 9+16+24Cos ?  
Or, cos ? = 0; 
? = 90°
 
 
c)For R =7, 
49= 9+16+24Cos ?, 
Or cos ? = 1, 
And ? = 0°
 
 
Question 7 
A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a 
perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped." 
Find the displacement of the car. 
Solution 7 
AB = 2i + 0.5j + 4i = 6i + 0.5j 
As the car went forward, took a left and then a right. 
So, AB = (6
2
+0.5
2
)
1/2
= 6.02km 
And ? = tan
-1
(BE\AE) = tan
-1
(0.5/6) = tan
-1
(1/12)  
Question 8 
A carrom board (4 ft  4 ft square) has the queen at the center. The queen, hit by the striker moves 
to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of 
displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and 
(c) from the center to the hole. 
  
Solution 8 
  
In ?ABC, tan ? = x/2 and in ?DCF, tan ?= (2 - x)/4, So, (x/2) = (2 - x)/4. Solving, 
 4 - 2x = 4x 
Or, 6x = 4 
 Or, x = 2/3 ft 
a) In ?ABC, AC = (AB
2
+ BC
2
)
1/2
 =  ft 
b) In ?CFD, DF = 1 - (2/3) = 4/3 ft 
So, CF = 4 ft. Now, CD = (CF
2
+ FD
2
)
1/2
 = 4v10/3 ft 
c) In ?ADE, AE = (AE
2
+ ED
2
)
1/2
 = 2v2 ft.  
Question 9 
A mosquito net over a 7 ft 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through 
which a mosquito enters the bed. It flies and sits at the diagonally opposite upper corner of the net. (a) 
Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length 
of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components 
of the displacement vector.  
Solution 9 
Page 4


HC VERMA Solutions for Class 11 Physics Chapter 2 - Physics 
and Mathematics 
Question 1 
A vector  makes an angle of  and  makes an angle of  with the X-axis. The magnitudes of 
these vectors are 3 m and 4 m respectively. Find the resultant. 
Solution 1 
The angle between A and B from the x-axis are 20° and 110° respectively. 
Their magnitudes are 3 units and 4 units respectively. 
Thus the angle between A and B is = 110 - 20 = 90°  
Now, R
2
 = A
2
 + B
2
 + 2ABcos?  
 = 3
2
 + 4
2
 + 2.3.4 Cos(90) 
   = 5
2
 
 Or, R = 5 
Let ? is the angle between R and A, 
Then tan ? = , or ? = 53°. 
The resultant makes an angle of (53+20)° = 73° with the x axis. 
Question 2 
Let  and  be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at 
angles  and  respectively, find the resultant.  
Solution 2 
A and B are inclined at angles of 30 degrees and 60 degrees with respect to the x axis 
Angle between them = (60 -30) = 90 degrees 
Given that |A| = |B| = 10 units, we get 
R
2
 = A
2
 + B
2
 +2ABcos?  
 = 10
2
 + 10
2
 + 2.10.10 Cos(30) 
 R = 20cos15°
 
 
Or, R = 19.3 units 
And tan ? = , or ? = 15°. 
Therefore, this resultant makes an angle of (15+30) = 45 degrees with the x axis 
Question 3 
Add vectors  and  each having magnitude of 100 unit and inclined to the X-axis at 
angles  and  respectively. 
Solution 3 
Vectors A, B and C are oriented at 45°, 135° and 315° respectively. 
|A|=|B|=|C|=100 units 
Let A = Axi + A yj +Azk, B = Bxi + B yj +B zk, and C = Cxi + C yj +C zk, and we can write that, 
Ax= Cx=100cos(45°)=100/v2 , by considering their components 
Bx=-100/v2 
Now Ay = 100sin(45°)= 100/v2, 
By= 100sin(135
0
)= 100/v2 
Similarly, Cy= -100/v2 
Net x component = 100/v2+100/v2-100/v2=100/v2 
Net y component = 100/v2+100/v2-100/v2=100/v2 
R
2
=x
2
+y
2
=100
2
 
R=100 and tan ? = (100/v2)/( 100/v2)=1, and ? = 45°  
Question 4 
Let  and . (a) Find the magnitudes of (a) , (b) , (c) and (d) . 
Solution 4 
a = 4i+3j, b = 3i+4j 
|a|=|b|=
 
=5 
a+b = 7i+7j and a-b = i-j 
|a+b| 
= 
 
7v2 and |a-b| =  = v2 
 
Question 5 
Refer to the figure. (a) Find the magnitude, (b) x and y 
components and (c) the angle with the X-axis of the resultant of  and . 
  
  
Solution 5 
x component of OA = 2cos30° = v3 
x component of BC = 1.5 cos 120° = -0.75 
x component of DE = 1 cos 270° = 0 
y component of OA = 2 sin 30° = 1 
component of BC = 1.5 sin 120° = 1.3 
component of DE = 1 sin 270° = -1 
Rx= x component of resultant = 3 - 0.75 + 0 = 0.98 m 
Ry = resultant y component = 1 + 1.3 - 1 = 1.3 m 
So, R = (Rx
2
+R y
2
)
1/2
 = 1.6 m 
It makes and angle ? with positive x-axis then tan ? = R y/Rx = 1.32 ? = tan
-1
(1.32) 
Question 6 
Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them 
if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit. 
Solution 6 
|a| = 3 and |b| = 4 
Let ? be the angle between them. 
Then, using the relation R
2
 = A
2
 + B
2
 +2ABcos?, 
a) 
We get for R = 1, 
1 = 9+16+24Cos ?  
Or, ? = 180°
 
 
b) 
For, R = 5, we have 
25 = 9+16+24Cos ?  
Or, cos ? = 0; 
? = 90°
 
 
c)For R =7, 
49= 9+16+24Cos ?, 
Or cos ? = 1, 
And ? = 0°
 
 
Question 7 
A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a 
perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped." 
Find the displacement of the car. 
Solution 7 
AB = 2i + 0.5j + 4i = 6i + 0.5j 
As the car went forward, took a left and then a right. 
So, AB = (6
2
+0.5
2
)
1/2
= 6.02km 
And ? = tan
-1
(BE\AE) = tan
-1
(0.5/6) = tan
-1
(1/12)  
Question 8 
A carrom board (4 ft  4 ft square) has the queen at the center. The queen, hit by the striker moves 
to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of 
displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and 
(c) from the center to the hole. 
  
Solution 8 
  
In ?ABC, tan ? = x/2 and in ?DCF, tan ?= (2 - x)/4, So, (x/2) = (2 - x)/4. Solving, 
 4 - 2x = 4x 
Or, 6x = 4 
 Or, x = 2/3 ft 
a) In ?ABC, AC = (AB
2
+ BC
2
)
1/2
 =  ft 
b) In ?CFD, DF = 1 - (2/3) = 4/3 ft 
So, CF = 4 ft. Now, CD = (CF
2
+ FD
2
)
1/2
 = 4v10/3 ft 
c) In ?ADE, AE = (AE
2
+ ED
2
)
1/2
 = 2v2 ft.  
Question 9 
A mosquito net over a 7 ft 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through 
which a mosquito enters the bed. It flies and sits at the diagonally opposite upper corner of the net. (a) 
Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length 
of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components 
of the displacement vector.  
Solution 9 
The displacement vector is given by r= 7i+4j +3k 
Magnitude of displacement of the mosquitoes= (7
2
+4
2
+3
2
)
1/2
 = 74ft 
The components of displacement along x axis, y axis and z axis are 7 units, 4 units and 3 units 
respectively.  
Question 10 
Suppose  is a vector of magnitude 4.5 unit due north. What is the vector(a) 3 , (b) -4 ? 
Solution 10 
a = 4.5 n, where n is unit vector in north direction 
a) 3a = 4.5 X 3n = 13.5 in north direction 
b) -4a = 4.5 X -4n = 18 in south direction 
Question 11 
Two vectors have magnitudes 2 m and 3 m. The angle between them is . Find (a) the scalar product 
of the two vectors, (b) the magnitude of their vector product? 
Solution 11 
We have a = 2m, b = 3m. 
? = 60
0
 is the angle between the two vectors 
Scalar product between the two vectors = a.b = 
2×3×cos(60
0
)=3m
2
 
Vector product between the two vectors = a×b = 
2×3×sin(60
0
)=3v3m
2
 
Question 12 
Let  be a regular hexagon. Write the x-components of the vectors represented by 
the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that 
  
Use the known cosine values to verify the result. 
  
  
Solution 12 
From polygon law of vector addition, the resultant of the six vectors can be affirmed to be zero. Here 
their magnitudes are the same. 
That is, A = B = C = D = E = F. 
Rx = A cos? + A cos ?/3 + A cos 2?/3 + A cos 3?/3 + A cos 
4 ?/4 + A cos 5?/5 = 0 [As resultant is zero, x component of 
resultant is also 0] 
Now taking A common and putting Rx=0, 
cos ? + cos ?/3 + cos 2 ?/3 + cos 3 ?/3 + cos 4?/3 + cos 5?/3 = 0 
Page 5


HC VERMA Solutions for Class 11 Physics Chapter 2 - Physics 
and Mathematics 
Question 1 
A vector  makes an angle of  and  makes an angle of  with the X-axis. The magnitudes of 
these vectors are 3 m and 4 m respectively. Find the resultant. 
Solution 1 
The angle between A and B from the x-axis are 20° and 110° respectively. 
Their magnitudes are 3 units and 4 units respectively. 
Thus the angle between A and B is = 110 - 20 = 90°  
Now, R
2
 = A
2
 + B
2
 + 2ABcos?  
 = 3
2
 + 4
2
 + 2.3.4 Cos(90) 
   = 5
2
 
 Or, R = 5 
Let ? is the angle between R and A, 
Then tan ? = , or ? = 53°. 
The resultant makes an angle of (53+20)° = 73° with the x axis. 
Question 2 
Let  and  be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at 
angles  and  respectively, find the resultant.  
Solution 2 
A and B are inclined at angles of 30 degrees and 60 degrees with respect to the x axis 
Angle between them = (60 -30) = 90 degrees 
Given that |A| = |B| = 10 units, we get 
R
2
 = A
2
 + B
2
 +2ABcos?  
 = 10
2
 + 10
2
 + 2.10.10 Cos(30) 
 R = 20cos15°
 
 
Or, R = 19.3 units 
And tan ? = , or ? = 15°. 
Therefore, this resultant makes an angle of (15+30) = 45 degrees with the x axis 
Question 3 
Add vectors  and  each having magnitude of 100 unit and inclined to the X-axis at 
angles  and  respectively. 
Solution 3 
Vectors A, B and C are oriented at 45°, 135° and 315° respectively. 
|A|=|B|=|C|=100 units 
Let A = Axi + A yj +Azk, B = Bxi + B yj +B zk, and C = Cxi + C yj +C zk, and we can write that, 
Ax= Cx=100cos(45°)=100/v2 , by considering their components 
Bx=-100/v2 
Now Ay = 100sin(45°)= 100/v2, 
By= 100sin(135
0
)= 100/v2 
Similarly, Cy= -100/v2 
Net x component = 100/v2+100/v2-100/v2=100/v2 
Net y component = 100/v2+100/v2-100/v2=100/v2 
R
2
=x
2
+y
2
=100
2
 
R=100 and tan ? = (100/v2)/( 100/v2)=1, and ? = 45°  
Question 4 
Let  and . (a) Find the magnitudes of (a) , (b) , (c) and (d) . 
Solution 4 
a = 4i+3j, b = 3i+4j 
|a|=|b|=
 
=5 
a+b = 7i+7j and a-b = i-j 
|a+b| 
= 
 
7v2 and |a-b| =  = v2 
 
Question 5 
Refer to the figure. (a) Find the magnitude, (b) x and y 
components and (c) the angle with the X-axis of the resultant of  and . 
  
  
Solution 5 
x component of OA = 2cos30° = v3 
x component of BC = 1.5 cos 120° = -0.75 
x component of DE = 1 cos 270° = 0 
y component of OA = 2 sin 30° = 1 
component of BC = 1.5 sin 120° = 1.3 
component of DE = 1 sin 270° = -1 
Rx= x component of resultant = 3 - 0.75 + 0 = 0.98 m 
Ry = resultant y component = 1 + 1.3 - 1 = 1.3 m 
So, R = (Rx
2
+R y
2
)
1/2
 = 1.6 m 
It makes and angle ? with positive x-axis then tan ? = R y/Rx = 1.32 ? = tan
-1
(1.32) 
Question 6 
Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them 
if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit. 
Solution 6 
|a| = 3 and |b| = 4 
Let ? be the angle between them. 
Then, using the relation R
2
 = A
2
 + B
2
 +2ABcos?, 
a) 
We get for R = 1, 
1 = 9+16+24Cos ?  
Or, ? = 180°
 
 
b) 
For, R = 5, we have 
25 = 9+16+24Cos ?  
Or, cos ? = 0; 
? = 90°
 
 
c)For R =7, 
49= 9+16+24Cos ?, 
Or cos ? = 1, 
And ? = 0°
 
 
Question 7 
A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a 
perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped." 
Find the displacement of the car. 
Solution 7 
AB = 2i + 0.5j + 4i = 6i + 0.5j 
As the car went forward, took a left and then a right. 
So, AB = (6
2
+0.5
2
)
1/2
= 6.02km 
And ? = tan
-1
(BE\AE) = tan
-1
(0.5/6) = tan
-1
(1/12)  
Question 8 
A carrom board (4 ft  4 ft square) has the queen at the center. The queen, hit by the striker moves 
to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of 
displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and 
(c) from the center to the hole. 
  
Solution 8 
  
In ?ABC, tan ? = x/2 and in ?DCF, tan ?= (2 - x)/4, So, (x/2) = (2 - x)/4. Solving, 
 4 - 2x = 4x 
Or, 6x = 4 
 Or, x = 2/3 ft 
a) In ?ABC, AC = (AB
2
+ BC
2
)
1/2
 =  ft 
b) In ?CFD, DF = 1 - (2/3) = 4/3 ft 
So, CF = 4 ft. Now, CD = (CF
2
+ FD
2
)
1/2
 = 4v10/3 ft 
c) In ?ADE, AE = (AE
2
+ ED
2
)
1/2
 = 2v2 ft.  
Question 9 
A mosquito net over a 7 ft 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through 
which a mosquito enters the bed. It flies and sits at the diagonally opposite upper corner of the net. (a) 
Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length 
of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components 
of the displacement vector.  
Solution 9 
The displacement vector is given by r= 7i+4j +3k 
Magnitude of displacement of the mosquitoes= (7
2
+4
2
+3
2
)
1/2
 = 74ft 
The components of displacement along x axis, y axis and z axis are 7 units, 4 units and 3 units 
respectively.  
Question 10 
Suppose  is a vector of magnitude 4.5 unit due north. What is the vector(a) 3 , (b) -4 ? 
Solution 10 
a = 4.5 n, where n is unit vector in north direction 
a) 3a = 4.5 X 3n = 13.5 in north direction 
b) -4a = 4.5 X -4n = 18 in south direction 
Question 11 
Two vectors have magnitudes 2 m and 3 m. The angle between them is . Find (a) the scalar product 
of the two vectors, (b) the magnitude of their vector product? 
Solution 11 
We have a = 2m, b = 3m. 
? = 60
0
 is the angle between the two vectors 
Scalar product between the two vectors = a.b = 
2×3×cos(60
0
)=3m
2
 
Vector product between the two vectors = a×b = 
2×3×sin(60
0
)=3v3m
2
 
Question 12 
Let  be a regular hexagon. Write the x-components of the vectors represented by 
the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that 
  
Use the known cosine values to verify the result. 
  
  
Solution 12 
From polygon law of vector addition, the resultant of the six vectors can be affirmed to be zero. Here 
their magnitudes are the same. 
That is, A = B = C = D = E = F. 
Rx = A cos? + A cos ?/3 + A cos 2?/3 + A cos 3?/3 + A cos 
4 ?/4 + A cos 5?/5 = 0 [As resultant is zero, x component of 
resultant is also 0] 
Now taking A common and putting Rx=0, 
cos ? + cos ?/3 + cos 2 ?/3 + cos 3 ?/3 + cos 4?/3 + cos 5?/3 = 0 
Question 13 
Let  and . Find the angle between them. 
Solution 13 
a = 2i +3j+4k and b = 3i + 4j +5k 
Let angle between them is ?  
Then a.b =2.3 +3.4 + 4.5 = 38 
|a| =   
|b| =   
Now cos ? = a.b/|a||b| = 38/   
Question 14 
Prove that . 
Solution 14 
(A×B)= ABsin? ?, where ? is a unit vector perpendicular to both A and B. 
Now, A.(A×B) is basically a dot product between two vectors which are perpendicular to each other. 
Then cos90° = 0, and thus 
A.(A×B) = 0 
Question 15 
If  and , find . 
Solution 15 
A = 2i + 3j + 4k, B = 4i + 3j + 2k 
A×B =  = -6i + 12j-6k  
Question 16 
If  are mutually perpendicular, show that . Is the converse true?  
Solution 16 
A, B and C are mutually perpendicular vectors. Now, if we take cross product between any two vectors, 
the resultant vector will be in parallel to the third vector, as there are only three axis perpendicular to 
each other. 
So if we consider (A×B), then it is parallel to C, and so angle between the resultant vector and C is 0°, 
and sin(0°)=0. So, C×(A×B) = 0 
Question 17 
A particle moves on a given straight line with a constant v. At a certain time it is at a point P on its 
straight line path. O is a fixed point. Show that  is independent of the position P.  
Solution 17 
The particle moves from PP' in a straight line with a constant speed v. 
From the figure, we see that OP×v = (OP)v sin? ?, where ? is a unit vector perpendicular to v and 
OP. Now, 
We know, OQ= OP sin? = OP'sin?' 
So, the position of the particle may vary, but the magnitude and direction of OP×v will be constant. 
OP×v is independent of P.