Sheet on CH.1 (Problems with only final solution )
Problem 𝟏. 𝟏
̅ = 4𝑎𝑥 + 5𝑎𝑦 + 6𝑎𝑧 and 𝐁
̅ = 2𝑎𝑥 − 3𝑎𝑦 + 6𝑎𝑧 , find A + 𝐁 and A − B.
If 𝐀
Solution
̅−𝐁
̅ = 2𝒂𝑥 + 8𝒂𝑦
𝐀
Problem 𝟏. 𝟐
̅ = 3𝑎𝑥 + 𝑎𝑦 + 2𝑎𝑧 and B
̅×B
̅ = 𝑎𝑥 + 2𝑎𝑦 + 𝑎𝑧 , find A
̅.
If A
Solution
̅×𝐁
̅ = −3𝑎𝑥 − 𝑎𝑦 + 5𝑎𝑧
𝐀
Problem 𝟏. 𝟑
Given three vectors
̅ = 2𝑎𝑥 + 𝑎𝑦
𝐀
̅ = 2𝑎𝑥 + 2𝑎𝑦 − 2𝑎𝑧
𝐁
𝐂̅ = 2𝑎𝑦 + 2𝑎𝑧
̅×𝐁
̅×𝐁
̅ (ii) (𝐀
̅ ) × 𝐂̅
Find (i) 𝐀
Solution
̅×𝐁
̅ = −2𝑎𝑥 + 4𝑎𝑦 + 2𝑎𝑧
(i) 𝐀
̅×𝐁
̅ ) × 𝐂̅ = 4𝑎𝑥 + 4𝑎𝑦 = 4𝑎𝑧
(ii) (𝐀
Problem 1.4
̅ be (𝒕𝟐 , 𝐬𝐢𝐧⁡ 𝒕, 𝟒) while 𝐁
̅. 𝑩
̅×𝑩
̅ is (𝒆𝒕 , 𝒕, 𝐜𝐨𝐬⁡ 𝒕). Compute 𝑨
̅ and 𝑨
̅.
Let 𝐀
Solution
̅. 𝑩
̅ ⁡ = 𝑡 2 𝑒 𝑡 + 𝑡sin⁡ 𝑡 + 4cos⁡ 𝑡
𝑨
̅×𝑩
̅ == 𝑎𝑥 (sin⁡ 𝑡cos⁡ 𝑡 − 4𝑡) + 𝑎𝑦 (4𝑒 𝑡 − 𝑡 2 cos⁡ 𝑡) + 𝑎𝑧 (𝑡 3 − 𝑒 𝑡 sin⁡ 𝑡)
𝑨
Problem 𝟏. 𝟓
Given 𝐴̅ = 2𝑎𝑥 + 2𝑎𝑦 − 𝑎𝑧 , 𝐵 = 6𝑎𝑥 − 3𝑎𝑦 + 2𝑎𝑧 , Find the unit vector perpendicular to both 𝐴̅
and 𝐵̅.
1
2
18
̂𝑛 =
Solution 𝒂
𝑎 − 17 𝑎𝑦 − 5 17 𝑎𝑧.
5 17 𝑥
√
√
√
Problem 𝟏. 𝟔
Show that the vector directed from 𝑀(𝑥1 , 𝑦1 , 𝑧1 ) to 𝑁(𝑥2 , 𝑦2 , 𝑧2 ) in Fig. P.1.6 is given by
(𝑥2 − 𝑥1 )𝐚𝑥 + (𝑦2 − 𝑦1 )𝐚𝑦 + (𝑧2 − 𝑧1 )𝐚𝑧
Solution
Fig. P.1.6
̅ = (𝑥2 − 𝑥1 )𝐚𝑥 + (𝑦2 − 𝑦1 )𝐚𝑦 + (𝑧2 − 𝑧1 )𝐚𝑧
̅−𝐀
𝐁
Problem 𝟏. 𝟕
̅ directed from (2, −4,1) to (0, −2,0) in Cartesian coordinates and find the unit
Find the vector 𝐀
̅
vector along 𝐀.
Solution
̅
𝐀
2
2
1
𝐚̂𝐴 =
= − 𝐚𝑥 + 𝐚𝑦 − 𝐚𝑧
̅|
3
3
3
|𝐀
Problem 𝟏. 𝟖
Find the distance between (5,3𝜋/2,0) and (5, 𝜋/2,10) in cylindrical coordinates.
Solution
̅ = 10𝐚𝑦 + 10𝐚𝑧 and the required distance between the points is
̅−𝐀
𝐁
|𝐁 − 𝐀| = 10√2
Problem 𝟏. 𝟗
̅ = 4𝐚𝑥 − 2𝐚𝑦 − 𝐚𝑧 and 𝐁
̅ = 𝐚𝑥 + 4𝐚𝑦 − 4𝐚𝑧 are mutually perpendicular.
Show that 𝐀
Problem 𝟏. 𝟏𝟎
̅ = 2𝐚𝑥 + 4𝐚𝑦 and 𝐁
̅ = 6𝐚𝑦 − 4𝐚𝑧 , find the smaller angle between them using
Given 𝐀
(a) the cross product, (𝑏) the dot product.
Solution
(a) 𝜃 = 41.9∘
(b)⁡𝜃 = 41.9∘
Problem 𝟏. 𝟏𝟏
̅ = (𝑦 − 1)𝐚𝑥 + 2𝑥𝐚𝑦 , find the vector at (2,2,1) and its projection on 𝐁
̅ , where 𝐁
̅ = 5𝐚𝑥 −
Given 𝐀
𝐚𝑦 + 2𝐚𝑧 .
̅ on 𝐁
̅= 1
Solution Proj. 𝐀
√30
Problem 𝟏. 𝟏𝟐
̅ = 𝐚𝑥 + 𝐚𝑦 , 𝐁
̅×𝐁
̅ = 𝐚𝑥 + 2𝐚𝑧 and 𝐂̅ = 2𝐚𝑦 + 𝐚𝑧 , find (𝐀
̅ ) × 𝐂̅ and compare it with
Given 𝐀
̅
̅
̅
𝐀 × (𝐁 × 𝐂).
Solution
̅×𝐁
̅ ) × 𝐂̅ = −2𝐚𝑦 + 4𝐚𝑧
(𝐀
̅
̅ × 𝐂̅) = 2𝐚𝑥 − 2𝐚𝑦 + 3𝐚𝑧 .
𝐀 × (𝐁
Problem 𝟏. 𝟏𝟑
̅, 𝐁
̅⋅𝐁
̅×𝐁
̅ , and 𝐂̅ of Problem 1.12, find 𝐀
̅ × 𝐂̅ and compare it with 𝐀
̅ ⋅ 𝐂̅.
Using the vectors 𝐀
Solution
̅⋅𝐁
̅ × 𝐂̅ = −5
𝐀
̅
̅ ⋅ 𝐂̅ = −5
𝐀×𝐁
Problem 𝟏. 𝟏𝟒
Express the unit vector which points from 𝑧 = ℎ on the 𝑧 axis toward (𝑟, 𝜙, 0) in cylindrical
coordinates. See Fig. 1.14.
Fig. P.1.14
Solution
𝐚̂𝑅
=
𝑟𝐚𝑟 −ℎ𝐚𝑧
√𝑟 2 +ℎ 2
Problem 𝟏. 𝟏𝟓
Express the unit vector which is directed toward the origin from an arbitrary point on the plane
𝑧 = −5, as shown in Fig. P.1.15.
Fig. P.1.15
Solution
𝐚̂𝑅 =
−𝑥𝐚𝑥 − 𝑦𝐚𝑦 + 5𝐚𝑧
√𝑥 2 + 𝑦 2 + 25
Problem 𝟏. 𝟏𝟔
Use the spherical coordinate system to find the area of the strip 𝛼 ≤ 𝜃 ≤ 𝛽 on the spherical shell
of radius 𝑎 (Fig. P.1.16). What results when 𝛼 = 0 and 𝛽 = 𝜋 ?
𝑧
𝑦
𝑥
Fig. P.1.16
̅ = 2𝜋𝑎2 (cos⁡ 𝛼 − cos⁡ 𝛽)
Solution⁡𝑺
When 𝛼 = 0 and 𝛽 = 𝜋,⁡⁡S= 4𝜋𝑎2
Problem 𝟏. 𝟏𝟕
Obtain the expression for the volume of a sphere of radius ‘a’ from the differential volume.
4
Solution 𝑣 = 3 𝜋𝑎3
Problem 𝟏. 𝟏𝟖
Use the cylindrical coordinate system to find the area of the curved surface of a right circular
cylinder where 𝑟 = 2 m, ℎ = 5 m, and 30∘ ≤ 𝜙 ≤ 120∘ (see Fig. P.1.16).
Fig. P.1.16
Solution 5𝜋⁡𝑎̂𝜌 ⁡⁡m2
Problem 𝟏. 𝟏𝟗
Transform
̅ = 𝑦𝐚𝑥 + 𝑥𝐚𝑦 +
𝐀
𝑥2
√𝑥 2 + 𝑦 2
𝐚𝑧
from Cartesian to cylindrical coordinates.
Solution
̅ = 2𝑟sin⁡ 𝜙cos⁡ 𝜙𝐚𝑟 + (𝑟cos2 ⁡ 𝜙 − 𝑟sin2 ⁡ 𝜙)𝐚𝜙 + 𝑟cos2 ⁡ 𝜙𝐚𝑧
𝐀
Problem 𝟏. 𝟐𝟎
A vector of magnitude 10 points from (5,5𝜋/4,0) in cylindrical coordinates toward the origin
(Fig. Fig. P.1.20). Express the vector in Cartesian coordinates.
Fig. P.1.20
10
10
Solution 𝐴̅ = 𝐚𝑥 + 𝐚𝑦
√2
√2
Example 1.21
A point in cartesian coordinates is given by ( , , ). Express it in cylindrical coordinates.
Solution(2.236, 63.43∘ , 3)
Example 1.22
If a vector, A is 4
+2
+
, express it in cylindrical coordinate system.
Solution
= 4.46 +
Problem 1.23
Transform the vector 4 − 2
−3, = 4).
Solution ̅( , !" #) = −3.342
−4
$
into spherical coordinates at a point P( = −2,
+ 2.266
%
+ 4.4378
(
Problem 1.24
Find the location of the point (2, −1,3) in cylindrical and spherical co-ordinates.
Solution
(a) *+√5, 33∘ 26. , 3/
(b) P+√14, 36.7∘ , 333.4∘ /
Problem . 0
Transform vector
A=
1 23 2
1 23 23 2
−
1 23 2 3 2
to spherical coordinates.
Solution
A( , !, #)
= sin !(sin !cos # − cos : !sin #)
+sin! cos !(cos # + sin! sin#);%
−sin! sin# (
$
Problem . <
Find the volume of a sphere of radius ' ' from the differential volume.
>?@ A
Solution V =
B
Problem . C
Transform the vector
Solution
=
−
+
into spherical co-ordinate
= Dcos ! ED − Dcos! sin! E% − Dsin! E(
=