# CH.1 Problems

```Sheet on CH.1 (Problems with only final solution )
Problem π. π
Μ = 4ππ₯ + 5ππ¦ + 6ππ§ and π
Μ = 2ππ₯ − 3ππ¦ + 6ππ§ , find A + π and A − B.
If π
Solution
Μ−π
Μ = 2ππ₯ + 8ππ¦
π
Problem π. π
Μ = 3ππ₯ + ππ¦ + 2ππ§ and B
Μ&times;B
Μ = ππ₯ + 2ππ¦ + ππ§ , find A
Μ.
If A
Solution
Μ&times;π
Μ = −3ππ₯ − ππ¦ + 5ππ§
π
Problem π. π
Given three vectors
Μ = 2ππ₯ + ππ¦
π
Μ = 2ππ₯ + 2ππ¦ − 2ππ§
π
πΜ = 2ππ¦ + 2ππ§
Μ&times;π
Μ&times;π
Μ (ii) (π
Μ ) &times; πΜ
Find (i) π
Solution
Μ&times;π
Μ = −2ππ₯ + 4ππ¦ + 2ππ§
(i) π
Μ&times;π
Μ ) &times; πΜ = 4ππ₯ + 4ππ¦ = 4ππ§
(ii) (π
Problem 1.4
Μ be (ππ , π¬π’π§β‘ π, π) while π
Μ is (ππ , π, ππ¨π¬β‘ π). Compute π¨
Μ and π¨
Μ.
Let π
Solution
Μ β‘ = π‘ 2 π π‘ + π‘sinβ‘ π‘ + 4cosβ‘ π‘
π¨
Μ == ππ₯ (sinβ‘ π‘cosβ‘ π‘ − 4π‘) + ππ¦ (4π π‘ − π‘ 2 cosβ‘ π‘) + ππ§ (π‘ 3 − π π‘ sinβ‘ π‘)
π¨
Problem π. π
Given π΄Μ = 2ππ₯ + 2ππ¦ − ππ§ , π΅ = 6ππ₯ − 3ππ¦ + 2ππ§ , Find the unit vector perpendicular to both π΄Μ
and π΅Μ.
1
2
18
Μπ =
Solution π
π − 17 ππ¦ − 5 17 ππ§.
5 17 π₯
√
√
√
Problem π. π
Show that the vector directed from π(π₯1 , π¦1 , π§1 ) to π(π₯2 , π¦2 , π§2 ) in Fig. P.1.6 is given by
(π₯2 − π₯1 )ππ₯ + (π¦2 − π¦1 )ππ¦ + (π§2 − π§1 )ππ§
Solution
Fig. P.1.6
Μ = (π₯2 − π₯1 )ππ₯ + (π¦2 − π¦1 )ππ¦ + (π§2 − π§1 )ππ§
Μ−π
π
Problem π. π
Μ directed from (2, −4,1) to (0, −2,0) in Cartesian coordinates and find the unit
Find the vector π
Μ
vector along π.
Solution
Μ
π
2
2
1
πΜπ΄ =
= − ππ₯ + ππ¦ − ππ§
Μ|
3
3
3
|π
Problem π. π
Find the distance between (5,3π/2,0) and (5, π/2,10) in cylindrical coordinates.
Solution
Μ = 10ππ¦ + 10ππ§ and the required distance between the points is
Μ−π
π
|π − π| = 10√2
Problem π. π
Μ = 4ππ₯ − 2ππ¦ − ππ§ and π
Μ = ππ₯ + 4ππ¦ − 4ππ§ are mutually perpendicular.
Show that π
Problem π. ππ
Μ = 2ππ₯ + 4ππ¦ and π
Μ = 6ππ¦ − 4ππ§ , find the smaller angle between them using
Given π
(a) the cross product, (π) the dot product.
Solution
(a) π = 41.9β
(b)β‘π = 41.9β
Problem π. ππ
Μ = (π¦ − 1)ππ₯ + 2π₯ππ¦ , find the vector at (2,2,1) and its projection on π
Μ , where π
Μ = 5ππ₯ −
Given π
ππ¦ + 2ππ§ .
Μ on π
Μ= 1
Solution Proj. π
√30
Problem π. ππ
Μ = ππ₯ + ππ¦ , π
Μ&times;π
Μ = ππ₯ + 2ππ§ and πΜ = 2ππ¦ + ππ§ , find (π
Μ ) &times; πΜ and compare it with
Given π
Μ
Μ
Μ
π &times; (π &times; π).
Solution
Μ&times;π
Μ ) &times; πΜ = −2ππ¦ + 4ππ§
(π
Μ
Μ &times; πΜ) = 2ππ₯ − 2ππ¦ + 3ππ§ .
π &times; (π
Problem π. ππ
Μ, π
Μ⋅π
Μ&times;π
Μ , and πΜ of Problem 1.12, find π
Μ &times; πΜ and compare it with π
Μ ⋅ πΜ.
Using the vectors π
Solution
Μ⋅π
Μ &times; πΜ = −5
π
Μ
Μ ⋅ πΜ = −5
π&times;π
Problem π. ππ
Express the unit vector which points from π§ = β on the π§ axis toward (π, π, 0) in cylindrical
coordinates. See Fig. 1.14.
Fig. P.1.14
Solution
πΜπ
=
πππ −βππ§
√π 2 +β 2
Problem π. ππ
Express the unit vector which is directed toward the origin from an arbitrary point on the plane
π§ = −5, as shown in Fig. P.1.15.
Fig. P.1.15
Solution
πΜπ =
−π₯ππ₯ − π¦ππ¦ + 5ππ§
√π₯ 2 + π¦ 2 + 25
Problem π. ππ
Use the spherical coordinate system to find the area of the strip πΌ ≤ π ≤ π½ on the spherical shell
of radius π (Fig. P.1.16). What results when πΌ = 0 and π½ = π ?
π§
π¦
π₯
Fig. P.1.16
Μ = 2ππ2 (cosβ‘ πΌ − cosβ‘ π½)
Solutionβ‘πΊ
When πΌ = 0 and π½ = π,β‘β‘S= 4ππ2
Problem π. ππ
Obtain the expression for the volume of a sphere of radius ‘a’ from the differential volume.
4
Solution π£ = 3 ππ3
Problem π. ππ
Use the cylindrical coordinate system to find the area of the curved surface of a right circular
cylinder where π = 2 m, β = 5 m, and 30β ≤ π ≤ 120β (see Fig. P.1.16).
Fig. P.1.16
Solution 5πβ‘πΜπ β‘β‘m2
Problem π. ππ
Transform
Μ = π¦ππ₯ + π₯ππ¦ +
π
π₯2
√π₯ 2 + π¦ 2
ππ§
from Cartesian to cylindrical coordinates.
Solution
Μ = 2πsinβ‘ πcosβ‘ πππ + (πcos2 β‘ π − πsin2 β‘ π)ππ + πcos2 β‘ πππ§
π
Problem π. ππ
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 π΄Μ = ππ₯ + ππ¦
√2
√2
Example 1.21
A point in cartesian coordinates is given by ( , , ). Express it in cylindrical coordinates.
Solution(2.236, 63.43β , 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 Μ( , !&quot; #) = −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, 33β 26. , 3/
(b) P+√14, 36.7β , 333.4β /
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 . &lt;
Find the volume of a sphere of radius ' ' from the differential volume.
&gt;?@ A
Solution V =
B
Problem . C
Transform the vector
Solution
=
−
+
into spherical co-ordinate
= Dcos ! ED − Dcos! sin! E% − Dsin! E(
=
```