| Date post: | 07-Aug-2018 |
| Category: |
Documents |
| Upload: | john-luke-cruz |
| View: | 219 times |
| Download: | 0 times |
of 7
8/20/2019 PHY10 Lesson 1 Vectors Part 2
1/15
PHY10 Lesson 1
UNIT VECTORSAND
VECTOR PRODUCTS
8/20/2019 PHY10 Lesson 1 Vectors Part 2
2/15
UNIT VECTOR
A unit vector is a vector with a magnitude of exactly
1 and points in a particular direction.
+x
+y
+z
A vector in its “regular”form or “magnitude-angle”form, such as
can be expressed in “unitvector” form or “rectangular”
form.
î
jˆ
k ̂
xabovem A o += 30,00.10
jmim A ˆ)00.5(ˆ)66.8( +=
8/20/2019 PHY10 Lesson 1 Vectors Part 2
3/15
+x
+y
+z
VECTOR ADDITION
+x
+y
+z
i A x
ˆ
k A zˆ
j A yˆ
A
B
i B xˆ
k B Z ˆ
j B yˆ
k m jmim A ˆ
)6(ˆ
)8(ˆ
)10( ++=
k m jmim B ˆ
)4(ˆ)12(ˆ)8( ++−=
8/20/2019 PHY10 Lesson 1 Vectors Part 2
4/15
Find the resultant of the following vectors.
k m jmim A ˆ)6(ˆ)8(ˆ)10( ++=
k m jmim B ˆ)4(ˆ)12(ˆ)8( ++−=
k m jmimC ˆ)10(ˆ)20(ˆ)2( ++=
B AC +=k m jmim A
ˆ)6(ˆ)8(ˆ)10( ++=
k m jmim B ˆ)4(ˆ)12(ˆ)8( ++−=
8/20/2019 PHY10 Lesson 1 Vectors Part 2
5/15
VECTOR PRODUCTS: Scalar/Dot Prodct
Given the following vectors
!he scalar "or dot# product of these vectors is defined as
$here A is the magnitude of
B is the magnitude of
φ is the angle between and .
k A j Ai A A z y xˆˆˆ ++= k B j Bi B B z y x
ˆˆˆ ++=
φ cos AB B A =⋅
A
B
A B
8/20/2019 PHY10 Lesson 1 Vectors Part 2
6/15
VECTOR PRODUCTS: Scalar/Dot Prodct
φ cos AB B A =⋅
10cos)1)(1(ˆˆ ==⋅ii
090cos)1)(1(ˆˆ ==⋅ ji
090cos)1)(1(ˆˆ ==⋅k i
090cos)1)(1(ˆˆ ==⋅i j
10cos)1)(1(ˆˆ ==⋅ j j
090cos)1)(1(ˆˆ ==⋅k j
090cos)1)(1(ˆˆ ==⋅ik
090cos)1)(1(ˆˆ ==⋅ jk
10cos)1)(1(ˆˆ ==⋅k k
8/20/2019 PHY10 Lesson 1 Vectors Part 2
7/15
VECTOR PRODUCTS: Scalar/Dot Prodct
k B j Bi Bk A j Ai A B A z y x z y xˆˆˆˆˆˆ ++⋅++=⋅
k B j Bi B B z y xˆˆˆ ++=k A j Ai A A z y x
ˆˆˆ ++=
( ) ( ) ( ) z z y y x x B A B A B A B A ++=⋅
φ cos AB B A =⋅
8/20/2019 PHY10 Lesson 1 Vectors Part 2
8/15
Find the dot product of the following vectors
k m jmim A ˆ
)6(ˆ
)8(ˆ
)10( ++=
k m jmim B ˆ
)4(ˆ
)12(ˆ
)8( ++−=
( ) ( ) ( ) z z y y x x B A B A B A B A ++=⋅
)]4)(6[()]12)(8[()]8)(10[( mmmmmm B A ++−=⋅
222 249680 mmm B A ++−=⋅
240m B A =⋅
8/20/2019 PHY10 Lesson 1 Vectors Part 2
9/15
Find the angle between the following vectors
k m jmim A ˆ
)6(ˆ
)8(ˆ
)10( ++=
k m jmim B ˆ
)4(ˆ
)12(ˆ
)8( ++−=
φ cos AB B A =⋅
( ) ( ) φ cos22420040 =
224200
40cos 1−=φ
o11.79=φ
8/20/2019 PHY10 Lesson 1 Vectors Part 2
10/15
VECTOR PRODUCTS: !ector/Cross Prodct
Given the following vectors
!he vector "or cross# product of these vectors is definedas
$here A is the magnitude of
B is the magnitude of
φ is the angle between and .
k A j Ai A A z y xˆˆˆ ++= k B j Bi B B z y x
ˆˆˆ ++=
φ sin AB B A =×
A
B
A B
8/20/2019 PHY10 Lesson 1 Vectors Part 2
11/15
VECTOR PRODUCTS: Vector/Cross Prodct
φ sin AB B A =×
00sin)1)(1(ˆˆ ==× ii
k ji ˆ90sin)1)(1(ˆˆ ==×
jk i ˆ90sin)1)(1(ˆˆ −==×
k i j ˆ90sin)1)(1(ˆˆ −==×
00sin)1)(1(ˆˆ ==× j j
ik j ˆ90sin)1)(1(ˆˆ ==×
jik ˆ90sin)1)(1(ˆˆ ==×
i jk ˆ90sin)1)(1(ˆˆ −==×
00sin)1)(1(ˆˆ ==× k k
8/20/2019 PHY10 Lesson 1 Vectors Part 2
12/15
VECTOR PRODUCTS: Vector/Cross Prodct
k B j Bi Bk A j Ai A B A z y x z y x ˆˆˆˆˆˆ ++×++=×
k B j Bi B B z y xˆˆˆ ++=k A j Ai A A z y x
ˆˆˆ ++=
φ sin AB B A =×
( ) ( ) ( ) k B A B A j B A B Ai B A B A B A x y y x z x x z y z z yˆˆˆ −+−+−=×
8/20/2019 PHY10 Lesson 1 Vectors Part 2
13/15
Find the cross product of the following vectors
k m jmim A
ˆ
)6(ˆ
)8(ˆ
)10( ++=
k m jmim B ˆ
)4(ˆ
)12(ˆ
)8( ++−=
( ) ( ) ( ) k B A B A j B A B Ai B A B A B A x y y x z x x z y z z y ˆˆˆ −+−+−=×
( ) ( ) ( ) k ji B A ˆ)64(120ˆ4048ˆ7232 −−+−−+−=×
( ) ( ) ( ) k m jmim B A ˆ184ˆ88ˆ40 222 +−−=×
8/20/2019 PHY10 Lesson 1 Vectors Part 2
14/15
PRO"LE#:
8/20/2019 PHY10 Lesson 1 Vectors Part 2
15/15
PRO"LE#:
%n the figure at right, a cube isplaced so that one corner "point is at the origin and threeedges are along the 'x 'y, and'( axes of a coordinate system.
)se unit vectors to find the
"a# angle between the edgealong the (-axis "line &*# andthe diagonal of the cube "line&+#.
"b# angle between the diagonalof a face "line &# and line &+.
