cosbab.a
a
bSuppose the angle between two vectors a and b is
The scalar product is written as a . b and is defined as
The dot must NEVER be missed out.
The scalar product is sometimes called the “dot” product.
The result of the scalar product is a scalar quantity not a vector !
The Vector or x ProductIn C4 the scalar product was covered a . b = |a||b|cosq Pronounced a dot b
Using this definition:i . i = j . j = k . k = 1 as the angle between these unit vectors is zero and cos0 = 1
i . j = 0 j . k = 0 i . k = 0j . i = 0 k . j = 0 k . i = 0
So an answer is obtained when the components of the vector are in the same direction.
cosbab.a
SUMMARY
For the scalar product of 2 column vectors,
3
2
1
aaa
a
3
2
1
bbb
be.g. and
we multiply the “tops”,
and add the results. So,
332211. babababa
“middles” and “bottoms”
e.g.1 Find the scalar product of the vectors
31
2a
224
band
332211. babababa Solution:
224
31
2. .ba
)2)(3()2)(1()4)(2( 12628
Perpendicular Vectors
Then 0cosba
either a = 0 or cosθ=0
a = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must have
cosθ = 090θ
The vectors are perpendicular.
b = 0 or
If a.b = 0
the product of 2 vectors divided by
Finding Angles between Vectors
cosbab.a ba
b.acos
The scalar product can be rearranged to find the angle between the vectors.
Notice how careful we must be with the lines under the vectors.
The r.h.s. is
the product of the 2 magnitudes of the vectors
Solution:
)1)(1()1)(1()2)(1(. ba 2
,3111 222 a 6112 222 b
ba
b.acos
e.g. Find the angle between kjia kjib 2
and
63
2cos 961
( 3s.f. )
Tip: If at this stage you get zero, STOP. The vectors are perpendicular.
)3,1,2(,)0,1,1(,)3,1,2( CBA
When solving problems, we have to be careful to use the correct vectors.e.g. The triangle ABC is given by
Solution: ( any shape triangle will do )
We always sketch and label a triangle
A
B
C
ba
b.acos
BUT the a and b of the formula are not the a and b of the question.
We need the vectors and AB CB
Find the cosine of angle ABC.
)3,1,2(,)0,1,1(,)3,1,2( CBA
321
31
2
011
abAB
321
31
2
011
cbCB
�⃗�𝐁=𝐁 𝐬−𝐀 𝐬𝐀𝐁𝐁𝐀𝐑𝐮𝐥𝐞
,14321 222 AB
321
AB
321
CB
14321 222 CB
1414
4cos
321
321
. .CBAB
4941
CBAB
CB.ABcos
7
2cos
14
4cos
2
7
Finding Angles between Lines
e.g.
With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.
a
We use the 2 direction vectors only since these define the angle.
( If the obtuse angle is found, subtract from 180 . )
ba
b.acos where
Solution:
sr23
1
tr
102
and
210
21
1
e.g. Find the acute angle, a, between the lines
sr23
1
tr
102
and
Solution:
ba
b.acos where
a and
210
210
21
1
e.g. Find the acute angle, a, between the lines
b
210
sr23
1
tr
102
and
Solution:
ba
b.acos where
a
21
1
and
21
1
210
e.g. Find the acute angle, a, between the lines
e.g. Find the acute angle, a, between the lines
sr23
1
tr
102
and
Solution:
ba
b.acos where
a
21
1
and
b
21
1
)2)(2()1)(1()1)(0(. ba 5
,521 22 a 6211 222 b
65
5cos
156 24
210
210
(nearest degree)Autograph
The Vector ProductThe vector product is defined as a b = |a| |b|sin is a unit vector perpendicular to both a and b.
To determine the direction of use the right hand rule where finger 1 is a, finger 2 is b and the thumb is a b
a b is out of the paper as the direction of is out
a b is into the paper as the direction of is in
Using this definition:i i = j j = k k = as the angle between these unit vectors is zero and sin0 = 0
ij = jk = ki =
ik = ji = kj =
a b = |a||b|sin
0
k i j
–j –k –i
a b = (a1i + a2j + a3k) (b1i + b2j + b3k)
= (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k
2 3 3 2
3 1 1 3
1 2 2 1
a b - a b
a b - a b
a b - a b
This result is on pg 4 of the formula book written as a column vector
= a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i
a b = |a||b|sin
Link to applet
Ex
2 0
1 1
2 3
1 1 2 3 3 2
2 2 3 1 1 3
3 3 1 2 2 1
a b a b - a b
a b a b - a b
a b a b - a b
Link to demo
5
6 5i 6j 2k
2
1 3 ( 2) 1
( 2) 0 2 3
2 1 1 0
So the vector is perpendicular to 5
6
2
2 0
1 and 1
2 3
This fact will be essential in understanding the equation of a plane
Link to worksheet