Definition: The Dot Product, aka the Scalar
Product
I
−
−
If →
v =< v1 , v2 > and →
w =< w1 , w2 >, then
def
→
−
−
v ·→
w = < v1 , v2 > · < w1 , w2 > = v1 w1 + v2 w2
I
−
−
If →
v =< v1 , v2 , v3 > and →
w =< w1 , w2 , w3 >, then
def
→
−
−
v ·→
w = < v1 , v2 , v3 > · < w1 , w2 , w3 > = v1 w1 + v2 w2 + v3 w3 .
I
−
−
In general, if →
v =< v1 , v2 , . . . , vn > and →
w =< w1 , w2 , . . . , wn >,
then
def
→
−
−
v ·→
w = < v1 , . . . , vn > · < w1 , . . . , wn > = v1 w1 + v2 w2 + · · · + vn wn .
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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Important Fact: The Dot/Scalar Product
def
→
−
−
v ·→
v = v1 v1 + v2 v2 + · · · + vn vn
= v12 + v22 + · · · + vn2
2
q
2
2
v1 + · · · + vn
=
def
−
= k→
v k2
The dot product of a vector with itself is equal to the square of the length
of that vector.
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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In Class Work
→
−
−
−
Let →
a =< 3, 5 >, b =< 7, 2 >, and →
c =< 1, 3, −2 >.
→
−
−
1. Find →
a · b , if possible. If it’s not possible, why not?
→
− −
2. Find b · →
c , if possible. If it’s not possible, why not?
→
−
−
3. Is the angle between →
a and b acute, orthogonal, or obtuse? (Recall:
→
−
−
this implies the smaller of the two angles between →
a and b .)
Do (3) without calculating the angle.
→
−
−
4. Find the angle between →
a and b .
→
−
5. Find two vectors perpendicular to b .
−
6. Find three vectors perpendicular to →
c.
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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Solutions:
→
−
−
−
Let →
a =< 3, 5 >, b =< 7, 2 >, and →
c =< 1, 3, −2 >.
→
−
−
1. Find →
a · b , if possible. If it’s not possible, why not?
→
−
→
−
a · b =< 3, 5 > · < 7, 2 >= 21 + 10 = 31.
→
− −
2. Find b · →
c , if possible. If it’s not possible, why not?
→
−
−
Because b sits in 2-dimensions and →
c sits in 3, this is not possible.
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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Solutions:
→
−
−
−
Let →
a =< 3, 5 >, b =< 7, 2 >, and →
c =< 1, 3, −2 >.
→
−
−
3. Is the angle between →
a and b acute, orthogonal, or obtuse?
Do (5) without calculating the angle.
We know that
→
−
→
−
a · b = kak kbk cos(θ).
→
−
→
−
−
−
a · b depends
Since the magnitudes k→
a k, k b k are positive, the sign of →
completely on the sign of cos(θ) ... which in turn depends completely on
whether the angle is acute or obtuse.
→
−
−
Thus since →
a · b > 0, θ must be acute.
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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Solutions:
→
−
−
−
Let →
a =< 3, 5 >, b =< 7, 2 >, and →
c =< 1, 3, −2 >.
→
−
−
4. Find the angle between →
a and b .
→
−
→
−
a · b = kak kbk cos(θ)
√
√
31 =
9 + 25 49 + 4 cos(θ)
31
√ √
= cos(θ)
34 53
31
θ = arccos √
≈ arccos(0.73027)
1802
θ = .752 radians
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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Solutions:
→
−
−
−
Let →
a =< 3, 5 >, b =< 7, 2 >, and →
c =< 1, 3, −2 >.
→
−
5. Find two vectors perpendicular to b .
→
−
In order to be perpendicular to b , the dot product must be zero.
< x, y > · < 7, 2 >= 0 ⇒ 7x + 2y = 0.
The vectors < −2, 7 > and < 2, −7 > seem like the most obvious choices.
Any other choice will turn out to be a multiple of these two, since there
are only two directions perpendicular to this vector in 2-space.
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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Solutions:
→
−
−
−
Let →
a =< 3, 5 >, b =< 7, 2 >, and →
c =< 1, 3, −2 >.
−
6. Find three vectors perpendicular to →
c.
< x, y , z > · < 1, 3, −2 >= 0 ⇒ 1x + 3y − 2z = 0.
I
If x = 1 and y = 1, then 1 + 3 − 2z = 0, so z = 2. Thus < 1, 1, 2 >
−
is orthogonal to →
c
I
If x = 1 and z = 1, then 1 + 3y − 2 = 0, so y = 31 . Thus < 1, 13 , 1 >
−
is orthogonal to →
c as well.
I
If y = 1 and z = 1, then x + 3 − 2 = 0, so x = −1. Thus
−
< −1, 1, 1 > is a third vector orthogonal to →
c.
Notice: none of these three are multiples of any of the others. In space,
there are an infinite number of directions perpendicular to any given vector.
Math 236-Multi (Sklensky)
In-Class Work
February 6, 2013
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