#$&*
course Mth 277
6/15 7 pm
qa 09_3The dot product of two vectors is equal to the sum of the products of their components:
If A = a_1 i + a_2 j + a_3 k andB = b_1 i + b_2 j + b_3 k, then A dot B = a_1 b_1 + a_2 b_2 + a_3 b_3.
The dot product is also equal to || A || * || B || cos(theta), where theta is the angle between the vectors. So
A dot B = a_1 b_1 + a_2 b_2 + a_3 b_3
A dot B = || A || * || B || cos(theta).
So for example if we know the components of A and B, we can easily find the dot product and the magnitudes of the two vectors. Having found the magnitudes and the dot product we can use the second relationship to get
cos(theta) = A dot B / || A || || B || so that
theta = arcCos( A dot B / || A || || B || ).
Two vectors are perpendicular to one another if the angle between them is 90 degrees. The cosine of 90 degrees is zero, and if the cosine of an angle between 0 and 180 degrees is zero the angle is 90 degrees. So two vectors are perpendicular if, and only if, their dot product is zero.
If the dot product of two vectors is zero we say that the vectors are orthogonal. In two or three dimensions, that means that the angle between the two vectors is 90 degrees.
Section 9.3
If you have a good Precalculus II background (or equivalent) with an appropriate introduction to vectors, or if you have worked through the recommended qa's on vectors, from Precalculus I, you should have little trouble with the query questions. So no q_a_ questions are included with this assignment.
For your convenience here's a listing of the Query questions:
1) 9.3.4 Find v dot u when v =<1,-5,0> u =<0,-4,2>.
<1,-5,0> dot <0,-4,2) is equal to each element of the first multiplied by its corresponding element in the second, added together.
(1)(0) + (-5)(-4) + (0)(2) = 0 + 20 + 0 = 20
2) 9.3.6 Find v dot w when v = 4i + j and w =3i + 2k.
Same as above:
(tricky, there's some components missing)
(4)(3) + (1)(0) + (0)(2) = 12 + 0 = 12
3) 9.3.10 Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal.
The vectors are orthogonal if their dot product is 0.
(5)(8) + (-5)(-10) + (5)(-2) = 40 + 50 - 10 = 80.
Therefore, the vectors are not orthogonal.
4) 9.3.12 Let v = 4i - 2j + k and w = -2i + j - k. Evaluate (v dot w) * w.
v dot w = (4)(-2) + (-2)(1) + (-2)(1) = -8 - 2 - 2 = -12.
v dot w * w = 24i - 12j + 12k
5) 9.3.16 Find the angle between v = 2i +3 k and w = -j + 4k.
v dot w = |v||w|cos(theta)
v dot w = (2)(0) + (0)(-1) + (3)(4) = 12
|v| = sqrt ( 2^2 + 3^2 )
|v| = sqrt ( 13 )
|w| = sqrt ( -1^2 + 4^2)
|w| = sqrt ( 1 + 16)
|w| = sqrt ( 17 )
Therefore:
sqrt(13) * sqrt(17) * cos(theta) = 12
theta = acos(12/sqrt(221))
theta ~= 36.18 degrees
6) 9.3.20 Find the scalar and vector projections of v = i - 2j onto w = j - 2k.
The scalar projection is equal to the the dot product between the two vectors divided by the magnitude of the second vector.
v dot w = (1)(0) + (-2)(1) + (0)(-2) = -2
magnitude v = sqrt ( 0^2 + 1^2 + (-2)^2 )
= sqrt ( 5 )
scalar projection of v onto w = -2/sqrt(5)
The vector proejction is equal to the dot product divided by the magnitude of the second vector squared, all multiplied by the second vector.
v dot w = -2
magnitude v = sqrt (5 )
vector projection of v onto w
= (-2/5)*(j-2k)
= -2/5 j+ 4/5 k
7) 9.3.24 Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k.
The vectors <2,0,1> and <4,0,2> are both orthogonal to both `v and `w, since their dot products result in a value of 0.
@&
Neither of these is a unit vector.
*@
8) 9.3.30 Find x so that v = 2i - xj + 3k and w = -2i + j + xk are orthogonal.
The dot product of these two vectors is equal to -4 + 2x. Therefore, for the two vectors to be orthogonal, it must equal 0.
The only value for x that would result in the dot product equalling 0 is 2.
9) 9.3.32 Give the direction cosines and direction angles of v = i - 4j.
Each direction cosine is equal to the coefficient of each unit vector divided by the magnitude of the whole vector.
Magnitude = sqrt (1^2 + 0^2 + (-4)^2)
= sqrt (17)
Cosine i = 1/sqrt(17)
Cosine j = 0
Cosine k = -4/sqrt(17)
10) 9.3.36(b,c,d) Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w
Cos(theta) is equal to the dot product over the magnitudes multiplied together.
v dot w = (1)(-1) + (-1)(3) + (4)(2) = 3
|v| = sqrt (1^2 + (-1)^2 + 4^2 ) = sqrt (18)
|w| = sqrt ((-1)^2 + 3^2 + 2^2) = sqrt (14)
Cos(theta) = 3/sqrt(252)
Vectors are orthogonal iff their dot product is 0.
sv - w = (s+1)i + (s-4)j + (4s-2)k
v dot (sv - w):
s+1 + (4-s) + (16s-8) = 16s -3
For this to equal 0, s must equal 3/16.
v - tw = (1 + t)i + (-1-3t)j + (4-2t)k
v dot (v - tw):
-1-t + (1+3t) + (16-8t) = -6t + 16
For this value to equal 0, t must equal 16/6 or 8/3.
11) 9.3.40 Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11).
Work equals the dot product of the force and distance moved.
Displacement = -7i - 14j - 7k
Force dot displacement:
Work = (-42/11)i + (28/11)j - (42/11)k"
@&
Very well done.
However check my note on #7. If after revision you aren't 100% sure of your solution, submit a copy of that problem and your solution and/or questions and I'll be glad to give you more feedback.
*@