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course Mth 277
6/16 9 pm
qa 09_04Section 9.4
The cross product of two vectors A and B is a vector perpendicular to the two vectors, the direction determined by the right-hand rule, and its magnitude is || A || * || B || * sin(theta), where theta is the angle between the two.
The formula for the cross product is a little more complicated than the simple formula for the dot product, but it's not bad:
A `X B is the determinant of the matrix whose first row is [ i, j, k ], whose second row is [ A_1, A_2, A_3 ], the coefficients of the A vector, and whose third row is [ B_1, B_2, B_3 ], the coefficients of the B vector.
Experience shows that students who have succeeded on Assignments 1-3 typically do not require a q_a_ for this assignment, and may move directly to the Query.
For your convenience here's a listing of the Query questions:
1) 9.4.6 Find v X w when v = 10i - 2j + 4k and w =-i -(1/2)j - 3k. (Where X denotes the cross product)
[ i j k ]
[ 10 -2 4 ]
[ -1 -1/2 -3 ]
= [ -2 4 ]i + [ 10 4 ]j + [ 10 -2 ]k
[ -1/2 -3 ] [ -1 -3 ] [ -1 -1/2 ]
V x W = 8i - 34j - 7k
2) 9.4.10 Find v X w when v = sin(theta)i + cos(theta)j and w = -cos(theta)i + sin(theta)j (theta is any angle).
[ i j ]
[ sin(theta) cos(theta)]
[ -cos(theta) sin(theta)]
V x W = sin^2(theta) + cos^2(theta)
V x W = 1
3) 9.4.12 Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k.
The angle between two vectors can be calculated as:
acos( (v dot w)/(|v|*|w|) )
= acos ( sqrt(1 + 1 + 0)/(sqrt(2)*2))
= acos (sqrt(2)/2*sqrt(2))
= acos (1/2)
= pi/3 = 60 degrees
sin(pi/3) = sqrt(3)/2
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You could also have used the fact that the magnitude of the cross product is the product of the magnitudes of the two vectors, multiplied by the sine of the angle.
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4) 9.4.16 Find a unit vector which is orthogonal to both v = -i + 3j and w = i - j - k
< 3,1,2 > is orthogonal to both vectors:
x dot v = -3 + 3 + 0 = 0
x dot w = 3 - 1 - 2 = 0
5) 9.4.18 Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k.
< 1,2,-2 > is orthogonal to both vectors:
x dot v = 2 - 2 - 0 = 0
x dot w = 2 - 0 - 2 = 0
6) 9.4.22 Find the area of the parallelogram determined by the vectors v = 4i + k and w = 4j - k.
The area of a parallelogram determined by 2 vectors is equal to the cross product of the two vectors.
v X w =
[ 0 1 ]i + [ 4 1]j + [ 4 0 ]k
[ 4 -1 ] [ 0 -1] [ 0 4 ]
= -5 - 5 + 16
= 6 square units
7) 9.4.24 Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2).
To find the area of a triangle, we need the altitude and the base. The altitude can be measured from the midpoint of QR to P:
The midpoint of QR is ( (1+3)/2, (1+1)/2, (-1+2)/2 )
(2, 1, 0.5)
Our altitude is the the vector from P to this midpoint:
( (2-2) , (1-0), (0.5-0) )
( 0,1,0.5 )
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The altitude does not necessarily go to the midpoint of the opposite segment.
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And the vector between P and Q is our base:
( (3-1),(1-1),(2-(-1)) )
( 2, 0, 3 )
The formula for the area of a triangle is the magnitude of the base cross the magnitude divided by 2.
|(0,1,0.5) X (2,0,3)|
= |0,1,0.5)|*|(2,0,3)|*sin(theta)
= sqrt(0 + 1^2 + 0.5^2) * sqrt(2^2 + 0 + 3^2)
= sqrt(1.25) * sqrt (13)
= sqrt(16.25)
In this case, theta is 90, since we're visualizing the triangle as half a parallelogram.
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I don't think that the triangle has been established as a right triangle.
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The area of the triangle is approximately equal to 2.0155 units squared.
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The parallelogram you mention would have sides congruent to the vectors PQ and PR.
If PQ is taken as the base, and the angle between PQ and PR is theta, then the altitude of the triangle is PR sin(theta).
The area of the triangle is therefore || PQ || * || PR || * sin(theta).
The cross product PQ X PR has magnitude || PQ || * || PR || * sin(theta).
Check this against your result.
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8) 9.4.28 Determine if each of the following products is a vector, scalar, or not defined at all. Explain why. u X (v X w) , u dot (v dot w), (u X v) dot (w X r).
u X (v X w) - Vector. Cross products take two vectors and create a resultant vector.
u dot (v dot w) - Undefined. Dot products take two vectors and yield a scalar. Therefore, u dot (resultant scalar) is impossible.
(u X v) dot (w X r) - Scalar. Both cross products yield a vector, and the dot product of those two vectors produces a scalar value.
9) 9.4.30 Find the area of the parallelepiped determined by u = i - j, v = i - 2k, and w = 4k
The volume of a parallelepiped is equal to the dot product of any of the vectors and the cross product of the other two vectors.
v X w = | i j k | = | 0 -2 |i + | 1 -2 |j + | 1 0 |k
| 1 0 -2 | | 0 4 | | 0 4 | | 0 0 |
| 0 0 4 |
z = 0i, 4j, 0k = 4j
u dot z = 0 + -4 + 0 = -4 units^3
10) 9.4.38 Find a number t such that the vectors -i - j, i - (1/2) j + (1/2)k and -2i -2j - 2tk all lie in the same plane.
Any three vectors are coplanar iff q X r dot s = 0.
<-1,-1,0> X <1,-1/2,1/2>:
| -1 0 |i + |-1 0 |j + |-1 -1 |k
| -1/2 1/2| |-1/2 1/2 | |1 -1/2|
<-1/2 i - 1/2 j + 3/2 k> dot s:
1 + 1 - 3t = 0
Therefore, t must equal 2/3.
11) 9.4.39 u = 2i + 2j, v= i-(1/2)j+ (1/2)k, w = i. Compute (u X v) X w and u X (v X w). What does this say about the associativity of the cross product?
The first calculations comes to <0,-4,-1>. The second comes to <-1,1,1>. Therefore, cross products are not associative and order matters when calculating them."
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Nicely done, but with a couple of errors. See my notes.
I'll be glad to check revisions or answer questions.
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