The Dot and Cross Product
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The Dot Product Definition v. w = ac + bd |
Notice that the dot product of two vectors is a number and not a vector. For 3 dimensional vectors, we define the dot product similarly:
Dot Product in R3 If v = ai + bj + ckand w = di + ej + fk then |
Examples:
If
v= 2i + 4j
and
w= i + 5j
then
v .w = (2)(1) + (4)(5) = 22
Exercise
Find the dot product of
2i + j - k and i + 2j
The Angle Between Two Vectors
We define the angle theta between two vectors v and w by the formulav. w
cos q =
||v|| ||w||
so that
v. w = ||v|| ||w|| cos q |
Two vectors are called orthogonal if their angle is a right angle.
We see that angles are orthogonal if and only if
v. w = 0
Example
To find the angle between
v = 2i + 3j + k
and
w = 4i + j + 2k
we compute:
and
and
v. w = 8 + 3 + 2 = 13
Hence
Direction Angles
Definition of Direction Cosines
|
Definition Let u and v be a vectors. Then ucan be broken up into two components, r and s such that r is parallel to v and s is perpendicular to v. ris called the projection ofuontov and s is called the component of u perpendicular to v. |
We see that
||u|| ||v|| ||projvu ||
u.v = ||u|| ||v|| cos q =
||u||
= ||v|| ||projvu ||
hence
u . v |
We can calculate the projection of u onto v by the formula:
u . v |
Notice that this works since if we take magnitudes of both sides we get that
u . v
| |projvu|| = ||v||
||v||2
and the right hand side simplifies to the formula above. The direction is correct since the right hand side of the formula is a constant multiple of v so the projection vector is in the direction of v as required.
To find the vector s, notice from the diagram that
projvu + s = u
so that
s = u - projv u
The work done by a constant force F along PQ is given by| W = F.PQ |
Example
Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7)?
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Solution
We have that the force vector is
F = ma = (10)(-9.8j) = -98j
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and the displacement vector is
v = (5 - 2) i + (7 - 3) j = 3i + 4j
The work is the dot product
W = F. v = (-98j) . (3i + 4j)
= (0)(3) + (-98)(4) = -392
Notice the negative sign verifies that the work is done against gravity. Hence, it takes 392 J of work to move the baby.
Suppose you are skiing and have a terrible fall. Your body spins around and you ski stays in place (do not try this at home). With proper bindings your bindings will release and your ski will come off. The bindings recognize that a force has been applied. This force is called torque. To compute it we use the cross produce of two vectors which not only gives the torque, but also produces the direction that is perpendicular to both the force and the direction of the leg.
The Cross Product Between Two Vectors
Definition |
We can compute this determinant as
= (bf - ce) i + (cd - af) j + (ae - bd) k
Example
Find the cross product u x v if
u = 2i + j - 3k v = 4j + 5k
Solution
We calculate World of tanks blitz 3 6 624 download free.
= 17i - 10j + 8k
If you need more help see the lecture notes for Math 103 B on matrices.
Exercises
Find u x v when
u = 3i + j - 2k, v = i - k
u = 2i - 4j - k, v = 3i - j + 2k
Notice that since switching the order of two rows of a determinant changes the sign of the determinant, we have
u x v = -v x u
Geometry and the Cross Product
Let u and vbe vectors and consider the parallelogram that the two vectors make. Then
||u x v|| = Area of the Parallelogram
and the direction of u x v is a right angle to the parallelogram that follows the right hand rule
Note:Fori x j the magnitude is 1 and the direction isk, hence i x j = k.
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Exercise
Find j x k and i x k
Torque Revisited
We define the torque (or the moment M of a force F about a point Q) asM = PQ x F |
Example
A 20 inch wrench is at an angle of 30 degrees with the ground. A force of 40 pounds that makes and angle of 45 degrees with the wrench turns the wrench. Find the torque.
Solution
We can write the wrench as the vector
20 cos 30 i + 20 sin 30 j = 17.3 i + 10 j
and the force as
-40 cos 75 i - 40 sin 75 j = -10.3 i - 38.6 j
hence, the torque is the magnitude of their cross product:
= -564 inch pounds
Definition of Direction Cosines
|
Definition Let u and v be a vectors. Then ucan be broken up into two components, r and s such that r is parallel to v and s is perpendicular to v. ris called the projection ofuontov and s is called the component of u perpendicular to v. |
We see that
||u|| ||v|| ||projvu ||
u.v = ||u|| ||v|| cos q =
||u||
= ||v|| ||projvu ||
hence
u . v |
We can calculate the projection of u onto v by the formula:
u . v |
Notice that this works since if we take magnitudes of both sides we get that
u . v
| |projvu|| = ||v||
||v||2
and the right hand side simplifies to the formula above. The direction is correct since the right hand side of the formula is a constant multiple of v so the projection vector is in the direction of v as required.
To find the vector s, notice from the diagram that
projvu + s = u
so that
s = u - projv u
The work done by a constant force F along PQ is given by| W = F.PQ |
Example
Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7)?
Wonderpen 1 2 123movies
Solution
We have that the force vector is
F = ma = (10)(-9.8j) = -98j
Wonderpen 1 2 1 2 Be Frozen
and the displacement vector is
v = (5 - 2) i + (7 - 3) j = 3i + 4j
The work is the dot product
W = F. v = (-98j) . (3i + 4j)
= (0)(3) + (-98)(4) = -392
Notice the negative sign verifies that the work is done against gravity. Hence, it takes 392 J of work to move the baby.
Suppose you are skiing and have a terrible fall. Your body spins around and you ski stays in place (do not try this at home). With proper bindings your bindings will release and your ski will come off. The bindings recognize that a force has been applied. This force is called torque. To compute it we use the cross produce of two vectors which not only gives the torque, but also produces the direction that is perpendicular to both the force and the direction of the leg.
The Cross Product Between Two Vectors
Definition |
We can compute this determinant as
= (bf - ce) i + (cd - af) j + (ae - bd) k
Example
Find the cross product u x v if
u = 2i + j - 3k v = 4j + 5k
Solution
We calculate World of tanks blitz 3 6 624 download free.
= 17i - 10j + 8k
If you need more help see the lecture notes for Math 103 B on matrices.
Exercises
Find u x v when
u = 3i + j - 2k, v = i - k
u = 2i - 4j - k, v = 3i - j + 2k
Notice that since switching the order of two rows of a determinant changes the sign of the determinant, we have
u x v = -v x u
Geometry and the Cross Product
Let u and vbe vectors and consider the parallelogram that the two vectors make. Then
||u x v|| = Area of the Parallelogram
and the direction of u x v is a right angle to the parallelogram that follows the right hand rule
Note:Fori x j the magnitude is 1 and the direction isk, hence i x j = k.
Wonderpen 1 2 1
Exercise
Find j x k and i x k
Torque Revisited
We define the torque (or the moment M of a force F about a point Q) asM = PQ x F |
Example
A 20 inch wrench is at an angle of 30 degrees with the ground. A force of 40 pounds that makes and angle of 45 degrees with the wrench turns the wrench. Find the torque.
Solution
We can write the wrench as the vector
20 cos 30 i + 20 sin 30 j = 17.3 i + 10 j
and the force as
-40 cos 75 i - 40 sin 75 j = -10.3 i - 38.6 j
hence, the torque is the magnitude of their cross product:
= -564 inch pounds
Wonderpen 1 2 1 10
wWonderpen 1 2 1 2 In Fraction
, we find the triple product:Volume = u. (v x w) |
This can be found by computing the determinate of the three vectors:
Example
Find the volume of the parallelepiped spanned by the vectors
u = <1,0,2> v = <0,2,3> w = <0,1,3>
Solution
We find
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