The Dot and Cross Product The Dot Product
Examples:
If
Exercise The Angle Between Two Vectors
We define the angle theta between two vectors v and
w by the formula
Example
Direction Angles
Projections and Components Suppose that a car is stopped on a steep hill, and let g be the force of gravity acting on it. We can split the vector g into the component that is pushing the car down the road and the component that is pushing the car onto the road. We define
We see that ||u|| ||v|| ||proj_{v}u || u ^{.} v = ||u|| ||v|| cos q = ||u|| = ||v|| ||proj_{v}u || hence
Notice that this works since if we take magnitudes of both sides we get that
u ^{. }v 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 proj_{v}u + s = u so that s = u - proj_{v} u
The work done by a constant force F along PQ is given by
Example Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7)? Solution We have that the force vector is F = ma = (10)(-9.8j) = -98j 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
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
= 17i - 10j + 8k
If you need more help see the lecture notes for Math 103 B on matrices.
Exercises
Notice that since switching the order of two rows of a determinant changes
the sign of the determinant, we have Geometry and the Cross Product
Let u and v be vectors and consider the parallelogram that
the two vectors make. Then Note: For i x j the magnitude is 1 and the direction is k, hence i x j = k.
Exercise Torque Revisited
We define the torque (or the moment M of a force F about a point
Q) as
Example
Solution
= -564
inch pounds
To find the volume of the parallelepiped spanned by three vectors u,
v, and w, we find the triple product:
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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