Section 11-3

Here's more than you probably want, especially about u x ( u x ( u x v) ) dot w. Due to time contraints it's not complete but it should give you the flavor the variery of approaches that can work for these situations.

For the x and y coordinates we get

x(t) = v0 cos(alpha) * t and

y(t) = s0 + v0 sin(alpha) * t - 1/2 g t^2.

The projectile strikes the ground when y(t) = 0. Using the quadratic formula to solve y(t) = 0 we obtain

t = (-v0 sin(alpha) + - sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / (-g). Since the square root is greater than v0 sin(alpha), only the - of the +- yields a positive result for t. Thus

t = (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g.

Substituting this into the expression for x(t) we obtain

x( (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g )

= v0 cos(alpha) (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g

= v0^2 cos(alpha)sin(alpha) + v0 / g * cos(alpha) sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) .

To maximize R as a function of alpha we take the derivative with respect to alpha and set it equal to zero:

v0^2 cos(alpha)sin(alpha) + v0 / g * cos(alpha) sqrt( v0^2 sin^2(alpha) + 2 s0 g) )

v0^2 (cos^2(alpha) - sin^2(alpha) ) + v0 / g * ( cos(alpha) * v0^2 * 2 sin(alpha) cos(alpha) / sqrt(v0^2 sin^2(alpha) + 2 s0 g) - sin(alpha) sqrt(v0^2 sin^2(alpha) + 2 s0 g) ) = 0.

Dividing through by v0^2 we get

cos^2(alpha) - sin^2(alpha) + v0 / g sin(alpha) cos^2(alpha) / sqrt( v0^2 sin^2*alpha + 2 s0 g) - sin(alpha) sqrt(sin^2(alpha) + 2 s0 g / v0) = 0.

At this point the algebra is running together on the page and I don't have time to write it down on paper. The identitiy cos^2(alpha) = 1 - sin^2(alpha) can reduce this to an expression in just sin(alpha), and I think it will come out

1 - 2 sin^2(alpha) + v0 / (g sqrt( v0^2 sin^2*alpha + 2 s0 g) )(sin(alpha) - sin^3(alpha)) - sin(alpha) sqrt(sin^2(alpha) + 2 s0 g / v0) = 0, but I'm getting lost in the parentheses etc. and it's late.

So I'll revert to DERIVE to obtain their expression for the derivative, which is

COS(alpha)·(v0·‹(v0^2·SIN(alpha)^2 + 2·g·s0)/g + v0^2·SIN(alpha)/g)COS(alpha)^2·(v0^3·SIN(alpha)/(g·‹(v0^2·SIN(alpha)^2 + 2·g·s0)) + v0^2/g) - v0·SIN(alpha)·‹(v0^2·SIN(alpha)^2 + 2·g·s0)/g - v0^2·SIN(alpha)^2/g

and their solution for alpha, which makes no sense.

In any case you get the idea. The solution should be

alpha = v0 / sqrt( v0^2 + 2 g s0)

and plugging this in to the expression for x(t) gives you the maximum range. You would also need to do a first- or second-derivative test to verify that this is a max, but that's pretty straightforward.

The details of the above aren't quite right--something off in my bookkeeping--but the procedure is as I outline it and if everything is done carefully and correctly the specified results fall out of the equations.

Problems 41 and 42 follow very easily from Problem 40. Once we understand Problem 40, we're just about home free.

u_r and u_theta are respectively the unit vectors which define a coordinate system that moves with the object, with u_r pointing away from the object and u_theta prependicular to u_r.

If the position of the object is r(t) = x(t) i + y(t) j then the vector in the direction of r is (x i + y j) / sqrt(x^2 + y^2) = x / sqrt(x^2 + y^2) i + y / sqrt(x^2 + y^2) j = cos(theta) i + sin(theta) j, where theta is the angle made by the vector r with the positive x axis.

A unit vector perpendicular to this vector is either -sin(theta) i + cos(theta) j or sin(theta) i - cos(theta) j. To form the coordinate system we desire, u_theta must be perpendicular to u_r, and if chosen so that u_r and u_theta form a right-handed coordinate system then u_theta = -sin(theta) i + cos(theta)j. It is straightforward to show that du_r / dtheta = d(cos(theta)) / dtheta) * i + d(sin(theta) / dtheta) * j= - sin(theta) i + cos(theta) j = u_theta, and similarly that du_theta / dtheta = - u_r. Note that if theta is a function of t, then du_r / dt = dtheta/dt * u_theta and du_theta / dt = dtheta / dt * (- u_r).

In terms of u_r and u_theta, r(t) = r * u_r with no u_theta component (r = sqrt(x^2 + y^2) so r u_r = r ( cos(theta) i + sin(theta) j) = r cos(theta) i + r sin(theta) j = x i + y j; that is, r(t) can be expressed either as r u_r or as x i + y j, with r = sqrt(x^2 + y^2) ).

Now

v(t) = dr/dt = d ( r u_r ) / dt

= dr/dt * u_r + r * d(u_r)/dtheta by the chain rule

= dr/dt * u_r + r dtheta / dt * (u_theta),

as in problem 40. Also as in Problem 40, we take another derivative with respect to t, finding that

A(t) = dv/dt

= d/dt ( dr/dt * u_r) + d/dt ( r dtheta / dt * u_theta)

= (r ' u_r) ' + (r theta ' u_theta) ', using ' to indicate derivative with respect to t

= r '' u_r + r' (u_r)' + (r theta ') ' u_theta + r theta ' ( u_theta) '

= r '' u_r + r' theta' u_theta + (r ' theta ' + r theta '') u_theta + r theta ' ( - theta' u_r)

= (r'' - r (theta')^2) u_r + (r theta '' + 2 r ' theta ' ) u_theta.

This is the result of Problem 40.

In problem 41 we multiply the acceleration by the mass to obtain the force function (i.e., we apply Newton's Second Law), which will be

F(t) = m A(t)

= m [ (r'' - r (theta')^2) u_r + (r theta '' + 2 r ' theta ' ) u_theta ]

= m (r'' - r (theta')^2) u_r + m (r theta '' + 2 r ' theta ' ) u_theta by the distributive law, which can be written

F_r u_r + F_theta u_theta with

F_r = m (r'' - r (theta')^2) u_r and

F_theta = m (r theta '' + 2 r ' theta ' ) u_theta.

Thus problem 41 is just a straightforward application of the distributive law.

Problem 42 asks us to show that r F_theta(t) = d ( m r^2 dtheta/dt) / dt. Using the prime notation this is

r F_theta = (m r^2 theta ') ' = m (r^2) ' theta ' + m r^2 theta ' ' = 2 m r r' + m r^2 theta'.

This is exactly what we get when we multiply F_theta = m (r theta '' + 2 r ' theta ' ) u_theta by r and use the distributive law.

We have throughout regarded the mass m as constant. Things get much more interesting if that is not the case, but the mathematics isn't that bad--just another product rule; it's the interpretation and application to actual physical systems that becomes a notch more challenging.

A couple of quick notes on u x (u x ( u x v)):

If we just let u = u1 i + u2 j + u3 k, v = v1 i + v2 j + v3 k and w = w1 i + w2 j + w3 k we can compute both sides of the equation (u x (u x (u x v))) dot w = ||u||^2 u dot(v x w) and verify the identity without understanding anything about the geometry. The bookkeeping is a mess, but if done carefully it works out. Using DERIVE, with obvious notation, the product u x (u x (u x v) ) is

i·(u1^2·(u2·v3 - u3·v2) + (u2^2 + u3^2)·(u2·v3 - u3·v2)) - j·(u1^3·v3 - u1^2·u3·v1 + u1·(u2^2·v3 - 2·u2·u3·v2 - u3^2·v3) + u3·v1·(u2^2 + u3^2)) + k·(u1^3·v2 - u1^2·u2·v1 - u1·(u2^2·v2 + 2·u2·u3·v3 - u3^2·v2) + u2·v1·(u2^2 + u3^2))

and the dot product with w is

w1·(u1^2·(u2·v3 - u3·v2) + (u2^2 + u3^2)·(u2·v3 - u3·v2)) - w2·(u1^3·v3 - u1^2·u3·v1 + u1·(u2^2·v3 - 2·u2·u3·v2 - u3^2·v3) + u3·v1·(u2^2 + u3^2)) + w3·(u1^3·v2 - u1^2·u2·v1 - u1·(u2^2·v2 + 2·u2·u3·v3 - u3^2·v2) + u2·v1·(u2^2 + u3^2)), which completely expanded is

u1^3·v2·w3 - u1^3·v3·w2 - u1^2·u2·v1·w3 + u1^2·u2·v3·w1 + u1^2·u3·v1·w2 - u1^2·u3·v2·w1 - u1·u2^2·v2·w3 - u1·u2^2·v3·w2 + 2·u1·u2·u3·v2·w2 - 2·u1·u2·u3·v3·w3 + u1·u3^2·v2·w3 + u1·u3^2·v3·w2 + u2^3·v1·w3 + u2^3·v3·w1 - u2^2·u3·v1·w2 - u2^2·u3·v2·w1 + u2·u3^2·v1·w3 + u2·u3^2·v3·w1 - u3^3·v1·w2 - u3^3·v2·w1.

v x w = i·(v2·w3 - v3·w2) + j·(v3·w1 - v1·w3) + k·(v1·w2 - v2·w1) so u dot (v x w) = u1·(v2·w3 - v3·w2) + u2·(v3·w1 - v1·w3) + u3·(v1·w2 - v2·w1). Multiplying this expression by || u || ^2 = u1^2 + u2^2 + u3^2 we get

u1^3·v2·w3 - u1^3·v3·w2 - u1^2·u2·v1·w3 + u1^2·u2·v3·w1 + u1^2·u3·v1·w2 - u1^2·u3·v2·w1 + u1·u2^2·v2·w3 - u1·u2^2·v3·w2 + u1·u3^2·v2·w3 - u1·u3^2·v3·w2 - u2^3·v1·w3 + u2^3·v3·w1 + u2^2·u3·v1·w2 - u2^2·u3·v2·w1 - u2·u3^2·v1·w3 + u2·u3^2·v3·w1 + u3^3·v1·w2 - u3^3·v2·w1

These two expressions should be identical, except perhaps for order.

They aren't, and one should actually be the negative of the other. I'll leave the correction and the rest of the bookkeeping to you, if you want to pursue it.

The geometrical interpretation of these expressions is much more revealing and more interesting.

u x v is perpendicular to u and to v, and therefore normal to the plane defined by u and v.

u x (u x v) is therefore perpendicular to that normal vector and to u. It's worth thinking for a minute about what this might mean. Note first that this means that || u x ( u x v) || = || u || || u x v || = || u || ^2 || v || sin(theta), in any case (theta being the angle between u and v). Also, in any case, u x (u x v) is perpendicular to u and to a vector normal to the plane of u and v, and must therefore be in the plane of u and v so that u x ( u x ( u x v)) is again normal to the plane of u and v. Except for the original product u x v, the vector u is always perpendicular to the vector with which it is being crossed, so what we end up with is a vector normal to the plane of u and v, having magnitude || u || ^ 3 || v || sin(theta).

The dot product of this vector with w is equal to this magnitude, multiplied by the component of w perpendicular to the plane of u and v.

The cross product of v and w is perpendicular to v (as is the vector u x ( u x ( u x w)) ), but unless w is in the plane of u and v the cross product will not be normal to the u-v plane and will therefore have a nonzero dot product with u. If alpha is the angle between v x w and the normal to the u-v plane, i.e., the angle between the v-w and u-v planes, then it is because the component of w perpendicular to the u-v plane is w sin(alpha) ... then a miracle occurs ... and everything comes together geometrically.

Actually there is no miracle, it's pretty straightforward at this point, but I'm out of time for the moment. I'll try to come back to this later tonight, probably after I break down and draw a picture instead of trying to hold it all in my aging head.

Worth noting:

If u and v are perpendicular in the first place, then || u x v || = || u || || v ||, and || u x ( u x v) || = || u || ^ 2 || v ||. Then of course u is still perpendicular to u x ( u x v) so the magnitude of u x ( u x ( u x v) ) is || u ||^3 || v ||

If u and v are perpendicular, then w can be expressed in the form w_u * u + w_v * v + w_k * k, where w_u = w dot u / || u ||, w_v = w dot v / || v || and w_k = w dot k / || k ||, k being normal to the plane of u and v (e.g., k could be u x v, or u x (u x ( u x v))).

If u and v aren't perpendicular, then we can express v in terms of components parallel to and perpendicular to u, both components being in the plane of u and v. The component parallel to u is v dot u / || u ||. The vector perpendicular to u might as well be u x ( u x v), so the component of v perpendicular to u is v_perp = v dot (u x ( u x v) ) / || u x ( u x v) || = v dot ( u x ( u x v) ) / (|| u || ^2 ||v|| sin(theta) ).

Now all vectors can be expressed in terms of u, v_perp and k; even better, we can divide each of these vectors by its magnitude and obtain an orthogonal basis of three unit vectors, one parallel to u, one perpendicular to u in the plane of u x v and the other perpendicular to this plane. The cross product of any two of these vectors will be either the third, or the negative of the third. Dot products are just the sums of component-by-component products. The entire result will just fall into our hands, no miracle required.

end of document

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Section 11-3

Here's more than you probably want, especially about u x ( u x ( u x v) ) dot w. Due to time contraints it's not complete but it should give you the flavor the variery of approaches that can work for these

situations.

For the x and y coordinates we get

x(t) = v0 cos(alpha) * t and

y(t) = s0 + v0 sin(alpha) * t - 1/2 g t^2.

The projectile strikes the ground when y(t) = 0. Using the quadratic formula to solve y(t) = 0 we obtain

t = (-v0 sin(alpha) + - sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / (-g). Since the square root is greater than v0 sin(alpha), only the - of the +- yields a positive result for t. Thus

t = (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g.

Substituting this into the expression for x(t) we obtain

x( (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g )

= v0 cos(alpha) (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g

= v0^2 cos(alpha)sin(alpha) + v0 / g * cos(alpha) sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) .

To maximize R as a function of alpha we take the derivative with respect to alpha and set it equal to zero:

v0^2 cos(alpha)sin(alpha) + v0 / g * cos(alpha) sqrt( v0^2 sin^2(alpha) + 2 s0 g) )

v0^2 (cos^2(alpha) - sin^2(alpha) ) + v0 / g * ( cos(alpha) * v0^2 * 2 sin(alpha) cos(alpha) / sqrt(v0^2 sin^2(alpha) + 2 s0 g) - sin(alpha) sqrt(v0^2 sin^2(alpha) + 2 s0 g) ) = 0.

Dividing through by v0^2 we get

cos^2(alpha) - sin^2(alpha) + v0 / g sin(alpha) cos^2(alpha) / sqrt( v0^2 sin^2*alpha + 2 s0 g) - sin(alpha) sqrt(sin^2(alpha) + 2 s0 g / v0) = 0.

At this point the algebra is running together on the page and I don't have time to write it down on paper. The identitiy cos^2(alpha) = 1 - sin^2(alpha) can reduce this to an expression in just sin(alpha), and I think it

will come out

1 - 2 sin^2(alpha) + v0 / (g sqrt( v0^2 sin^2*alpha + 2 s0 g) )(sin(alpha) - sin^3(alpha)) - sin(alpha) sqrt(sin^2(alpha) + 2 s0 g / v0) = 0, but I'm getting lost in the parentheses etc. and it's late.

So I'll revert to DERIVE to obtain their expression for the derivative, which is

COS(alpha)·(v0·‹(v0^2·SIN(alpha)^2 + 2·g·s0)/g + v0^2·SIN(alpha)/g)COS(alpha)^2·(v0^3·SIN(alpha)/(g·‹(v0^2·SIN(alpha)^2 + 2·g·s0)) + v0^2/g) - v0·SIN(alpha)·‹(v0^2·SIN(alpha)^2 + 2·g·s0)/g -

v0^2·SIN(alpha)^2/g

and their solution for alpha, which makes no sense.

In any case you get the idea. The solution should be

alpha = v0 / sqrt( v0^2 + 2 g s0)

and plugging this in to the expression for x(t) gives you the maximum range. You would also need to do a first- or second-derivative test to verify that this is a max, but that's pretty straightforward.

The details of the above aren't quite right--something off in my bookkeeping--but the procedure is as I outline it and if everything is done carefully and correctly the specified results fall out of the equations.

Problems 41 and 42 follow very easily from Problem 40. Once we understand Problem 40, we're just about home free.

u_r and u_theta are respectively the unit vectors which define a coordinate system that moves with the object, with u_r pointing away from the object and u_theta prependicular to u_r.

If the position of the object is r(t) = x(t) i + y(t) j then the vector in the direction of r is (x i + y j) / sqrt(x^2 + y^2) = x / sqrt(x^2 + y^2) i + y / sqrt(x^2 + y^2) j = cos(theta) i + sin(theta) j, where theta is the angle

made by the vector r with the positive x axis.

A unit vector perpendicular to this vector is either -sin(theta) i + cos(theta) j or sin(theta) i - cos(theta) j. To form the coordinate system we desire, u_theta must be perpendicular to u_r, and if chosen so that u_r

and u_theta form a right-handed coordinate system then u_theta = -sin(theta) i + cos(theta)j. It is straightforward to show that du_r / dtheta = d(cos(theta)) / dtheta) * i + d(sin(theta) / dtheta) * j= - sin(theta) i +

cos(theta) j = u_theta, and similarly that du_theta / dtheta = - u_r. Note that if theta is a function of t, then du_r / dt = dtheta/dt * u_theta and du_theta / dt = dtheta / dt * (- u_r).

In terms of u_r and u_theta, r(t) = r * u_r with no u_theta component (r = sqrt(x^2 + y^2) so r u_r = r ( cos(theta) i + sin(theta) j) = r cos(theta) i + r sin(theta) j = x i + y j; that is, r(t) can be expressed either as r u_r

or as x i + y j, with r = sqrt(x^2 + y^2) ).

Now

v(t) = dr/dt = d ( r u_r ) / dt

= dr/dt * u_r + r * d(u_r)/dtheta by the chain rule

= dr/dt * u_r + r dtheta / dt * (u_theta),

as in problem 40. Also as in Problem 40, we take another derivative with respect to t, finding that

A(t) = dv/dt

= d/dt ( dr/dt * u_r) + d/dt ( r dtheta / dt * u_theta)

= (r ' u_r) ' + (r theta ' u_theta) ', using ' to indicate derivative with respect to t

= r '' u_r + r' (u_r)' + (r theta ') ' u_theta + r theta ' ( u_theta) '

= r '' u_r + r' theta' u_theta + (r ' theta ' + r theta '') u_theta + r theta ' ( - theta' u_r)

= (r'' - r (theta')^2) u_r + (r theta '' + 2 r ' theta ' ) u_theta.

This is the result of Problem 40.

In problem 41 we multiply the acceleration by the mass to obtain the force function (i.e., we apply Newton's Second Law), which will be

F(t) = m A(t)

= m [ (r'' - r (theta')^2) u_r + (r theta '' + 2 r ' theta ' ) u_theta ]

= m (r'' - r (theta')^2) u_r + m (r theta '' + 2 r ' theta ' ) u_theta by the distributive law, which can be written

F_r u_r + F_theta u_theta with

F_r = m (r'' - r (theta')^2) u_r and

F_theta = m (r theta '' + 2 r ' theta ' ) u_theta.

Thus problem 41 is just a straightforward application of the distributive law.

Problem 42 asks us to show that r F_theta(t) = d ( m r^2 dtheta/dt) / dt. Using the prime notation this is

r F_theta = (m r^2 theta ') ' = m (r^2) ' theta ' + m r^2 theta ' ' = 2 m r r' + m r^2 theta'.

This is exactly what we get when we multiply F_theta = m (r theta '' + 2 r ' theta ' ) u_theta by r and use the distributive law.

We have throughout regarded the mass m as constant. Things get much more interesting if that is not the case, but the mathematics isn't that bad--just another product rule; it's the interpretation and application to

actual physical systems that becomes a notch more challenging.

A couple of quick notes on u x (u x ( u x v)):

If we just let u = u1 i + u2 j + u3 k, v = v1 i + v2 j + v3 k and w = w1 i + w2 j + w3 k we can compute both sides of the equation (u x (u x (u x v))) dot w = ||u||^2 u dot(v x w) and verify the identity without

understanding anything about the geometry. The bookkeeping is a mess, but if done carefully it works out. Using DERIVE, with obvious notation, the product u x (u x (u x v) ) is

i·(u1^2·(u2·v3 - u3·v2) + (u2^2 + u3^2)·(u2·v3 - u3·v2)) - j·(u1^3·v3 - u1^2·u3·v1 + u1·(u2^2·v3 - 2·u2·u3·v2 - u3^2·v3) + u3·v1·(u2^2 + u3^2)) + k·(u1^3·v2 - u1^2·u2·v1 - u1·(u2^2·v2 + 2·u2·u3·v3 - u3^2·v2) +

u2·v1·(u2^2 + u3^2))

and the dot product with w is

w1·(u1^2·(u2·v3 - u3·v2) + (u2^2 + u3^2)·(u2·v3 - u3·v2)) - w2·(u1^3·v3 - u1^2·u3·v1 + u1·(u2^2·v3 - 2·u2·u3·v2 - u3^2·v3) + u3·v1·(u2^2 + u3^2)) + w3·(u1^3·v2 - u1^2·u2·v1 - u1·(u2^2·v2 + 2·u2·u3·v3 - u3^2·v2)

+ u2·v1·(u2^2 + u3^2)), which completely expanded is

u1^3·v2·w3 - u1^3·v3·w2 - u1^2·u2·v1·w3 + u1^2·u2·v3·w1 + u1^2·u3·v1·w2 - u1^2·u3·v2·w1 - u1·u2^2·v2·w3 - u1·u2^2·v3·w2 + 2·u1·u2·u3·v2·w2 - 2·u1·u2·u3·v3·w3 + u1·u3^2·v2·w3 + u1·u3^2·v3·w2 + u2^3·v1·w3 +

u2^3·v3·w1 - u2^2·u3·v1·w2 - u2^2·u3·v2·w1 + u2·u3^2·v1·w3 + u2·u3^2·v3·w1 - u3^3·v1·w2 - u3^3·v2·w1.

v x w = i·(v2·w3 - v3·w2) + j·(v3·w1 - v1·w3) + k·(v1·w2 - v2·w1) so u dot (v x w) = u1·(v2·w3 - v3·w2) + u2·(v3·w1 - v1·w3) + u3·(v1·w2 - v2·w1). Multiplying this expression by || u || ^2 = u1^2 + u2^2 + u3^2 we

get

u1^3·v2·w3 - u1^3·v3·w2 - u1^2·u2·v1·w3 + u1^2·u2·v3·w1 + u1^2·u3·v1·w2 - u1^2·u3·v2·w1 + u1·u2^2·v2·w3 - u1·u2^2·v3·w2 + u1·u3^2·v2·w3 - u1·u3^2·v3·w2 - u2^3·v1·w3 + u2^3·v3·w1 + u2^2·u3·v1·w2 -

u2^2·u3·v2·w1 - u2·u3^2·v1·w3 + u2·u3^2·v3·w1 + u3^3·v1·w2 - u3^3·v2·w1

These two expressions should be identical, except perhaps for order.

They aren't, and one should actually be the negative of the other. I'll leave the correction and the rest of the bookkeeping to you, if you want to pursue it.

The geometrical interpretation of these expressions is much more revealing and more interesting.

u x v is perpendicular to u and to v, and therefore normal to the plane defined by u and v.

u x (u x v) is therefore perpendicular to that normal vector and to u. It's worth thinking for a minute about what this might mean. Note first that this means that || u x ( u x v) || = || u || || u x v || = || u || ^2 || v ||

sin(theta), in any case (theta being the angle between u and v). Also, in any case, u x (u x v) is perpendicular to u and to a vector normal to the plane of u and v, and must therefore be in the plane of u and v so that u

x ( u x ( u x v)) is again normal to the plane of u and v. Except for the original product u x v, the vector u is always perpendicular to the vector with which it is being crossed, so what we end up with is a vector

normal to the plane of u and v, having magnitude || u || ^ 3 || v || sin(theta).

The dot product of this vector with w is equal to this magnitude, multiplied by the component of w perpendicular to the plane of u and v.

The cross product of v and w is perpendicular to v (as is the vector u x ( u x ( u x w)) ), but unless w is in the plane of u and v the cross product will not be normal to the u-v plane and will therefore have a nonzero

dot product with u. If alpha is the angle between v x w and the normal to the u-v plane, i.e., the angle between the v-w and u-v planes, then it is because the component of w perpendicular to the u-v plane is w

sin(alpha) ... then a miracle occurs ... and everything comes together geometrically.

Actually there is no miracle, it's pretty straightforward at this point, but I'm out of time for the moment. I'll try to come back to this later tonight, probably after I break down and draw a picture instead of trying to

hold it all in my aging head.

Worth noting:

If u and v are perpendicular in the first place, then || u x v || = || u || || v ||, and || u x ( u x v) || = || u || ^ 2 || v ||. Then of course u is still perpendicular to u x ( u x v) so the magnitude of u x ( u x ( u x v) ) is || u ||^3 || v ||

If u and v are perpendicular, then w can be expressed in the form w_u * u + w_v * v + w_k * k, where w_u = w dot u / || u ||, w_v = w dot v / || v || and w_k = w dot k / || k ||, k being normal to the plane of u and v (e.g., k

could be u x v, or u x (u x ( u x v))).

If u and v aren't perpendicular, then we can express v in terms of components parallel to and perpendicular to u, both components being in the plane of u and v. The component parallel to u is v dot u / || u ||. The

vector perpendicular to u might as well be u x ( u x v), so the component of v perpendicular to u is v_perp = v dot (u x ( u x v) ) / || u x ( u x v) || = v dot ( u x ( u x v) ) / (|| u || ^2 ||v|| sin(theta) ).

Now all vectors can be expressed in terms of u, v_perp and k; even better, we can divide each of these vectors by its magnitude and obtain an orthogonal basis of three unit vectors, one parallel to u, one perpendicular

to u in the plane of u x v and the other perpendicular to this plane. The cross product of any two of these vectors will be either the third, or the negative of the third. Dot products are just the sums of

component-by-component products. The entire result will just fall into our hands, no miracle required.