MAT1341C, Winter 2017, University of Ottawa - Extra Questions and comments

Geometry behind the dot and cross products

A version in $$\mathbb{R}^2$$:

In $$\mathbb{R}^2$$, you can really see why the geometric identity for the dot product $\vec{u} \cdot \vec{v} = \Vert \vec{u} \Vert \Vert \vec{v} \Vert \cos(\varphi)$ (where $$\varphi$$ is the angle between u and v) holds:

Consider a vector $$\vec{v} = (v_1,v_2)$$ in $$\mathbb{R}^2$$. If its norm is $$\Vert \vec{v}\Vert$$ and it makes angle $$\theta$$ with the $$x$$-axis, then we can picture it as

So using trigonometry, we have that $v_1 = \Vert \vec{v}\Vert \cos(\theta) \quad \text{and}\quad v_2 = \Vert \vec{v}\Vert\sin(\theta).$

Similarly, we can write $$\vec{u} = (u_1,u_2) = (\Vert \vec{u}\Vert \cos(\theta'), \Vert \vec{u}\Vert\sin(\theta'))$$ where $$\theta'$$ is the angle $$\vec{u}$$ makes with the positive $$x$$-axis.

So then the angle between $$\vec{u}$$ and $$\vec{v}$$ is $$\varphi = \vert \theta - \theta'\vert$$. Now let's calculate the dot product: $\vec{u}\cdot \vec{v} = u_1v_1+u_2v_2 = \Vert \vec{u}\Vert \cos(\theta) \Vert \vec{v}\Vert \cos(\theta') + \Vert \vec{u}\Vert \sin(\theta) \Vert \vec{v}\Vert \sin(\theta')$

which simplifies, using the difference of angles formula from trigonometry, to $= \Vert \vec{u}\Vert \Vert \vec{v}\Vert \left(\cos(\theta) \cos(\theta') - \sin(\theta)\sin(-\theta')\right) = \Vert \vec{u}\Vert \Vert \vec{v}\Vert \cos(\theta - \theta')$ which is what we wanted to show.

Higher dimensions:

In higher dimensions, there are some nice geometric arguments using the cosine law that explain the dot product formula above; for example, here is an Oxford Math Centre site with good pictures.

What about the cross product? Actually the formula $\Vert \vec{u}\times \vec{v} \Vert = \Vert \vec{u}\Vert \Vert \vec{v}\Vert \sin(\varphi)$

is easy to see directly. Remember that $$\vec{u}\times \vec{v} = (yz'-zy', -(xz'-x'z), xy'-y'x)$$; so you can write out a formula for $\Vert \vec{u}\times \vec{v} \Vert^2$

(We square it because it is nicer that way.) The neat trick is to observe that, as expressions in the variables $$x,y,z,x',y',z'$$ we have $\Vert \vec{u}\times \vec{v} \Vert^2 + (\vec{u}\cdot \vec{v})^2 = \Vert \vec{u}\Vert^2 \Vert \vec{v}\Vert^2$

which then gives you the identity you want by replacing $$(\vec{u}\cdot \vec{v})^2$$ with $$\Vert \vec{u}\Vert^2 \Vert \vec{v}\Vert^2\cos^2(\varphi)$$ and using a standard trigonometric identity.

More practice with row reduction

We will keep practicing on Thursday, and will also take the time to examine all the different kinds of cases. For now:

For each of the following problems, set up a system of linear equations, then write down the augmented matrix, and perform row operations (adding a multiple of one row to another, interchanging two rows, or scaling a row by a nonzero constant) to put the system into RREF. Then determine the solution.

• Show that the following planes intersect in just one point: $$x+y+z=3, 2x-y+z = 4, 3x-2z=6$$.
• Show that the following planes intersect in a line (and find parametric equations for the line): $$x+y+z=3, 2x-y+z=4, 4x+y+3z=10$$
• Show that the following planes do not intersect (that is, the system is inconsistent): $$x+y+z=3, 2x-y+z = 4, 3x-2z=6, x-y-z=7$$.