I have started writing a schedule so that you can follow which lecture covers which section of the textbook. The homework list is organized with respect to the textbook sections.

Convention. From now on, unless we state it otherwise, we'll write the components of a vector with the same letter, and lower indices. That is, after we've introduced a vector \(\mathbf u\), it will be understood that \(u_1,u_2,u_3\) are the components of \(\mathbf u\).

Products of vectors

Dot products

Definition. Let \(\mathbf u=\left(\begin{smallmatrix}u_1\\u_2\\u_3\end{smallmatrix}\right)\) and \(\mathbf v=\left(\begin{smallmatrix}v_1\\v_2\\v_2\end{smallmatrix}\right)\) be two space vectors. Then their dot product is \[ \mathbf u\cdot\mathbf v=\left(\begin{smallmatrix}u_1 & u_2 & u_2\end{smallmatrix}\right)\cdot\left(\begin{smallmatrix}v_1\\v_2\\v_3\end{smallmatrix}\right)=u_1v_1+u_2v_2+u_3v_3. \] That is we add up the products of the components.

Note that when writing the formula, I have written \(\mathbf u\) in row vector form. This way, the dot product looks exactly like matrix multiplication, which we'll learn in Chapter 3. For another example of notation, in Quantum Mechanics, vectors are written as \(|u_1,u_2,u_3\rangle\), and scalar products are written as \(\langle u_1,u_2,u_3|v_1,v_2,v_3\rangle\).

Exercise. Calculate the following dot products. \[ \left(\begin{smallmatrix}3&2\end{smallmatrix}\right)\left(\begin{smallmatrix}2\\-1\end{smallmatrix}\right)\quad\left(\begin{smallmatrix}-3&0&1\end{smallmatrix}\right)\left(\begin{smallmatrix}2\\3\\2\end{smallmatrix}\right)\quad\left(\begin{smallmatrix}-3&0&1\end{smallmatrix}\right)\left(\begin{smallmatrix}2\\3\\6\end{smallmatrix}\right). \]

Theorem. Dot products satisfy the following arithmetic rules. \[\begin{align*} \mathbf u\mathbf v=&\mathbf v\mathbf u\\ c(\mathbf u\mathbf v)=&(c\mathbf u)\mathbf v=\mathbf u(c\mathbf v)\\ \mathbf u(\mathbf v+\mathbf w)=&\mathbf u\mathbf v+\mathbf u\mathbf w\\ \mathbf u\mathbf u=&\|\mathbf u\|^2. \end{align*}\]

Again, these rules follow from the arithmetic rules for scalars. For example, let's check the third one. \[ \left(\begin{smallmatrix}u_1&u_2&u_3\end{smallmatrix}\right)\left(\left(\begin{smallmatrix}v_1\\v_2\\v_3\end{smallmatrix}\right)+\left(\begin{smallmatrix}w_1\\w_2\\w_3\end{smallmatrix}\right)\right)= \left(\begin{smallmatrix}u_1&u_2&u_3\end{smallmatrix}\right)\left(\begin{smallmatrix}v_1+w_1\\v_2+w_2\\v_3+w_3\end{smallmatrix}\right)= u_1(v_1+w_1)+u_2(v_2+w_2)+u_3(v_3+w_3)=(u_1v_1+u_2v_2+u_3v_3)+(u_1w_1+u_2w_2+u_3w_3)= \left(\begin{smallmatrix}u_1&u_2&u_3\end{smallmatrix}\right)\left(\begin{smallmatrix}v_1\\v_2\\v_3\end{smallmatrix}\right)+ \left(\begin{smallmatrix}u_1&u_2&u_3\end{smallmatrix}\right)\left(\begin{smallmatrix}w_1\\w_2\\w_3\end{smallmatrix}\right). \]

Theorem. Let \(\mathbf u=\vec{OP}\) and \(\mathbf v=\vec{OQ}\) be nonzero vectors. Let \(\theta=\measuredangle POQ\). Then we have \[ \mathbf u\mathbf v=\cos\theta\|\mathbf u\|\,\|\mathbf v\|. \] That is, we can use dot products to calculate angles between nonzero vectors when represented with arrows with the same origin: \[ \cos\theta=\frac{\mathbf u\mathbf v}{\|\mathbf u\|\,\|\mathbf v\|}. \]

Proposition. Let \(\mathbf u\) and \(\mathbf v\) be two nonzero vectors.

  1. They have an acute angle if and only if \(\mathbf u\mathbf v>0\).

  2. They are orthogonal if and only if \(\mathbf u\mathbf v=0\).

  3. They have an obtuse angle if and only if \(\mathbf u\mathbf v<0\).

Note that to decide between these, we don't have to divide by the lengths. Yay!

Exercise. Consider the points \(P(1,3),Q(2,0),R(2,-1)\). Letting \(\theta=\measuredangle PQR\), find \(\cos\theta\).

Cross products

Let \(\mathbf u\) and \(\mathbf v\) be two space vectors. The dot product \(\mathbf u\cdot\mathbf v\) we have just defined is a scalar. The cross product \(\mathbf u\times\mathbf v\) is a space vector. We will define them using determinants, because I find that formula the easiest to remember, and also that ties it in to Chapter 4.

Warning. As I keep saying, the vector operations we have seen last time, and the dot product makes sense for vectors of any dimension. On the other hand, cross products only make sense for space vectors.

Definition. A \(2\times2\) determinant is \[ \left|\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right|=ad-bc. \] A \(3\times3\) determinant is \[ \left|\begin{smallmatrix}a&b&c\\p&q&r\\x&y&z\end{smallmatrix}\right|=a\left|\begin{smallmatrix}q&r\\y&z\end{smallmatrix}\right|-p\left|\begin{smallmatrix}b&c\\y&z\end{smallmatrix}\right|+x\left|\begin{smallmatrix}b&c\\q&r\end{smallmatrix}\right|. \]

Exercise. Find the following determinants. \[ \left|\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right|\quad\left|\begin{smallmatrix}2&0\\1&7\end{smallmatrix}\right|\quad\left|\begin{smallmatrix}1&2&0\\-2&3&1\\2&1&4\end{smallmatrix}\right| \]

Definition. Let \(\mathbf u\) and \(\mathbf v\) be space vectors. Then their cross product is \[ \mathbf u\times\mathbf v=\left|\begin{smallmatrix}\mathbf i&u_1&v_1\\\mathbf j&u_2&v_2\\\mathbf k&u_3&v_3\end{smallmatrix}\right|=\left(\begin{smallmatrix}u_2v_3-u_3v_2\\-u_1v_3+u_3v_1\\u_1v_2-u_2v_1\end{smallmatrix}\right). \]

Exercise. Find the cross product \(\left(\begin{smallmatrix}2\\-1\\4\end{smallmatrix}\right)\times\left(\begin{smallmatrix}0\\3\\2\end{smallmatrix}\right)\).

Theorem. Let \(\mathbf u\) and \(\mathbf v\) be space vectors. Then the cross product \(\mathbf u\times\mathbf v\) is orthogonal to both \(\mathbf u\) and \(\mathbf v\).

Let's prove this for \(\mathbf u\). We need to show that their dot product is zero: \[ \mathbf u(\mathbf u\times\mathbf v)=\left(\begin{smallmatrix}u_1&u_2&u_3\end{smallmatrix}\right)\left(\begin{smallmatrix}u_2v_3-u_3v_2\\-u_1v_3+u_3v_1\\u_1v_2-u_2v_1\end{smallmatrix}\right)= u_1(u_2v_3-u_3v_2)+u_2(-u_1v_3+u_3v_1)+u_3(u_1v_2-u_2v_1)=u_1(u_2v_3-u_3v_2-u_2v_3+u_3v_2)+v_1(u_2u_3-u_2u_3)=0. \] Duh.

Remark. Note that this still leaves two possible directions for the cross product \(\mathbf u\times\mathbf w\). To decide between these, you can use that the vectors \(\mathbf u,\mathbf v,\mathbf u\times\mathbf v\) satisfy the right-hand rule. This basically means that you can put them on your thumb, index finger and middle finger without breaking your middle finger.

Theorem. The cross product satifies the following arithmetic properties. \[\begin{align*} \mathbf v\times\mathbf u=&-(\mathbf u\times\mathbf v)\\ c(\mathbf u\times\mathbf v)=&(c\mathbf u)\times\mathbf v=\mathbf u\times(c\mathbf v)\\ \mathbf u\times(\mathbf v+\mathbf w)=&(\mathbf u\times\mathbf v)+(\mathbf u\times\mathbf w)\\ \mathbf u\times\mathbf u=&\boldsymbol0. \end{align*}\]

Warning. Note that the first property shows that the cross product is not commutative. It is not associative either, as the following exericise shows.

Exercise. Find \((\mathbf i\times\mathbf i)\times\mathbf j\) and \(\mathbf i\times(\mathbf i\times\mathbf j)\).

Theorem. (a) Let \(\mathbf u=\vec{OP}\) and \(\mathbf v=\vec{OQ}\) and \(\theta=\measuredangle POQ\). Then the area of the parallelogram with adjacent edges \(\overline{OP}\) and \(\overline{OQ}\) is \[ \|\mathbf u\times\mathbf v\|=\sin\theta\|\mathbf u\|\,\|\mathbf v\|. \] (b) Let in addition \(\mathbf w=\vec{OR}\). Then the area of the parallelepiped with adjacent edges \(\overline{OP},\,\overline{OQ}\) and \(\overline{OR}\) is \[ |\mathbf u(\mathbf v\times\mathbf w)|. \] Corollary. Let \(P,Q,R\) be points. Then the area of the triangle \(\triangle PQR\) is \(\frac12\|\vec{PQ}\times\vec{PR}\|\).

Remark. The scalar \(\mathbf u(\mathbf v\times\mathbf w)\) is called a triple product.

Exercise. Find the area of the triangle with vertices \(P(2,0,3),Q(-1,3,0),R(-2,1,2)\).