Final review 1-2

General instructions

Refer to the hints I've given before Midterm 1.

It's very important to regularly check how much time you have left! If you can't see a clock, ask a proctor to write the remaining time on the board regularly.

Make sure to practice timing yourself using the practice finals on OWL. Set aside 3 hours, and use no notes, textbooks, calculators.

Geometry with vectors

For \(n\)-vectors \(\mathbf u\) and \(\mathbf v\), the dot product is the matrix product if \(\mathbf u\) is represented as a row vector, and \(\mathbf v\) is represented as a column vector: \[ \mathbf u\cdot\mathbf v=\begin{pmatrix} u_1 & \dots & u_n\end{pmatrix}\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}=u_1v_1+\dotsb+u_nv_n. \] The length of an \(n\)-vector \(\mathbf u\) is the square root of its dot product with itself: \[ \|\mathbf u\|=\sqrt{\mathbf u\cdot\mathbf u}=\sqrt{u_1^2+\dotsb+u_n^2}. \]

Example. Let \(\mathbf u=(2,-1,3)\). Then the unit vector with the same direction as \(\mathbf u\) is \[ \frac{1}{\|\mathbf u\|}\mathbf u=\frac{1}{\sqrt{14}}\begin{pmatrix}2\\-1\\3\end{pmatrix}. \]

The "distance of \(\mathbf u\) and \(\mathbf v\)" is the distance of \(P\) and \(Q\), where \(\mathbf u=\vec{OP}\) and \(\mathbf v=\vec{OQ}\). That is, \[ d(\mathbf u,\mathbf v)=\|\mathbf u-\mathbf v\|=\sqrt{(u_1-v_1)^2+\dotsb+(u_n-v_n)^2}. \]

Example. The distance of \(\mathbf u=(2,-1,3)\) and \(\mathbf v=(0,-2,1)\) is \[ \|\mathbf u-\mathbf v\|=\|(2,1,2)\|=3. \]

Suppose that \(\boldsymbol0\ne\mathbf u,\mathbf v\), and let \(\alpha\) be the angle of \(\mathbf u\) and \(\mathbf v\). Then we have \(\alpha=\frac{\mathbf u\cdot\mathbf v}{\|\mathbf u\|\cdot\|\mathbf v\|}\). If \(\cos\alpha=1\), then we say \(\mathbf u\) and \(\mathbf v\) have the same direction. If \(\cos\alpha=-1\), then we say \(\mathbf u\) and \(\mathbf v\) have the opposite direction. If \(\cos\alpha=\pm1\), then we say that \(\mathbf u\) and \(\mathbf v\) are parallel. Is \(\cos\alpha=0\), then we say that \(\mathbf u\) and \(\mathbf v\) are orthogonal.

Method. We have 3 ways of deciding whether \(\mathbf u\) and \(\mathbf v\) are parallel.

  1. Check whether \(\mathbf u\cdot\mathbf v=\pm\|\mathbf u\|\cdot\|\mathbf v\|\).

  2. Suppose that \(u_j\ne0\). Then check whether \(\frac{v_j}{u_j}\mathbf u=\mathbf v\).

  3. Check whether the matrix with rows \(\mathbf u\) and \(\mathbf v\) has rank 1.

Example. Let \(\mathbf u=(2,-1,3)\) and \(\mathbf v=(4,-2,-6)\). Then by the reduction step \[ \begin{pmatrix}2 & -1 & 3 \\ 4 & -2 & -6\end{pmatrix}\xrightarrow{R_2-2R_1}\begin{pmatrix}2 & -1 & 3 \\ 0 & 0 & -12\end{pmatrix} \] we get an echelon for matrix and thus see that the matrix with rows \(\mathbf u\) and \(\mathbf v\) has rank 2. Thus, \(\mathbf u\) and \(\mathbf v\) are not parallel.

Example. Let \(\mathbf u=(2,-1,3)\) and \(\mathbf v=(4,k,6)\). Then \(\mathbf u\) and \(\mathbf v\) are orthogonal, precisely when \[ 0=\mathbf u\cdot\mathbf v=26-k. \] Therefore, \(\mathbf u\) and \(\mathbf v\) are orthogonal, if \(k=26\).

The cross product of \(\mathbf u\) and \(\mathbf v\) is \(\left|\begin{smallmatrix}\mathbf i & u_1 & v_1 \\ \mathbf j & u_2 & v_2 \\ \mathbf k & u_3 & v_3 \end{smallmatrix}\right|\). You can use any expansion formula you like for this determinant.

Warning. Do not use any row or column operation involving multiplication or division by \(\mathbf i,\mathbf j\) or \(\mathbf k\) as that will give a wrong answer or not even make sense! The easiest way is probably to expand along \(C_1\).

Example. We have \[ \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\times\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}=\begin{vmatrix} \mathbf i & 2 & 0 \\ \mathbf j & -1 & -2 \\ \mathbf k & 3 & 1 \end{vmatrix}=\mathbf i\begin{vmatrix} -1 & -2 \\ 3 & 1 \end{vmatrix}-\mathbf j\begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} +\mathbf k\begin{vmatrix} 2 & 0 \\ -1 & -2 \end{vmatrix}=\begin{pmatrix} 5 \\ -2 \\ -4 \end{pmatrix}. \]

Lines and planes

Example. Consider the line with standard equation \[ 2x-3y=4. \] You can consider this as a SLE. Solving it as such, we can write its solution set as the 1-parameter family \[\begin{align*} x&=(3/2)t+2\\ y&=t \end{align*}\]

That is, we have a parametrization \[ \mathbf r(t)=\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}(3/2)t+2\\t\end{pmatrix}=\begin{pmatrix}2 \\ 0 \end{pmatrix}+t\begin{pmatrix}3/2 \\ 1\end{pmatrix} \] Note that the 1-parameter family of solutions has given a point-parallel form equation for the line! You can see that it contains the point \(P(2,0)\), and that a direction vector is \((3/2,1)\).

Of course, there's nothing wrong with using our method of chapter 1: if \((a,b)\) is a direction vector, then \(\pm(-b,a)\) are normal vectors, and vice versa.

Example. Consider the line with point-parallel form equation \[ \mathbf r(t)=\begin{pmatrix}2\\-1\end{pmatrix}+t\begin{pmatrix}-3\\1\end{pmatrix}. \] We can see that the line contains the point \(P(2,-1)\), and that a direction vector is \(\mathbf u=(-3,1)\). Therefore, a normal vector is \(\mathbf n=(1,3)\), and from the point-normal form equation \[ \left(\begin{pmatrix}x & y\end{pmatrix}-\begin{pmatrix}2 & -1\end{pmatrix}\right)\cdot\begin{pmatrix}1\\3\end{pmatrix}=0 \] we get the standard form equation \[ x+3y=1. \]

Note that a standard equation of a hyperplane in \(\mathbf R^n\) gives a \(1\times n\) matrix with one nonzero row. Therefore, the rank is 1, and you get an (\(n-1\))-parameter family of solutions. This is why in \(\mathbf R^2\), from one normal vector you get one directional vector (unique up multiplication by a nonzero scalar).

Example. Consider the plane in \(\mathbf R^3\) with standard equation \(2x-y+z=2\). Solving the SLE, you get the 2-parameter family of solutions \[\begin{align*} x&=s/2-t/2+2\\ y&=s\\ z&=t \end{align*}\]

That is, we have the parametrization \[ \mathbf r(s,t)=\begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} s/2-t/2+2 \\ s \\ t \end{pmatrix}=\begin{pmatrix}2\\0\\0\end{pmatrix}+\begin{pmatrix}1/2\\1\\0\end{pmatrix}+\begin{pmatrix}-1/2\\0\\1\end{pmatrix}. \]

You can see that a plane is determined by one point and two nonparallel direction vectors. To get the normal vector back from the two direction vectors, you need to find a nonzero vector that is orthogonal to both. There are two ways you can do it:

  1. Take the cross product
  2. Solve the homogeneous system of linear equations with coefficient matrix the rows the direction vectors as row vectors.

Example. Let \(\Pi\) be the plane with points \(P(1,0,2),Q(0,-3,1),R(-1,1,-2)\). The direction vectors are two nonparallel vectors between two pairs of these points. For example, we can take \[ \mathbf u=\vec{PQ}=\begin{pmatrix}-1\\-3\\-1\end{pmatrix}\text{ and }\vec{PR}=\begin{pmatrix} -2\\1\\-4\end{pmatrix}. \]

Now one way to get a normal vector is to take the cross product: \[ \mathbf u\times\mathbf v=\begin{vmatrix}-3 & 1 \\ -1 & -4 \end{vmatrix}\mathbf i+\begin{vmatrix}-1 & -4 \\-1 & -2 \end{vmatrix}\mathbf j+\begin{vmatrix}-1 & -2 \\ -3 & 1 \end{vmatrix}\mathbf k=\begin{pmatrix}13\\-2\\-7\end{pmatrix}. \]

Another way is to solve the HSLE with coefficient matrix having \(\mathbf u\) and \(\mathbf v\) as rows: \[ \begin{pmatrix} -1 & -3 & -1 \\ -2 & 1 & -4 \end{pmatrix} \xrightarrow{-R_1;R_2+2R_1}\begin{pmatrix} 1 & 3 & 1 \\ 0 & 7 & -2 \end{pmatrix}\xrightarrow{(1/7)R_2;R_1-3R_2}\begin{pmatrix} 1 & 0 & 13/7 \\ 0 & 1 & -2/7 \end{pmatrix} \] This tells us that \((x,y,z)\) is a normal vector precisely when \((x,y,z)=(-\frac{13}{7}t,\frac27t,t)\) for some \(t\). Note that you get the cross product for \(t=-7\).

Let \(\Pi\) be a hyperplane (note that this includes a line in \(\mathbf R^2\) and a plane in \(\mathbf R^3\)) with normal vector \(\mathbf n\) and point \(P\). Let \(Q\) be another point. Then the distance of \(Q\) from \(\Pi\) is \[ d(Q,\Pi)=\frac{|\vec{PQ}\cdot\mathbf n|}{|\mathbf n|}. \]

Warning. Don't forget to take the absolute value in the nominator. Distances can't be negative.

Example. Let \(\Pi\) be the plane \(-2x+3y+z=4\), and \(Q(-1,0,-1)\) some other point. To calculate the distance of \(Q\) from \(\Pi\), we need to find a point \(P\) on \(\Pi\), and a normal vector. Setting \(y=z=0\), we see that we need \(x=-2\) to get a solution to the standard equation. That is, \(P(-2,0,0)\) is a point on \(\Pi\). A normal vector \(\mathbf n=(-2,3,4)\) can be read off the standard equation. We have \(\vec{PQ}=(1,0,-1)\), therefore the distance is \[ d(Q,\Pi)=\frac{|\mathbf n\cdot\vec{PQ}|}{|\mathbf n|}=\frac{|-6|}{\sqrt{29}}=\frac{6}{29}. \]

Matrix multiplication

Example. Let \(A=\left(\begin{smallmatrix}2 & -1 \\ 3 & -2 \end{smallmatrix}\right)\). Let's find \(A^11\). Well, first let's calculate \[ A^2=\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} \] Since \(A^2=I\), we can simplify the calculation: \[ A^11=A^{10+1}=(A^2)^5A=IA=A. \]