# 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: $A^11=A^{10+1}=(A^2)^5A=IA=A.$