Midterm #1 Solutions

Professor Gross
Spring 2006

Green Version

1. Let
   

   $$\mathbf{r} = \mathbf{i}+4\mathbf{j}+3\mathbf{k}$$

   $$\mathbf{s} = 2\mathbf{i}+\mathbf{j}-\mathbf{k}$$

   
   Compute:
   1. $3\mathbf{r}-2\mathbf{s}$.
      

      $$3\mathbf{r}-2\mathbf{s} = 3(\mathbf{i}+4\mathbf{j}+3\mathbf{k})-2(2\mathbf{i}+\mathbf{j}-\mathbf{k})$$

      $$= 3\mathbf{i}+12\mathbf{j}+9\mathbf{k}-4\mathbf{i}-2\mathbf{j}+2\mathbf{k}$$

      $$= -\mathbf{i}+10\mathbf{j}+11\mathbf{j}$$

      
      

   2. $\mathbf{r}\cdot\mathbf{s}$.
      

      $$\mathbf{r}\cdot\mathbf{s} = (1)(2)+(4)(1)+(3)(-1)$$

      $$= 2+4-3$$

      $$= 3$$

      
      

   3. $\cos\theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{s}$.
      

      $$\mathbf{r}\cdot\mathbf{s} = \vert\mathbf{r}\vert\vert\mathbf{s}\vert\cos\theta$$

      $$3 = \sqrt{(1)^2+(4)^2+(3)^2}\sqrt{(2)^2+(1)^2+(-1)^2}\cos\theta$$

      $$3 = \sqrt{1+16+9}\sqrt{4+1+1}\cos\theta$$

      $$3 = \sqrt{26}\sqrt{6}\cos\theta$$

      $$3 = \sqrt{156}\cos\theta$$

      $$3 = 2\sqrt{39}\cos\theta$$

      $$\cos\theta = \frac{3}{2\sqrt{39}}$$

      $$\cos\theta = \frac{3\sqrt{39}}{2\cdot 39}$$

      $$\cos\theta = \frac{\sqrt{39}}{26}$$

      
      

   4. $\mathbf{r}\times\mathbf{s}$.
      

      $$\mathbf{r}\times\mathbf{s} = \left\vert\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\ 1&4&3\ 2&1&-1\end{array}\right\vert$$

      $$= \left\vert\begin{array}{cc}4&3\ 1&-1\end{array}\right\vert\mathb... ...mathbf{j} +\left\vert\begin{array}{cc}1&4\ 2&1\end{array}\right\vert\mathbf{k}$$

      $$= -7\mathbf{i}+7\mathbf{j}-7\mathbf{k}$$

      
      

2. 1. Give a parametric equation for the line passing through the two points
      $$P=(1,0,-1), Q=(2,1,3).$$
      1. We may use the formula $\mathbf{r}(t)=\mathbf{r}_0(1-t)+\mathbf{r}_ft$, where $\mathbf{r}_0$ is the initial point and $\mathbf{r}_f$ is the final point. This will give us:
         

         $$\mathbf{r}(t) = \langle 1,0,-1\rangle (1-t)+\langle 2,1,3\rangle t$$

         $$= \langle 1(1-t)+2t,0(1-t)+1t,-1(1-t)+3t\rangle$$

         $$= \langle 1+t,t,-1+4t\rangle,$$

         
         so our parametric equations are:
         $$\left\lbrace\begin{array}{rcl}x&=&1+t\ y&=&t\ z&=&-1+4t\end{array}\right..$$

      2. Another method is to use $P$ as the initial point and use $\overrightarrow{PQ}$ as the direction vector for the line:
         $$\overrightarrow{PQ}=\langle 2-1,1-0,3-(-1)\rangle=\langle 1,1,4\rangle,$$
         So we get the vector equation $\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{v}=\langle1,0,-1\rangle+t\langle 1,1,4\rangle$, which again gives the parametric equations
         $$\left\lbrace\begin{array}{rcl}x&=&1+t\ y&=&t\ z&=&-1+4t\end{array}\right..$$

   2. Find a parametric equation for the line which is the intersection of the planes
      $$2x+y+z=1$$ and $$y+2z=2.$$
      1. Method 1
         

         The above picture shows the two planes we are considering. We will call the planes $P_1$ and $P_2$. Specifically, $P_1: 2x+y+z=1$ and $P_2: y+2z=2$ We will use $l$ to represent the line formed by the intersection, and we will say that $\mathbf{v}$ is the direction vector of $l$. Now, $l$ lies in $P_1$, so if $\mathbf{n}_1$ is a normal vector to $P_1$, then $\mathbf{n}_1$ and $\mathbf{v}$ are orthogonal. Similarly, $\mathbf{n}_2$, a normal vector of $P_2$ will be orthogonal to $\mathbf{v}$. This means that if we know $\mathbf{n}_1$ and $\mathbf{n}_2$, then we can find a vector for $\mathbf{v}$.

         Now, just by looking at the coefficients, we can see that $\mathbf{n}_1=\langle 2,1,1\rangle$ and $\mathbf{n}_2=\langle 0,1,2\rangle$. Now we want $\mathbf{v}$ to be orthogonal to both of these vectors, so we will take the cross product and get:
         

         $$\mathbf{v} = \mathbf{n}_1\times\mathbf{n}_2$$

         $$= \left\vert\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\ 2&1&1\ 0&1&2\end{array}\right\vert$$

         $$= \left\vert\begin{array}{cc}1&1\ 1&2\end{array}\right\vert\mathbf... ...mathbf{j} +\left\vert\begin{array}{cc}2&1\ 0&1\end{array}\right\vert\mathbf{k}$$

         $$= \mathbf{i}-4\mathbf{j}+2\mathbf{k}$$

         
         So we have a direction vector. Now, to have the equation for a line, we also need an initial point, which means that we need to find a point on the line $l$. To find that point, notice that we can choose any value for $x$ and we will be able to find a $y$ and $z$ to satisfy the equations. So, we will just set $x=0$ and find what $y$ and $z$ are when $x=0$. But $x=0$ gives us $y+z=1$ from the first equation, so $z=1-y$, Plugging this in to the equation for the second plane, we get $y+2(1-y)=2\implies y+2-2y=2\implies 2-y=2\implies y=0$, but since we had $z=1-y$, we get that $z=1$, so that means that the point $(0,0,1)$ is on the line $l$, which can be verified by plugging it in to the equations of the planes. Now, we have a point on the line and a direction vector for the line, so we can write a vector equation for the line $l$:
         $$\mathbf{r}(t)=\langle 0,0,1\rangle+t\langle 1,-4,2\rangle,$$
         and changing this in to a parametric equation, we get
         $$\left\lbrace\begin{array}{rcl}x&=&0+1t\ y&=&0+(-4)t\ z&=&1+2t\end{array}\right.,$$
         or slightly simplified,
         $$\left\lbrace\begin{array}{rcl}x&=&t\ y&=&-4t\ z&=&1+2t\end{array}\right..$$

      2. Method 2: This is another method, but it seemed to be one which was prone to errors and misinterpretation, for which reason I suggest the previous method. The previous method is also easier to understand, I think; this method is more one of pushing symbols around until an answer falls out:

         We will start with the equations for the planes:

         $$2x+y+z=1$$ (1)

         
         

         $$y+2z=2,$$ (2)

         
         
         and from (), we get

         $$y=2-2z,$$ (3)

         
         
         which yields $2x+(2-2z)+z=1$ when plugged in to (). Simplifying, we get $z=2x+1.$ If we plug this in to (), we get $y=2-2(2x+1)=-4x.$ Now that we have everything in terms of $x$, we can set $x=t$ and get
         $$\left\lbrace\begin{array}{rcl} x&=&t\\ y&=&-4t\\ z&=&2t+1 \end{array}\right..$$

3. 1. Give Cartesian coordinates for the point whose cylindrical coordinates are
      $$(2,\pi,3/2).$$

      We know the following relations:

      $$\left\lbrace\begin{array}{rcl}x&=&r\cos\theta\ y&=&r\sin\theta\ z&=&z\end{array}\right.$$

      So:

      $$x=2\cos\pi=2(-1)=-2$$
      $$y=2\sin\pi=2\cdot 0=0$$
      $$z=3/2$$
      Thus in Cartesian coordinates, the point is $(-2,0,3/2)$.

   2. Give Cartesian coordinates for the point whose spherical coordinates are
      $$(2,\pi/3,\pi/4).$$

      We know the following relations:

      $$\left\lbrace\begin{array}{rcl}x&=&\rho\cos\theta \sin\varphi\ y&=&\rho\sin\theta \sin\varphi\ z&=&\rho\cos\varphi\end{array}\right.$$

      So:

      $$x=\rho\cos\theta \sin\varphi=2\cos(\pi/3)\sin(\pi/4)=2\cdot\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$$
      $$y=\rho\sin\theta \sin\varphi=2\sin(\pi/3)\sin(\pi/4)=2\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{2}$$
      $$z=\rho\cos\varphi=2\cos(\pi/4)=2\cdot\frac{\sqrt{2}}{2}=\sqrt{2}$$

      Thus in Cartesian coordinates, the point is $(\frac{\sqrt{2}}{2},\frac{\sqrt{6}}{2},\sqrt{2})$.

   3. Sketch the surface whose equation in spherical coordinates $(\rho,\theta,\varphi)$ is given by $\varphi=\pi/4.$

      Since there are no restriction on $\rho$ or $\theta$, we can look at any points for which $\varphi=\pi/4$. That is going to be any point on the cone shown below:

      

4. 1. Find a vector orthogonal to the plane containing the points $P$, $Q$, and $R$ with
      $$P=(2,1,0), Q=(-1,1,0), R=(1,0,2).$$

      

      $$\overrightarrow{PQ}=Q-P=\langle-3,0,0\rangle.$$
      $$\overrightarrow{PR}=R-P=\langle-1,-1,2\rangle,$$
      $$\mathbf{n}=\left\vert\begin{matrix}\mathbf{i}&\mathbf{j}&\mathbf{... ...-1&-1&2\end{matrix}\right\vert=\langle0,-(-6-0),3-0\rangle=\langle0,6,3\rangle.$$

   2. Find the area of the triangle $PQR$.

      

      The area of the shape above is equal to the magnitude of the cross product $\mathbf{n}=\overrightarrow{PQ}\times\overrightarrow{PR}$. This means that the area of the triangle is one half of that:

      $$A=\frac{\vert\mathbf{n}\vert}{2}=\frac{1}{2}\sqrt{0^2+6^2+3^2}=\frac{1}{2}\sqrt{36+9}=\frac{\sqrt{45}}{2}$$

5. Compute
   $$\lim_{t\rightarrow 0}\mathbf{r}(t),$$
   where
   $$\mathbf{r}(t)=\frac{\sin t}{t}\mathbf{i}+\frac{t^2-1}{t-1}\mathbf{j}+(\cos t-1)\mathbf{k}.$$
   $$\lim_{t\rightarrow 0}\frac{\sin t}{t}\mathop{=}^{\mbox{\begin{tiny}(L'H\^opital)\end{tiny}}}\lim_{t\rightarrow 0}\frac{\cos t}{1}=1,$$
   $$\lim_{t\rightarrow 0}\frac{t^2-1}{t-1}=\lim_{t\rightarrow 0}\frac{(t-1)(t+1)}{t-1}=\lim_{t\rightarrow 0}(t+1)=1,$$
   $$\lim_{t\rightarrow 0}(\cos t-1)=1-1=0,$$
   so
   $$\lim_{t\rightarrow 0}\mathbf{r}(t)=\langle1,1,0\rangle.$$

Important differences in the Yellow version

In the yellow version's problem 1(c), there is an error. $\varphi=-\pi/4$ should be changed to $\varphi=\pi/4$. As it is written, there is no solution becuase $0\leq\varphi\leq\pi$. If you assumed that you could have $\varphi=-\pi/4$, you should get the same answer as in the green version.

In the yellow version's problem 4(b), the dot product is zero. So, in 4(c), you will get that $\cos\theta=0$, however you should mention, or show that $\vert\mathbf{r}\vert\vert\mathbf{s}\vert\neq0$, that is $\mathbf{r}\neq 0$ and $\mathbf{s}\neq 0$. Otherwise, $\cos\theta$ is not defined.

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Midterm #1 Solutions

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