2. Vectors

Exercises

a. Definition, Magnitude and Direction

If the vector, \(\vec a=\left\langle 4,2 \right\rangle\) is located at \(P=(-2,3)\), where is its tip?

\(Q=(2,5)\)

\[\begin{aligned} Q&=P+\vec a =(-2,3)+\left\langle 4,2\right\rangle \\ &=(2,5) \end{aligned}\]

tj
Find the displacement vector from \(P=(-5,-3)\) to \(Q=(-2,-10)\)

\(\overrightarrow{PQ}=\left\langle 3,-7\right\rangle\)

Since \(P=(-5,-3)\) and \(Q=(-2,-10)\) \[\begin{aligned} \overrightarrow{PQ} &=\left\langle Q_1-P_1,Q_2-P_2\right\rangle \\ &=\left\langle -2-(-5),-10-(-3)\right\rangle \\ &=\left\langle 3,-7\right\rangle \end{aligned}\]

tj
Find the displacement vector from \(A=(2,-5)\) to \(B=(-4,6)\)

\(\overrightarrow{AB}=\left\langle -6,11\right\rangle\)

Since \(A=(2,-5)\) and \(B=(-4,6)\) \[\begin{aligned} \overrightarrow{AB} &=\langle B_1-A_1,B_2-A_2\rangle \\ &=\langle -4-2,6-(-5) \rangle \\ &=\langle -6,11 \rangle \end{aligned}\]

tj
Find the magnitude of vector \(\vec u=\left\langle -5,8\right\rangle\)

\(|\vec u|=\sqrt{(u_1)^2+(u_2)^2}\)

\(|\vec u|=\sqrt{89}\)

Since \(\vec u=\left\langle -5,8 \right\rangle\) \[\begin{aligned} |\vec u| &=\sqrt{(-5)^2+8^2}\\ &=\sqrt{25+64} \\ &=\sqrt{89} \end{aligned}\]

tj
Find the magnitude of vector \(\vec v=\left\langle 3,6\right\rangle\)

\(|\vec v|=\sqrt{(v_1)^2+(v_2)^2} \)

\(|\vec v|=3\sqrt{5}\)

Since \(\vec v=\langle 3,6 \rangle \) \[\begin{aligned} |\vec v| &=\sqrt{3^2+6^2} \\ &=\sqrt{9+36}=\sqrt{45}=3\sqrt{5} \end{aligned}\]

tj
Which of the following are unit vectors?

\(\left\langle\dfrac{5}{13},\dfrac{12}{13}\right\rangle\)

\(\left\langle0.4,0.3\right\rangle\)

\(\left\langle0.6,-0.8\right\rangle\)

\(\left\langle-\,\dfrac{4}{5},\dfrac{3}{5}\right\rangle\)

YES!
\( \begin{aligned} \left|\left\langle\dfrac{5}{13},\dfrac{12}{13}\right\rangle\right| &=\sqrt{\left(\dfrac{5}{13}\right)^2+\left(\dfrac{12}{13}\right)^2} \\ &=\sqrt{\dfrac{25+144}{169}}=1 \end{aligned} \)

NO!
\( \begin{aligned} \left|\left\langle 0.4,0.3 \right\rangle\right| &=\sqrt{0.4^2+0.3^2} \\ &=\sqrt{0.16+.09}=0.5 \neq 1 \end{aligned} \)

YES!
\( \begin{aligned} \left|\left\langle 0.6,-0.8 \right\rangle\right| &=\sqrt{0.6^2+(-0.8)^2} \\ &=\sqrt{0.36+0.64}=1 \end{aligned} \)

YES!
\( \begin{aligned} \left|\left\langle -\,\dfrac{4}{5},\dfrac{3}{5}\right\rangle\right| &=\sqrt{\left(\dfrac{4}{5}\right)^2+\left(\dfrac{3}{5}\right)^2} \\ &=\sqrt{\dfrac{16+9}{25}}=1 \end{aligned} \)
Find a unit vector in the same direction as vector \(\vec w=\left\langle-4,3\right\rangle\).

Remember, the unit vector in the direction of a given vector is that vector divided by its magnitude.

\(\hat w=\left\langle\dfrac{-4}{5},\dfrac{3}{5}\right\rangle\)

The length of \(\vec w\) is: \[ |\vec w|=\sqrt{16+9}=5 \] So its unit vector is: \[ \hat w=\dfrac{1}{5}\left\langle-4,3\right\rangle =\left\langle\dfrac{-4}{5},\dfrac{3}{5}\right\rangle \]

tj

You can check your answer by calculating the magnitude of your unit vector: \[\begin{aligned} |\hat w| &=\sqrt{\left(\dfrac{-4}{5}\right)^2+\left(\dfrac{3}{5}\right)^2} \\ &=\sqrt{\dfrac{16}{25}+\dfrac{9}{25}}=1 \end{aligned}\] Since \(|\hat w|=1\), our answer is a unit vector.
Find a unit vector in the direction opposite to the vector \(\vec v=\left\langle24,-7\right\rangle\)

If \(\hat v\) is the direction of \(\vec v\), then \(-\hat v\) is the unit vector opposite to \(\vec v\).

\(-\hat v=\left\langle-\,\dfrac{24}{25},\dfrac{7}{25}\right\rangle\)

Since \(\vec v=\left\langle24,-7\right\rangle\), its length is: \[\begin{aligned} |\vec v| &=\sqrt{ {v_1}^2+{v_2}^2} \\ &=\sqrt{576+49}=25 \end{aligned}\] So its unit vector is: \[ \hat v=\left\langle\dfrac{24}{25},-\,\dfrac{7}{25}\right\rangle \] And the unit vector in the opposite direction is: \[ -\hat v=\left\langle-\,\dfrac{24}{25},\dfrac{7}{25}\right\rangle \]

tj
Find a vector \(\vec v\) whose length is \(5\) and whose direction cosines are \(-.6\) and \(.8\).

\(\vec v=\left\langle-3,4\right\rangle\)

For the vector \(\vec v=\langle v_1,v_2\rangle\) to have length \(5\), we need \[ |\vec v|=\sqrt{(v_1)^2+(v_2)^2}=5 \] The direction of \(\vec v\) is the unit vector \[ \hat v=\dfrac{1}{|\vec v|}\langle v_1,v_2\rangle =\left\langle\dfrac{v_1}{5},\dfrac{v_2}{5}\right\rangle \] But the components of this unit vector must be the direction cosines. So: \[ \dfrac{v_1}{5}=-.6 \quad \text{and} \quad \dfrac{v_2}{5}=.8 \] Therefore, \(v_1=-.6\cdot5=-3\) and \(v_2=.8\cdot5=4\). So: \[ \vec v=\langle-3,4\rangle \]

tj

b. Vector Addition

Find the sum of vectors \(\vec a=\left\langle-5,3\right\rangle\) and \(\vec b=\left\langle9,-10\right\rangle\)

Add corresponding components of the vectors: \[ \left\langle u_1,u_2\right\rangle+\left\langle v_1,v_2\right\rangle =\left\langle u_1+v_1,u_2+v_2\right\rangle \]

\(\vec a+\vec b=\left\langle4,-7\right\rangle\)

Since \(\vec a=\left\langle-5,3\right\rangle\) and \(\vec b =\left\langle9,-10\right\rangle\) \[\begin{aligned} \vec a+\vec b &=\left\langle a_1+b_1,a_2+b_2\right\rangle \\ &=\left\langle-5+9,3+(-10)\right\rangle =\left\langle4,-7\right\rangle \end{aligned}\]

tj
Compute \(\vec v=\left\langle-3,2\right\rangle+\left\langle5,1\right\rangle\)

\(\vec v=\left\langle2,3\right\rangle\)

\[ \vec v=\left\langle-3+5,2+1\right\rangle=\left\langle2,3\right\rangle \]

tj
What is \(\vec u+\vec v+\vec w\) if \(\vec u=\langle3,1\rangle\), \(\vec v=\langle-5,2\rangle\), and \(\vec w=\langle-2,0\rangle\)?

\(\vec u+\vec v+\vec w=\left\langle-4,3\right\rangle\)

\[\begin{aligned} \vec u+\vec v+\vec w &=\langle3,1\rangle+\langle-5,2\rangle+\langle-2,0\rangle \\ &=\langle3-5-2,1+2+0\rangle =\langle-4,3\rangle \end{aligned}\]

tj
Consider the vectors \(\vec a\) and \(\vec b\) shown at the right. Which of the following is \(\vec a+\vec b\)?

\(+\) \(=\)

This is answer (C).
Prove: The Zero Vector, \(\vec0\), is Unique.
Equivalently: If \(\vec z\) is another vector which satisfies \(\vec v+\vec z=\vec v\) then \(\vec z=\vec0\)

Add \(-\vec v\) to both sides.

Start with:

\(\vec v+\vec z=\vec v\)

Use commutativity on the left:

\(\vec z+\vec v=\vec v\)

Add \(-\vec v\) to both sides:

\((\vec z+\vec v) -\vec v=\vec v\ -\vec v\)

Use associativity on the left:

\(\vec z+(\vec v -\vec v)=\vec v\ -\vec v\)

Use the negative property on both sides:

\(\vec z+\vec 0 =\vec 0\)

Use the zero property on the left:

\(\vec z =\vec 0\)

tj
Prove: The Negative, \(-\vec v\), is Unique.
Equivalently: Given a vector \(\vec v\), if there is a vector \(\vec w\) satisfying \(\vec v+\vec w=\vec0\) then \(\vec w=-\vec v\).

Start with:

\(\vec v+\vec w=\vec0\)

Use commutativity:

\(\vec w+\vec v=\vec0\)

Add \(-\vec v\) to both sides:

\((\vec w+\vec v)-\vec v=\vec0-\vec v\)

Use associativity on the left and
the zero property on the right:

\(\vec w+(\vec v-\vec v)=-\vec v\)

Use the negative property:

\(\vec w+\vec0=-\vec v\)

Finally, use the zero property again:

\(\vec w=-\vec v\)

tj
Prove: Additive Cancellation.
Specifically: If \(\vec u+\vec v=\vec w+\vec v\) then \(\vec u=\vec w\).

Start with:

\(\vec u+\vec v=\vec w+\vec v\)

Add \(-\vec v\) to both sides:

\((\vec u+\vec v) -\vec v=(\vec w+\vec v) - \vec v\)

Use associativity on both sides:

\(\vec u+(\vec v -\vec v)=\vec w+(\vec v - \vec v)\)

Use definition of \(-\vec v\):

\(\vec u+\vec 0=\vec w+\vec 0\)

Use definition of \(\vec 0\):

\(\vec u=\vec w\)

tj
A buoy is pulled by three tugboats but it is not moving. If the first two tugboats exert the forces \[ \vec F_1=\left\langle2,-4\right\rangle \qquad \text{and} \qquad \vec F_2=\left\langle-3,1\right\rangle \] find the force \(\vec F_3\) exerted by the third tugboat.

If the buoy is not moving, the sum of the forces acting on it must be zero.

\(\vec F_3=\left\langle1,3\right\rangle\)

Let \(\vec F_3=\left\langle a,b\right\rangle\). The sum of the forces must be \(\vec0\): \[\begin{aligned} \vec F_1+\vec F_2+\vec F_3=\vec0 \\ \left\langle2,-4\right\rangle+\left\langle-3,1\right\rangle+\left\langle a,b\right\rangle =\left\langle0,0\right\rangle \\ -1+a=0 \qquad \text{and} \qquad -3+b=0 \end{aligned}\] So \(a=1\) and \(b=3\) and so \(\vec F_3=\left\langle1,3\right\rangle\).

tj

PY: Add a plot to 17.

A buoy, with mass \(m=60\,\text{kg}\), is pulled by two tugboats which exert the forces \[ \vec F_1=\left\langle50,-35\right\rangle \qquad \text{and} \qquad \vec F_2=\left\langle40,5\right\rangle \] in units of \(\text{N}=\dfrac{\text{kg m}}{\text{sec}^2}\). Find the acceleration \(\vec a\) of the buoy in units of \(\dfrac{\text{m}}{\text{sec}^2}\).

Newton's Second Law says \(\vec F=m\vec a\). So \(\vec a=\dfrac{1}{m}\vec F\).

\(\vec a=\left\langle1.5,-.5\right\rangle \dfrac{\text{m}}{\text{sec}^2}\)

By Newton's Second Law, \(\vec F=m\vec a\), we compute: \[\begin{aligned} \vec a &=\dfrac{1}{m}(\vec F_1+\vec F_2) =\dfrac{1}{60}(\left\langle50,-35\right\rangle+\left\langle40,5\right\rangle) \\ &=\dfrac{1}{60}(\left\langle90,-30\right\rangle) =\left\langle1.5,-.5\right\rangle \dfrac{\text{m}}{\text{sec}^2} \end{aligned}\]

tj

PY: Add a plot to 18.

The lunar rover travels \(5\) km NW and then \(10\) km NE. What distance and in what direction (in degrees E or W of S) should the rover travel to get back to its start?

The rover must travel a distance of \(11.18\) m at \(18.4^\circ\) W of S.

The two movements are the vectors: \[\begin{aligned} \vec a&=\left\langle-5\cos45^\circ,5\sin45^\circ\right\rangle \\ &=\left\langle\dfrac{-5}{\sqrt{2}},\dfrac{5}{\sqrt{2}}\right\rangle \end{aligned}\] and \[\begin{aligned} \vec b&=\left\langle10\cos45^\circ,10\sin45^\circ\right\rangle \\ &=\left\langle\dfrac{10}{\sqrt{2}},\dfrac{10}{\sqrt{2}}\right\rangle \end{aligned}\]

The total movement is: \[ \vec c=\vec a+\vec b=\left\langle\dfrac{5}{\sqrt{2}},\dfrac{15}{\sqrt{2}}\right\rangle \] To return, the rover must travel along the vector: \[ -\vec c=\left\langle\dfrac{-5}{\sqrt{2}},\dfrac{-15}{\sqrt{2}}\right\rangle \] which is in the third quadrant. So the rover must travel a distance: \[\begin{aligned} |\vec c|&=\sqrt{\left(\dfrac{-5}{\sqrt{2}}\right)^2 +\left(\dfrac{-15}{\sqrt{2}}\right)^2} =\sqrt{\dfrac{25+225}{2}} \\ &=\sqrt{125} \approx11.18\,\text{m} \end{aligned}\] at an angle (Don't forget the \(+\pi\) for quadrant III.): \[ \theta=\arctan\left(\dfrac{15}{5}\right)+\pi \approx4.39\,\text{rad} \] counterclockwise from E. We convert to degrees: \[ 4.39\,\text{rad}=4.39\,\text{rad} \dfrac{180^\circ}{\pi\,\text{rad}} =251.6^\circ \] Finally, we convert to and angle W of S: \[ 270^\circ-251.6^\circ=18.4^\circ \] In summary, the rover must travel a distance of \(11.18\) m at \(18.4^\circ\) W of S.

tj

There is an easy way to check the distance. Since NW and NE are perpendicular, we can simply use the Pythagorean Theorem: \[ |\vec c|=\sqrt{5^2+10^2}=11.18\,\text{m} \]

PY: Improve plot of rover in 19.

c. Scalar Multiplication

Find the vector \(\vec w=c\vec u\) if \(c=3\) and \(\vec u=\left\langle4,1\right\rangle\).

\(\vec w=\left\langle12,3\right\rangle\)

\[\begin{aligned} \vec w&=c\vec u = \langle cu_1,cu_2 \rangle \\ &=3\langle 4,1 \rangle = \langle 3\cdot4,3\cdot1 \rangle \\ &= \langle 12,3 \rangle \end{aligned}\]

tj
For the vector \(\vec a=\left\langle4,3\right\rangle\), compute each of the following vectors and determine if it is in the same or opposite direction as \(\vec a\).
1. \(\vec b=5\vec a\)
  
  \(\vec b=\left\langle20,15\right\rangle\)
  Yes, \(\vec b\) and \(\vec a\) point in the same direction because their unit vectors are the same: \[ \hat b=\left\langle\dfrac{4}{5},\dfrac{3}{5}\right\rangle=\hat a \]
2. \(\vec c=-5\vec a\)
  
  \(\vec c=\left\langle-20,-15\right\rangle\)
  \(\vec c\) points in the opposite direction from \(\vec a\) because their unit vectors are negatives: \[ \hat c=\left\langle-\,\dfrac{4}{5},-\,\dfrac{3}{5}\right\rangle \qquad \text{while} \qquad \hat a=\left\langle\dfrac{4}{5},\dfrac{3}{5}\right\rangle \]
3. \(\vec d=-\vec c\)
  
  Notice that \(\vec d\) is defined in terms of \(\vec c\) rather than \(\vec a\).
  
  \(\vec d=\left\langle20,15\right\rangle=\vec b\) which is in the same direction as \(\vec a\).
Write \(\vec v=\left\langle6,8\right\rangle\) as the product of its magnitude and its direction.

\(\vec v=10\left\langle\dfrac{3}{5},\dfrac{4}{5}\right\rangle\)

\[ |\vec v|=\sqrt{6^2+8^2}=10 \] \[ \hat v=\dfrac{1}{10}\left\langle6,8\right\rangle =\left\langle\dfrac{3}{5},\dfrac{4}{5}\right\rangle \] So \[ \vec v=|\vec v|\hat v=10\left\langle\dfrac{3}{5},\dfrac{4}{5}\right\rangle \]

tj

We check: \(10\left\langle\dfrac{3}{5},\dfrac{4}{5}\right\rangle =\left\langle6,8\right\rangle\)
Consider the vector \(\vec a\) shown at the right. Which of the following is \(3\vec a\)?

\(3\times\) \(=\) \(=\)

This is bullet (C).

tj
What is \(\vec c-\vec d+\vec e\) if \(\vec c=\left\langle1,2\right\rangle\), \(\vec d=\left\langle2,3\right\rangle\), and \(\vec e=\left\langle3,4\right\rangle\)?

\(\vec c-\vec d+\vec e=\left\langle 2,3 \right\rangle\)

\[\begin{aligned} \vec c-\vec d+\vec e &=\left\langle 1,2 \right\rangle - \left\langle 2,3 \right\rangle + \left\langle 3,4 \right\rangle \\ &= \left\langle 1-2+3,2-3+4 \right\rangle \\ &= \left\langle 2,3 \right\rangle \end{aligned}\]

tj
Compute \(\vec u=\left\langle-1,2\right\rangle +2\left\langle2,-3\right\rangle-\left\langle3,4\right\rangle\)

\(\vec u=\left\langle0,-8\right\rangle\)

\[\begin{aligned} \vec u &=\left\langle -1,2 \right\rangle + \left\langle 2\cdot 2,2\cdot (-3) \right\rangle - \left\langle 3,4 \right\rangle \\ &= \left\langle -1+4-3,2-6-4 \right\rangle \\ &= \left\langle 0,-8 \right\rangle \end{aligned}\]

tj
What is \(-\vec u+3\vec v+4\vec w\) if \(\vec u=\left\langle-1,3\right\rangle\), \(\vec v=\left\langle6,-5\right\rangle\), and \(\vec w=\left\langle3,2\right\rangle\)?

\(-\vec u+3\vec v+4\vec w=\left\langle31,-10\right\rangle\)

\[\begin{aligned} -\vec u+3\vec v+4\vec w &=-\left\langle -1,3 \right\rangle + 3\left\langle 6,-5 \right\rangle + 4\left\langle 3,2 \right\rangle \\ &=\left\langle 1,-3 \right\rangle + \left\langle 18,-15 \right\rangle + \left\langle 12,8 \right\rangle \\ &=\left\langle 1+18+12,-3-15+8 \right\rangle \\ &=\left\langle 31,-10 \right\rangle \end{aligned}\]

tj
Compute \(\vec w=\left\langle5,1\right\rangle-2\left\langle0,-3\right\rangle-3\left\langle2,2\right\rangle\)

\(\vec w=\left\langle-1,1\right\rangle\)

\[\begin{aligned} \vec w &=\left\langle 5,1 \right\rangle -2\left\langle 0,-3 \right\rangle -3\left\langle 2,2 \right\rangle \\ &=\left\langle 5,1 \right\rangle + \left\langle 0,6 \right\rangle + \left\langle -6,-6 \right\rangle \\ &=\left\langle 5+0-6,1+6-6 \right\rangle \\ &=\left\langle -1,1 \right\rangle \end{aligned}\]

tj
Compute \(\vec v =-6\left\langle-2,-5\right\rangle+\left\langle0,-4\right\rangle -2\left\langle-9,3\right\rangle\).

\(\vec v=\left\langle30,20\right\rangle\)

\[\begin{aligned} \vec v &=-6\left\langle -2,-5 \right\rangle + \left\langle 0,-4 \right\rangle -2\left\langle -9,3 \right\rangle \\ &=\left\langle 12,30 \right\rangle + \left\langle 0,-4 \right\rangle + \left\langle 18,-6 \right\rangle \\ &=\left\langle 12+0+18,30-4-6 \right\rangle \\ &=\left\langle 30,20 \right\rangle \end{aligned}\]

tj
Write the vector \(\vec u=\left\langle3,6\right\rangle\) as a linear combination of the vectors \(\vec v=\left\langle3,2\right\rangle\) and \(\vec w=\left\langle3,-2\right\rangle\). In other words, find numbers \(a\) and \(b\) satisfying, \(\vec u=a\vec v+b\vec w\).

Using variables \(a\) and \(b\), write out the equation \(\vec u=a\vec v+b\vec w\) and equate \(x\) and \(y\) components.

\(\vec u=2\vec v-\vec w\)
\(\left\langle3,6\right\rangle=2\left\langle3,2\right\rangle-\left\langle3,-2\right\rangle\)

We first write out the equation \(\vec u=a\vec v+b\vec w\): \[ \left\langle3,6\right\rangle =a\left\langle3,2\right\rangle+b\left\langle3,-2\right\rangle =\left\langle3a+3b,2a-2b\right\rangle \] This becomes the two equations: \[\begin{aligned} 3a+3b&=3 \\ 2a-2b&=6 \end{aligned}\] The first equation implies \(b=1-a\). We substitute this into the other equation which becomes: \[ 2a-2+2a=6 \] which says \(a=2\) and hence \(b=-1\). We conclude: \[ \vec u=2\vec v-\vec w \]

tj

We check: \[ 2\vec v-\vec w=2\left\langle3,2\right\rangle-\left\langle3,-2\right\rangle =\left\langle3,6\right\rangle=\vec u \]
Write the vector \(\vec v=\left\langle4,10\right\rangle\) as a linear combination of the vectors \(\vec a=\left\langle2,2\right\rangle\) and \(\vec b=\left\langle1,-2\right\rangle\). In other words, find numbers \(p\) and \(q\) satisfying \(\vec v=p\vec a+q\vec b\).

\(\vec v=3\vec a-2\vec b\)
\(\left\langle4,10\right\rangle =3\left\langle2,2\right\rangle-2\left\langle1,-2\right\rangle\)

We first write out the equation \(\vec v=p\vec a+q\vec b\): \[ \left\langle4,10\right\rangle =p\left\langle2,2\right\rangle+q\left\langle1,-2\right\rangle \] This becomes the two equations: \[\begin{aligned} 4&=2p+q \\ 10&=2p-2q \end{aligned}\] The first equation implies \(q=4-2p\). We substitute this into the other equation which becomes: \[\begin{aligned} 10&=2p-8+4p=6p-8 \end{aligned}\] This equation then says \(p=3\) and hence \(q=-2\). We conclude: \[ \vec v=3\vec a-2\vec b \]

tj

We check: \[\begin{aligned} 3\vec a-2\vec b &=3\left\langle2,2\right\rangle-2\left\langle1,-2\right\rangle \\ &=\left\langle4,10\right\rangle=\vec v \end{aligned}\]
Write the vector \(\vec w=\langle5,11\rangle\) as a linear combination of the vectors \(\vec a=\langle2,3\rangle\), \(\vec b=\langle1,2\rangle\) and \(\vec c=\langle1,1\rangle\). In other words, find numbers \(p\), \(q\) and \(r\) satisfying \(\vec w=p\vec a+q\vec b+r\vec c\).

There may be more than one solution to this problem.

\(\vec w=(6-t)\vec a+t\vec b+(-7+t)\vec c\) for any choice of \(t\).
\(\langle5,11\rangle=(6-t)\langle2,3\rangle+t\langle1,2\rangle+(-7+t)\langle1,1\rangle\) for any choice of \(t\).
In particular, if \(t=1\), then
\(\vec w=5\vec a+\vec b-6\vec c\)
\(\langle5,11\rangle=5\langle2,3\rangle+1\langle1,2\rangle-6\langle1,1\rangle\)

We first write out the equation \(\vec w=p\vec a+q\vec b+r\vec c\): \[ \left\langle5,11\right\rangle =p\left\langle2,3\right\rangle+q\left\langle1,2\right\rangle +r\left\langle1,1\right\rangle \] This becomes the two equations: \[\begin{aligned} 5&=2p+q+r \\ 11&=3p+2q+r \end{aligned}\] The first equation implies \(r=5-2p-q\). We substitute this into the other equation which become: \[ 11=3p+2q+(5-2p-q)=5+p+q \quad \text{or} \quad p+q=6 \] We can pick \(q\) arbitrarily. Suppose we pick \(q=1\). Then \(p=5\) and \(r=5-2p-q=5-2(5)-1=-6\). So \[ \vec w=5\vec a+1\vec b-6\vec c \] More generally, suppose we pick \(q=t\) Then \(p=6-t\) and \(r=5-2p-q=5-2(6-t)-t=-7+t\). So \[ \vec w=(6-t)\vec a+t\vec b+(-7+t)\vec c \] for any choice of \(t\).

tj

We check: \[\begin{aligned} 5\vec a+\vec b-6\vec c &=5\langle2,3\rangle+\langle1,2\rangle-6\langle1,1\rangle \\ &=\langle10+1-6,15+2-6\rangle \\ &=\langle5,11\rangle=\vec w \end{aligned}\]
Write the vector \(\vec w=\langle6,11\rangle\) as a linear combination of the vectors \(\vec a=\langle2,3\rangle\) and \(\vec b=\langle4,6\rangle\). In other words, find numbers \(p\) and \(q\) satisfying \(\vec w=p\vec a+q\vec b\).

So there is no solution because \(\vec b=2\vec a\) and \(\vec w\) is not a multiple of \(\vec a\)!

We first write out the equation \(\vec w=p\vec a+q\vec b\): \[ \left\langle6,11\right\rangle =p\left\langle2,3\right\rangle+q\left\langle4,6\right\rangle \] This becomes the two equations: \[\begin{aligned} 6&=2p+4q \\ 11&=3p+6q \end{aligned}\] The first equation implies \(p=3-2q\). We substitute this into the other equation which become: \[ 11=9-6q+6q=9 \] This is impossible. So there is no solution.

tj

d. Dot Product

1

2

3

4

5

6

7
Compute \(\langle 3,2\rangle\cdot\langle -3,1\rangle\).

By definition, \[ \langle u_1,u_2\rangle\cdot\langle v_1,v_2\rangle =u_1v_1+u_2v_2 \]

\(-7\)

\[\begin{aligned} \left\langle 3,2\right\rangle\cdot\left\langle -3,1\right\rangle &=3\cdot(-3)+2\cdot1 \\ &=-9+2 =-7 \end{aligned}\]

tj
Find \(\vec a\cdot\vec b\), if \(\vec a=\left\langle -2,4,\right\rangle\) and \(\vec b=\left\langle 0,-1\right\rangle\).

\(\vec a\cdot\vec b=-4\)

\[\begin{aligned} \vec a\cdot\vec b &=a_1b_1+a_2b_2 \\ &=(-2)\cdot0+4\cdot(-1) \\ &=0-4 =-4 \end{aligned}\]

tj
Compute \(\vec u\cdot\vec v\), if \(|\vec u|=2\) and \(|\vec v|=3\) and the angle between the vectors is \(60^\circ\).

By the geometric formula for the dot product, \[ \vec u\cdot\vec v=|\vec u|\,|\vec v|\cos\theta \]

\(\vec u\cdot\vec v=3\)

\[\begin{aligned} \vec u\cdot\vec v &=|\vec u|\,|\vec v|\cos\theta \\ &=2\cdot3\cdot\cos60^\circ=6\cdot\dfrac{1}{2}=3 \\ \end{aligned}\]

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Compute \(\vec p\cdot\vec q\), if \(|\vec p|=7\) and \(|\vec q|=10\) and the angle between the vectors is \(\dfrac{\pi}{4}\).

\(\vec p\cdot\vec q=35\sqrt{2}\)

\[\begin{aligned} \vec p\cdot\vec q &=|\vec p|\,|\vec q|\cos\theta =7\cdot10\cdot\cos\left(\dfrac{\pi}{4}\right) \\ &=70\cdot\dfrac{\sqrt{2}}{2}=35\sqrt{2} \end{aligned}\]

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Find \(\vec w\cdot\vec w\), if \(\vec w=\left\langle 5,-6\right\rangle\).

\(\vec w\cdot\vec w=61\)

\[\begin{aligned} \vec w\cdot\vec w &={w_1}^2+{w_2}^2 =5^2+(-6)^2 \\ &=25+36=61 \end{aligned}\]

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Evaluate \(\vec u\cdot(\vec v+c\vec w)\), if \(\vec u=\left\langle 2,0\right\rangle\), \(\vec v=\left\langle 3,3\right\rangle\), \(\vec w=\left\langle -1,-2\right\rangle\) and \(c=2\).

\(\vec u\cdot(\vec v+c\vec w)=2\)

\[\begin{aligned} \vec u\cdot(\vec v+c\vec w) &=\left\langle 2,0\right\rangle\cdot(\left\langle 3,3\right\rangle +2\left\langle -1,-2\right\rangle) \\ &=\left\langle 2,0\right\rangle\cdot\left\langle 1,-1\right\rangle \\ &=2\cdot1+0\cdot(-1)=2+0=2 \end{aligned}\]

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Compute the angle between the vectors \(\vec u=\left\langle 0,2\right\rangle\) and \(\vec v=\left\langle 4,4\right\rangle\).

\(\theta=\dfrac{\pi}{4}=45^\circ\)

We compute: \[ |\vec u|=\sqrt{0+4}=2 \qquad |\vec v|=\sqrt{16+16}=\sqrt{32}=4\sqrt{2} \] and \[ \vec u\cdot\vec v=0+8=8 \] So: \[ \cos\theta =\dfrac{\vec u\cdot\vec v}{|\vec u|\,|\vec v|} =\dfrac{8}{2\cdot4\sqrt{2}}=\dfrac{1}{\sqrt{2}} \] and \[ \theta=\dfrac{\pi}{4}=45^\circ \]

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Compute the angle between the vectors \(\vec u=\left\langle 2,4\right\rangle\) and \(\vec v=\left\langle 3,1\right\rangle\).

\(\theta=\dfrac{\pi}{4}=45^\circ\)

We compute: \[ |\vec u|=\sqrt{4+16}=2\sqrt{5} \qquad |\vec v|=\sqrt{9+1}=\sqrt{10} \] and \[ \vec u\cdot\vec v=6+4=10 \] So: \[ \cos\theta =\dfrac{\vec u\cdot\vec v}{|\vec u|\,|\vec v|} =\dfrac{10}{2\sqrt{5}\cdot\sqrt{10}}=\dfrac{1}{\sqrt{2}} \] and \[ \theta=\dfrac{\pi}{4}=45^\circ \]

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Find the angle between the vectors \(\vec a=\left\langle 1+2\sqrt{3},2-\sqrt{3}\right\rangle\) and \(\vec b=\left\langle 1-2\sqrt{3},2+\sqrt{3}\right\rangle\).

\(\theta=\dfrac{2\pi}{3} =120^\circ\)

We compute: \[\begin{aligned} |\vec a|&=\sqrt{(1+4\sqrt{3}+12)+(4-4\sqrt{3}+3)} =\sqrt{20}=2\sqrt{5} \\ |\vec b|&=\sqrt{(1-4\sqrt{3}+12)+(4+4\sqrt{3}+3)} =\sqrt{20}=2\sqrt{5} \end{aligned}\] and \[ \vec a\cdot\vec b =(1+2\sqrt{3})(1-2\sqrt{3})+(2-\sqrt{3})(2+\sqrt{3}) =(1-12)+(4-3)=-10 \] So: \[ \cos\theta =\dfrac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|} =\dfrac{-10}{2\sqrt{5}\cdot2\sqrt{5}}=-\dfrac{1}{2} \] So \[ \theta=\arccos\left(-\dfrac{1}{2}\right) =\dfrac{2\pi}{3}\,\text{rad}=120^\circ \]

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Find the angle between the vectors \(\vec p=\left\langle 3,2\right\rangle\) and \(\vec q=\left\langle 4,-6\right\rangle\).

\(\theta=\dfrac{\pi}{2}=90^\circ\)

We compute: \[ |\vec p|=\sqrt{9+4}=\sqrt{13} \qquad |\vec q|=\sqrt{16+36}=\sqrt{52} \] and \[ \vec p\cdot\vec q=12-12=0 \] Since their dot product is zero, \(\vec p\) and \(\vec q\) are perpendicular and the angle between them is \(\theta=\dfrac{\pi}{2}=90^\circ\). Notice, it was not necessary to compute \(|\vec p|\) and \(|\vec q|\), but we didn't know that beforehand.

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Determine whether the vectors \(\vec p=\left\langle 3,-1\right\rangle\) and \(\vec q=\left\langle 2,6\right\rangle\) are orthogonal. If not, find the angle between them.

Yes, they are orthogonal.

To be orthogonal, the dot product must be zero. In fact: \[ \vec p\cdot\vec q=6-6=0 \] Therefore, the vectors are orthogonal to each other.

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Determine whether the vectors \(\vec u=\left\langle 5,-2\right\rangle\) and \(\vec v=\left\langle -1,-3\right\rangle\) are orthogonal. If not, find the angle between them.

No, they are not orthogonal.
The angle between them is \(\theta=\arccos\left(\dfrac{1}{\sqrt{290}}\right) \approx1.51\,\text{rad}\approx86.6^\circ\).

To be orthogonal, the dot product must be zero. In fact: \[ \vec u\cdot\vec v=-5+6=1 \] So, the vectors are not orthogonal. We compute the angle: \[ |\vec u|=\sqrt{25+4}=\sqrt{29} \qquad |\vec v|=\sqrt{1+9}=\sqrt{10} \] So: \[ \cos\theta =\dfrac{\vec u\cdot\vec v}{|\vec u|\,|\vec v|} =\dfrac{1}{\sqrt{29}\cdot\sqrt{10}}=\dfrac{1}{\sqrt{290}} \] and: \[ \theta=\arccos\left(\dfrac{1}{\sqrt{290}}\right) \approx1.51\,\text{rad}\approx86.6^\circ \] They are almost perpendicular but not quite!

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Find each of the following if \(\vec u=\left\langle 4,3\right\rangle\) and \(\vec v=\left\langle -5,12\right\rangle\).
1. \(\text{proj}_{\vec u}\vec v\), the vector projection of \(\vec v\) onto \(\vec u\).
  
  \(\text{proj}_{\vec u}\vec v=\left\langle \dfrac{64}{25},\dfrac{48}{25}\right\rangle\)
  
  \[\begin{aligned} \text{proj}_{\vec u}\vec v &=\dfrac{\vec u\cdot\vec v}{|u|^2}\vec u =\dfrac{-20+36}{16+9}\left\langle 4,3\right\rangle \\ &=\dfrac{16}{25}\left\langle 4,3\right\rangle =\left\langle \dfrac{64}{25},\dfrac{48}{25}\right\rangle \end{aligned}\]
  
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2. \(\text{proj}_{\perp\vec u}\vec v\), the vector projection of \(\vec v\) perpendicular to \(\vec u\).
  
  \(\text{proj}_{\perp\vec u}\vec v=\left\langle -\dfrac{189}{25},\dfrac{252}{25}\right\rangle\)
  
  \[\begin{aligned} \text{proj}_{\perp\vec u}\vec v &=\vec v-\text{proj}_{\vec u}\vec v \\ &=\left\langle -5,12\right\rangle-\left\langle \dfrac{64}{25},\dfrac{48}{25}\right\rangle =\left\langle -\dfrac{189}{25},\dfrac{252}{25}\right\rangle \end{aligned}\]
  
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  We check \(\text{proj}_{\perp\vec u}\vec v\) is perpendicular to \(\vec u\): \[ \vec u\cdot \text{proj}_{\perp\vec u}\vec v =\left\langle 4,3\right\rangle\cdot\left\langle -\dfrac{189}{25},\dfrac{252}{25}\right\rangle =\dfrac{756}{25}-\dfrac{756}{25}=0 \]
3. \(\text{comp}_{\vec u}\vec v\), the scalar projection of \(\vec v\) onto \(\vec u\).
  
  \(\text{comp}_{\vec u}\vec v=\dfrac{16}{5}\)
  
  \[ \text{comp}_{\vec u}\vec v =\dfrac{\vec u\cdot\vec v}{|u|} =\dfrac{-20+36}{\sqrt{16+9}} =\dfrac{16}{5} \]
  
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  We check \(\text{comp}_{\vec u}\vec v=\pm|\text{proj}_{\vec u}\vec v|\): \[ |\text{proj}_{\vec u}\vec v|=\sqrt{\dfrac{4096}{625}+\dfrac{2304}{625}}=\dfrac{16}{5} \]
4. \(\text{comp}_{\perp\vec u}\vec v\), the scalar projection of \(\vec v\) perpendicular to \(\vec u\).
  
  \(\text{comp}_{\perp\vec u}\vec v=\dfrac{63}{5}\)
  
  \[ \text{comp}_{\perp\vec u}\vec v =\sqrt{|\vec v|^2-\left(\text{comp}_{\vec u}\vec v\right)^2} =\sqrt{169-\dfrac{256}{25}} =\sqrt{\dfrac{3969}{25}} =\dfrac{63}{5} \]
  
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  We check \(\text{comp}_{\perp\vec u}\vec v=|\text{proj}_{\perp\vec u}\vec v|\): \[ |\text{proj}_{\perp\vec u}\vec v|=\sqrt{\dfrac{35721}{625}+\dfrac{63504}{625}}=\dfrac{63}{5} \]
5. \(\text{proj}_{\vec v}\vec u\), the vector projection of \(\vec u\) onto \(\vec v\).
  
  \(\text{proj}_{\vec v}\vec u =\left\langle -\dfrac{80}{169},\dfrac{192}{169}\right\rangle\)
  
  \[\begin{aligned} \text{proj}_{\vec v}\vec u &=\dfrac{\vec v\cdot\vec u}{|v|^2}\vec v =\dfrac{-20+36}{25+144}\left\langle -5,12\right\rangle \\ &=\dfrac{16}{169}\left\langle -5,12\right\rangle =\left\langle -\dfrac{80}{169},\dfrac{192}{169}\right\rangle \end{aligned}\]
  
  tj
  
  Notice that \(\text{proj}_{\vec v}\vec u\) is very different from \(\text{proj}_{\vec u}\vec v\).
Find each of the following if \(\vec a=\left\langle 1,-1\right\rangle\) and \(\vec b=\left\langle -4,0\right\rangle\).
1. \(\text{proj}_{\vec a}\vec b\), the vector projection of \(\vec b\) onto \(\vec a\).
  
  \(\text{proj}_{\vec a}\vec b=\left\langle -2,2\right\rangle\)
  
  \[\begin{aligned} \text{proj}_{\vec a}\vec b &=\dfrac{\vec a\cdot\vec b}{|a|^2}\vec a =\dfrac{-4+0}{1+1}\left\langle 1,-1\right\rangle \\ &=-2\left\langle 1,-1\right\rangle =\left\langle -2,2\right\rangle \end{aligned}\]
  
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2. \(\text{proj}_{\perp\vec a}\vec b\), the vector projection of \(\vec b\) perpendicular to \(\vec a\).
  
  \(\text{proj}_{\perp\vec a}\vec b=\left\langle -2,-2\right\rangle\)
  
  \[\begin{aligned} \text{proj}_{\perp\vec a}\vec b &=\vec b-\text{proj}_{\vec a}\vec b \\ &=\left\langle -4,0\right\rangle-\left\langle -2,2\right\rangle =\left\langle -2,-2\right\rangle \end{aligned}\]
  
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  We check \(\text{proj}_{\perp\vec a}\vec b\) is perpendicular to \(\vec a\): \[ \vec a\cdot \text{proj}_{\perp\vec a}\vec b =\left\langle 1,-1\right\rangle\cdot\left\langle -2,-2\right\rangle =-2+2=0 \]
3. \(\text{comp}_{\vec a}\vec b\), the scalar projection of \(\vec b\) onto \(\vec a\).
  
  \(\text{comp}_{\vec a}\vec b=-2\sqrt{2}\)
  
  \[ \text{comp}_{\vec a}\vec b =\dfrac{\vec a\cdot\vec b}{|a|} =\dfrac{-4+0}{\sqrt{1+1}} =-2\sqrt{2} \]
  
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  We check \(\text{comp}_{\vec a}\vec b=\pm|\text{proj}_{\vec a}\vec b|\): \[ |\text{proj}_{\vec a}\vec b|=\sqrt{4+4}=2\sqrt{2} \] The angle is obtuse because \(\vec a\cdot\vec b=-4 \lt 0\). So: \[ \text{comp}_{\vec a}\vec b=-2\sqrt{2} \]
4. \(\text{comp}_{\perp\vec a}\vec b\), the scalar projection of \(\vec b\) perpendicular to \(\vec a\).
  
  \(\text{comp}_{\perp\vec a}\vec b=2\sqrt{2}\)
  
  \[ \text{comp}_{\perp\vec a}\vec b =\sqrt{|\vec b|^2-\left(\text{comp}_{\vec a}\vec b\right)^2} =\sqrt{16-8} =2\sqrt{2} \]
  
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  We check \(\text{comp}_{\perp\vec a}\vec b=|\text{proj}_{\perp\vec a}\vec b|\): \[ |\text{proj}_{\perp\vec a}\vec b|=\sqrt{4+4}=2\sqrt{2} \]
5. \(\text{proj}_{\vec b}\vec a\), the vector projection of \(\vec a\) onto \(\vec b\).
  
  \(\text{proj}_{\vec b}\vec a=\left\langle 1,0\right\rangle\)
  
  \[\begin{aligned} \text{proj}_{\vec b}\vec a &=\dfrac{\vec b\cdot\vec a}{|b|^2}\vec b =\dfrac{-4+0}{16}\left\langle -4,0\right\rangle \\ &=\dfrac{-1}{4}\left\langle -4,0\right\rangle =\left\langle 1,0\right\rangle \end{aligned}\]
  
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  Notice that \(\text{proj}_{\vec b}\vec a=\left\langle 1,0\right\rangle\) is very different from \(\text{proj}_{\vec a}\vec b=\left\langle -2,2\right\rangle\).
Find \(\text{proj}_{\vec p}\vec q\), the vector projection of \(\vec q=\left\langle -2,1\right\rangle\) onto \(\vec p=\left\langle 1,5\right\rangle\).

\(\text{proj}_{\vec p}\vec q =\left\langle\dfrac{3}{26},\dfrac{15}{26}\right\rangle\)

\[\begin{aligned} \text{proj}_{\vec p}\vec q &=\dfrac{\vec p\cdot\vec q}{|p|^2}\vec p =\dfrac{-2+5}{1+25}\left\langle 1,5\right\rangle \\ &=\dfrac{3}{26}\left\langle 1,5\right\rangle =\left\langle\dfrac{3}{26},\dfrac{15}{26}\right\rangle \end{aligned}\]

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Compute \(\text{proj}_{\perp \vec u}\vec v\), the vector projection of \(\vec v=\left\langle -1,2\right\rangle\) perpendicular to \(\vec u=\left\langle 4,-3\right\rangle\).

\(\text{proj}_{\perp \vec u}\vec v =\left\langle \dfrac{3}{5},\dfrac{4}{5}\right\rangle\)

We first need to find \(\text{proj}_{\vec u}\vec v\): \[\begin{aligned} \text{proj}_{\vec u}\vec v &=\dfrac{\vec u\cdot\vec v}{|u|^2}\vec u =\dfrac{-4-6}{16+9}\left\langle 4,-3\right\rangle \\ &=\dfrac{-2}{5}\left\langle 4,-3\right\rangle =\left\langle\dfrac{-8}{5},\dfrac{6}{5}\right\rangle \end{aligned}\] Now we can find \(\text{proj}_{\perp \vec u}\vec v\): \[\begin{aligned} \text{proj}_{\perp\vec u}\vec v &=\vec v-\text{proj}_{\vec u}\vec v \\ &=\left\langle -1,2\right\rangle -\left\langle\dfrac{-8}{5},\dfrac{6}{5}\right\rangle \\ &=\left\langle \dfrac{3}{5},\dfrac{4}{5}\right\rangle \end{aligned}\]

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We check \(\text{proj}_{\perp\vec u}\vec v\) is perpendicular to \(\vec u\): \[ \vec u\cdot \text{proj}_{\perp\vec u}\vec v =\left\langle 4,-3\right\rangle\cdot \left\langle\dfrac{3}{5},\dfrac{4}{5}\right\rangle =\dfrac{12-12}{5}=0 \]
Compute \(\text{comp}_{\vec b}\vec a\), the scalar projection of \(\vec a=\left\langle 2,0\right\rangle\) onto \(\vec b=\left\langle 2,2\right\rangle\). Is the angle between \(\vec a\) and \(\vec b\) acute or obtuse?

\(\text{comp}_{\vec b}\vec a=\sqrt{2}\)
The angle is acute.

\[\begin{aligned} \text{comp}_{\vec b}\vec a &=\dfrac{\vec a\cdot\vec b}{|b|} =\dfrac{4+0}{\sqrt{4+4}} \\ &=\dfrac{4}{2\sqrt{2}} =\sqrt{2} \end{aligned}\] Since this is positive, the angle is acute.

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Find the scalar and vector projections of \(\vec a=\left\langle 2,0,5\right\rangle\) onto \(\vec b=\left\langle 3,5,2\right\rangle\).

\(\text{comp}_{\vec b}\vec a=\dfrac{16}{\sqrt{38}}\)

\(\text{proj}_{\vec b}\vec a=\left\langle \dfrac{24}{19},\dfrac{40}{19},\dfrac{16}{19}\right\rangle\)

Since \(\vec a=\left\langle 2,0,5\right\rangle\) and \(\vec b=\left\langle 3,5,2\right\rangle\), we have: \[ \vec a\cdot\vec b=6+0+10=16 \qquad |\vec b|=\sqrt{9+25+4}=\sqrt{38} \] \[ \text{comp}_{\vec b}\vec a =\dfrac{\vec a\cdot\vec b}{|\vec b|} =\dfrac{16}{\sqrt{38}} \] \[ \text{proj}_{\vec b}\vec a =\dfrac{\vec a\cdot\vec b}{|b|^2}\vec b =\dfrac{16}{38}\left\langle 3,5,2\right\rangle =\left\langle \dfrac{24}{19},\dfrac{40}{19},\dfrac{16}{19}\right\rangle \]

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A weight is pulled up a vertical \(30\) m track by a rope which exerts a \(20\) N force at \(45^\circ\) from vertical. Find the work done.

\(W=300\sqrt{2}\,\text{J}\)

The work is \[\begin{aligned} W&=\vec F\cdot\vec D =|\vec F|\,|\vec D|\cos\theta \\ &=20\,\text{N}\cdot30\,\text{m}\cdot\cos45^\circ =300\sqrt{2}\,\text{J} \end{aligned}\]

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A barge is pulled from \((0,0)\) to \((4,2)\) by two tugboats exerting the forces: \[ \vec F_1=\left\langle 1,6\right\rangle \qquad \text{and} \qquad \vec F_2=\left\langle 5,-1\right\rangle \] Here, distances are given in kilometers and forces are given in kiloNewtons. The positive \(x\)-axis points East and the positive \(y\)-axis points North. Find the total work done by the tugboats.

\(W=34\) megaJoules

The total force is: \[ \vec F=\vec F_1+\vec F_2 =\left\langle 1,6\right\rangle+(5,-1) =\left\langle 6,5\right\rangle \] The displacement is \(\vec D=\left\langle 4,2\right\rangle\). So the work is: \[ W=\vec F\cdot\vec D =\left\langle 6,5\right\rangle\cdot\left\langle 4,2\right\rangle =34\,\text{megaJoules} \]

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A boy slides a book \(2\) meters across a table by pulling it with a force of \(10\) N from an angle of \(45^\circ\) above horizontal. Find the work done by the boy on the book.

\[ W=10\sqrt{2}\,\text{J}\approx14.14\,\text{J} \]

Using the geometric formula for the dot product, the work is: \[\begin{aligned} W&=\vec F\cdot\vec D =|\vec F|\,|\vec D|\cos\theta \\ &=10\,\text{N}\cdot2\,\text{m}\cdot\cos45^\circ \\ &=\dfrac{20}{\sqrt{2}}\,\text{N-m} =10\sqrt{2}\,\text{J} \\ &\approx14.14\,\text{J} \end{aligned}\]

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e. Parametric Curves

Find an equation of the line through the point \(P=(3,2)\) with direction vector \(\vec v=\langle 1,5\rangle\),
1. in parametric form.
  
  The equation of a line in parametric form is \(X=P+t\vec v\).
  
  \((x,y)=(3+t,2+5t)\)
  
  The general parametric equation of a line is: \[ X=P+t \vec v \] Here \(P=(3,2)\) and \(\vec v=\langle 1,52\rangle\). So: \[ (x,y)=(3+t,2+5t) \]
  
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2. in standard form.
  
  Solve the two parts of the parametric equations for \(t\) and equate them. Then rearrange the equation to have \(x\) and \(y\) on one side and a constant on the other.
  
  \(5x-y=13\)
  
  The parametric equations are \(x=3+t\) and \(y=2+5t\). To get the standard form, we solve each equation for \(t\) and equate: \[ t=x-3=\dfrac{1}{5}y-\dfrac{2}{5} \] We now clear the denominator, take the variables to one side and the constants to the other side. \[\begin{aligned} 5x-15&=y-2 \\ 5x-y&=13 \\ \end{aligned}\] This is the standard form.
  
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Find an equation of the line through the point \(P=(-2,4)\) with tangent vector \(\vec v=\langle -1,2\rangle\),
1. in parametric form.
  
  \(\vec r(t)=\left\langle -2-t,4+2t\right\rangle\)
  
  The general parametric equation of a line is: \[ X=P+t \vec v \] Here \(P=(-2,4)\) and \(\vec v=\langle -1,2\rangle\). So: \[ (x,y)=(-2-t,4+2t) \]
  
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2. in standard form
  
  \(2x+y=0\)
  
  The parametric equations are \(x=-2-t\) and \(y=4+2t\). To get the standard form, we solve each equation for \(t\) and equate: \[ t=-x-2=\dfrac{1}{2}y-2 \] We now clear the denominator, take the variables to one side and the constants to the other side. \[\begin{aligned} 2x+4&=-y+4 \\ 2x+y&=0 \\ \end{aligned}\] This is the standard form.
  
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A curve is given parametrically as \(x=2+t\) and \(y=3-2t^2\). Write the curve as a function \(y=f(x)\) and describe its shape.

Solve the \(x\) equation for \(t\) and plug it into the \(y\) equation.

\( y=3-2(x-2)^2 \)
This is a parabola with vertex at \((2,3)\) opening down.

We solve the \(x\) equation for \(t\): \[ t=x-2 \] and plug into the \(y\) equation: \[ y=3-2(x-2)^2 \] This is a parabola with vertex at \((2,3)\) opening down.

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A curve is given parametrically as \(x=-1+2\cos\theta\) and \(y=2+3\sin\theta\). Write the curve as an equation in \(x\) and \(y\) and describe its shape.

Solve the \(x\) equation for \(\cos\theta\). Solve the \(y\) equation for \(\sin\theta\). Then use the Pythagorean identity.

\( \left(\dfrac{x+1}{2}\right)^2+\left(\dfrac{y-2}{3}\right)^2=1 \)
This is an ellipse centred at \((-1,2)\) with \(x\)-radius \(2\) and \(y\)-radius \(3\).

We solve the \(x\) equation for \(\cos\theta\) and the \(y\) equation for \(\sin\theta\): \[ \cos\theta=\dfrac{x+1}{2} \qquad \sin\theta=\dfrac{y-2}{3} \] We substitute these into the Pythagorean identity \(\cos^2\theta+\sin^2\theta=1\) to get \[ \left(\dfrac{x+1}{2}\right)^2+\left(\dfrac{y-2}{3}\right)^2=1 \] This is an ellipse centred at \((-1,2)\) with \(x\)-radius \(2\) and \(y\)-radius \(3\).

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A curve is given parametrically as \((x,y)=(t^2,t^3)\). This curve has two branches, one with \(y \ge 0\) and one with \(y \lt 0\). Write each branch as a function with \(y\) as a function of \(x\). Describe its shape and plot it.

Solve for \(t\) as a function of \(x\) and plug into \(y\).

\(\displaystyle y=\pm x^{3/2}\)

Since \(x=t^2\), we find that \(t=\pm\sqrt{x}\).
So \(y=t^3=(\pm\sqrt{x})^3=\pm x^{3/2}\).
The upper branch is when \(y\ge0\). So \(y=x^{3/2}\).
The lower branch is when \(y\lt0\). So \(y=-x^{3/2}\). The two branches are plotted.

The curve \(y=x^{3/2}\) looks a lot like \(y=x^2\), but does not bend up as fast.

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Consider the position vector \(\vec{r}(t)=\langle t,t^2-1\rangle\).
1. Find the position at \(t=2\)
  
  Just substitute \(t=2\) into \(\vec{r}(t)=\langle t,t^2-1\rangle\).
  
  \(\vec{r}(2)=\langle 2,3\rangle\)
2. Plot the parametric curve for \(-3 \le t \le 3\).
  
  Express \(y\) as a function of \(x\) by setting \(t=x\).
  
  Since \(x=t\), we set \(t=x\) to get \(y=x^2-1\), which is a parabola with vertex at \((0,-1)\) opening upward. Its graph is:
  
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3. Add the plot of the position vector at \(t=2\).
  
  The position vector is the arrow from the origin to the point \(\langle 2,3\rangle\).
  
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Consider the position vector \(\vec{r}(t) =\left\langle 3\cos\left(t\dfrac{\pi}{4}\right), 3\sin\left(t\dfrac{\pi}{4}\right)\right\rangle\)
1. Find the position at \(t=3\).
  
  Just substitute \(t=3\) into \(\left\langle 3\cos\left(t\dfrac{\pi}{4}\right), 3\sin\left(t\dfrac{\pi}{4}\right)\right\rangle\).
  
  \(\begin{aligned} \vec{r}(3)&=\left\langle 3\cos\left(\dfrac{3\pi}{4}\right), 3\sin\left(\dfrac{3\pi}{4}\right)\right\rangle \\ &=\left\langle -\,\dfrac{3}{\sqrt{2}},\dfrac{3}{\sqrt{2}}\right\rangle \end{aligned}\)
  
  We subsitute \(t=3\) into \(\left\langle 3\cos\left(t\dfrac{\pi}{4}\right), 3\sin\left(t\dfrac{\pi}{4}\right)\right\rangle\) to get \[\begin{aligned} \vec{r}(3)&=\left\langle 3\cos\left(\dfrac{3\pi}{4}\right), 3\sin\left(\dfrac{3\pi}{4}\right)\right\rangle \\ &=\left\langle -\,\dfrac{3}{\sqrt{2}},\dfrac{3}{\sqrt{2}}\right\rangle \end{aligned}\] This is in the second quadrant.
  
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2. Plot the parametric curve for \(0 \le t \le 5\).
  
  The parametric curve is the part of a circle of radius \(3\) with angles between \(0\) and \(\dfrac{5\pi}{4}\).
  
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3. Add to the plot the position vector at \(t=3\)
  
  The position vector is the arrow from the origin to the point \(\left\langle 3\cos\left(\dfrac{3\pi}{4}\right), 3\sin\left(\dfrac{3\pi}{4}\right)\right\rangle\) which is at the angle \(\dfrac{3\pi}{4}\).
  
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PY: All Checked

Review Exercises

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2. Vectors

Exercises

a. Definition, Magnitude and Direction

b. Vector Addition

c. Scalar Multiplication

d. Dot Product

e. Parametric Curves

Review Exercises

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