Quiz

Hint

\(
12 \vec{u}+\vec{v}=12(8 \vec{i}-\vec{j}+3 \vec{k})+(7 \vec{j}-4 \vec{k})=(96 \vec{i}-12 \vec{j}+36 \vec{k})+(7 \vec{j}-4 \vec{k})=96 \vec{i}-5 \vec{j}+32 \vec{k}
\)

Hint

Of course, the first step here really should be to check and see if we are lucky enough to actually have a unit vector already. It’s unlikely we do have a unit vector but you never know until you check!
\(
\mid\vec{q}\mid=\sqrt{(1)^2+(3)^2+(9)^2}=\sqrt{91}
\)
Okay, as we pretty much had already guessed, this isn’t a unit vector (its magnitude isn’t one!) but we can use this to help find the answer.
Hide Step 27
Recall that all we need to do to turn any vector into a unit vector is divide the vector by its magnitude. Doing that for this vector gives,
\(
\vec{u}=\frac{\vec{q}}{\mid\vec{q}\mid}=\frac{1}{\sqrt{91}}(\vec{i}+3 \vec{j}+9 \vec{k})=\frac{1}{\sqrt{91}} \vec{i}+\frac{3}{\sqrt{91}} \vec{j}+\frac{9}{\sqrt{91}} \vec{k}
\)
As a quick check, not really required of course, we can compute a quick magnitude to verify that we do in fact have a unit vector.
\(
\mid\vec{u}\mid=\sqrt{\left(\frac{1}{\sqrt{91}}\right)^2+\left(\frac{3}{\sqrt{91}}\right)^2+\left(\frac{9}{\sqrt{91}}\right)^2}=\sqrt{\frac{91}{91}}=1
\)
So, we do have a unit vector

Hint

First, let’s check to see what the magnitude of this vector is.
\(
\mid\vec{c}\mid=\sqrt{(-1)^2+(4)^2}=\sqrt{17}
\)
The unit vector is,
\(
\vec{u}=\frac{\vec{c}}{\mid\vec{c}\mid}=\frac{1}{\sqrt{17}}\langle-1,4\rangle=\left\langle-\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right\rangle
\)

We know that scalar multiplication can change the magnitude of a vector. We’ve got a vector with magnitude of one that points in the correct direction. To convert this into a vector with magnitude of 10 all we need to do is multiply this new unit vector by 10 to get,
\(
\vec{v}=10 \vec{u}=10\left\langle-\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right\rangle=\left\langle-\frac{10}{\sqrt{17}}, \frac{40}{\sqrt{17}}\right\rangle
\)
Now, let’s verify that this does what we want it to do with a quick magnitude computation.
\(
\mid\vec{v}\mid=\sqrt{\left(-\frac{10}{\sqrt{17}}\right)^2+\left(\frac{40}{\sqrt{17}}\right)^2}=\sqrt{\frac{1700}{17}}=\sqrt{100}=10
\)
So, we do have a vector with magnitude 10 as predicted.

Hint

Recall that two vectors are parallel if they are scalar multiples of each other. In other words, these two vectors will be scalar multiples if we can find a number \(k\) such that,
\(
\vec{a}=k \vec{b}
\)

Let’s just take a look at the first component from each vector. It is obvious that \(6=2(3)\). So, to convert the first components we’d need to multiply \(\vec{a}\) by 2 . However, if we did that we’d get,
\(
2 \vec{a}=\langle 6,-10,2\rangle \neq \vec{b}
\)
This is clearly not \(\vec{b}\). The first component is correct and the third component is correct but the second isn’t correct. Therefore, there is no single number, \(k\), that we can use to convert \(\vec{a}\) into \(\vec{b}\) through scalar multiplication.
This in turn means that \(\vec{a}\) and \(\vec{b}\) cannot possibly be parallel.

Hint

\(
\vec{a} \cdot \vec{b}=(0)(2)+(4)(-1)+(-2)(7)=-18
\)

Hint

\(
\vec{a} \cdot \vec{b}=(5)\left(\frac{3}{7}\right) \cos \left(\frac{\pi}{12}\right)=2.0698
\)

Hint

All we really need to do is rewrite the formula from the geometric interpretation of the dot product as,
\(
\cos \theta=\frac{\vec{v} \cdot \vec{w}}{\mid\vec{v}\mid\mid\vec{w}\mid}
\)
This will allow us to quickly determine the angle between the two vectors.
We’ll first need the following quantities (we’ll leave it to you to verify the arithmetic involved in these computations….).
\(
\vec{v} \cdot \vec{w}=2 \quad\mid\vec{v}\mid=\sqrt{30} \quad\mid\vec{w}\mid=\sqrt{21}
\)
The angle between the vectors is then,
\(
\cos \theta=\frac{2}{\sqrt{30} \sqrt{21}}=0.07968 \quad \Rightarrow \quad \theta=\cos ^{-1}(0.07968)=1.49103 \text { radians }
\)

Hint

Based on a quick inspection of the components we can see that the first and second components of the two vectors have opposite signs and the third doesn’t. This means there is no possible way for these two vectors to be scalar multiples since there is no number that will change the sign on the first two components and leave the sign of the third component unchanged.
Therefore, we can quickly see that the two vectors are not parallel.
Let’s do a quick dot product on the two vectors next.
\(
\vec{q} \cdot \vec{p}=0
\)
Okay, the dot product is zero and we know from the notes that this in turn means that the two vectors must be orthogonal.
On a side note an alternate method for working this problem is to find the angle between the two vectors and using that to determine the answer.
Depending on which method you find easiest either will get you the correct answer.

Hint

Based on a quick inspection of the components we can see that the first components of the vectors have the same sign and the second have opposite signs. This means there is no possible way for these two vectors to be scalar multiples since there is no number that will change the sign on the second components and leave the sign of the first component unchanged.
Therefore, we can quickly see that the two vectors are not parallel.

Let’s do a quick dot product on the two vectors next.
\(
\vec{a} \cdot \vec{b}=2
\)
Okay, the dot product is not zero and we know from the notes that this in turn means that the two vectors are not orthogonal.
The answer to the problem is therefore the two vectors are neither parallel or orthogonal.

Hint

Based on a quick inspection is seems (hopefully) fairly clear that we have,
\(
\vec{v}=-3 \vec{w}
\)
Therefore, the two vectors are parallel.

Hint

All we really need to do here is use the formula from the notes. That will need the following quantities.
\(
\vec{a} \cdot \vec{b}=-6 \quad\mid\vec{a}\mid^2=68
\)
The projection is then,
\(
\operatorname{proj}_\vec{a} \vec{b}=\frac{-6}{68}\langle-8,2\rangle=\left\langle\frac{12}{17},-\frac{3}{17}\right\rangle
\)

Hint

All we really need to do here is use the formula from the notes. That will need the following quantities.
\(
\vec{u} \cdot \vec{w}=-25 \quad\mid\vec{w}\mid^2=65
\)
The projection is then,
\(
\operatorname{proj}_{\vec{w}} \vec{u}=\frac{-25}{65}(-2 \vec{i}+5 \vec{j}-6 \vec{k})=\frac{10}{13} \vec{i}-\frac{25}{13} \vec{j}+\frac{30}{13} \vec{k}
\)

Hint

All we really need to do here is use the formulas from the notes. That will need the following quantity.
\(
\mid\vec{r}\mid=\sqrt{\frac{161}{16}}=\frac{\sqrt{161}}{4}
\)
The direction cosines and angles are then,
\(
\begin{aligned}
& \cos \alpha=\frac{-3}{\sqrt{161} / 4}=-\frac{12}{\sqrt{161}} \quad \Rightarrow \quad \alpha=\cos ^{-1}\left(-\frac{12}{\sqrt{161}}\right)=2.8106 \text { radians } \\
& \cos \beta=\frac{-1 / 4}{\sqrt{161} / 4}=-\frac{1}{\sqrt{161}} \quad \Rightarrow \quad \beta=\cos ^{-1}\left(-\frac{1}{\sqrt{161}}\right)=1.6497 \text { radians } \\
& \cos \gamma=\frac{1}{\sqrt{161} / 4}=\frac{4}{\sqrt{161}} \quad \Rightarrow \quad \gamma=\cos ^{-1}\left(\frac{4}{\sqrt{161}}\right)=1.2501 \text { radians } \\
&
\end{aligned}
\)

Hint

\(
\vec{w} \times \vec{v}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
0 & 4 & -2 \\
3 & -1 & 5
\end{array}\right| \quad \begin{array}{cc}
\vec{i} & \vec{j} \\
0 & 4 \\
3 & -1
\end{array}
\)

\(
=20 \vec{i}-6 \vec{j}+0 \vec{k}-0 \vec{j}-2 \vec{i}-12 \vec{k}=18 \vec{i}-6 \vec{j}-12 \vec{k}
\)

Hint

We first need two vectors that are both parallel to the plane. Using the points that we are given (all in the plane) we can quickly get quite a few vectors that are parallel to the plane. We’ll use the following two vectors.
\(
\overrightarrow{P Q}=\langle 1,-2,0\rangle \quad \overrightarrow{P R}=\langle 2,3,-2\rangle
\)

Now we know that the cross product of any two vectors will be orthogonal to the two original vectors. Since the two vectors from Step 1 are parallel to the plane (they actually lie in the plane in this case!) we know that the cross product must then also be orthogonal, or normal, to the plane.
So, using the “trick” we used in the notes the cross product is,

\(
\overrightarrow{P Q} \times \overrightarrow{P R}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -2 & 0 \\
2 & 3 & -2
\end{array}\right| \quad \begin{array}{cc}
\vec{i} & \vec{j} \\
1 & -2 \\
2 & 3
\end{array}
\)

\(
=4 \vec{i}+0 \vec{j}+3 \vec{k}-(-2 \vec{j})-0 \vec{i}-(-4 \vec{k})=4 \vec{i}+2 \vec{j}+7 \vec{k}
\)

Hint

\(
\begin{aligned}
\vec{u} \cdot(\vec{v} \times \vec{w}) & =\left|\begin{array}{ccc}
1 & 2 & -4 \\
-5 & 3 & -7 \\
-1 & 4 & 2
\end{array}\right| \begin{array}{cl}
1 & 2 \\
-5 & 3 \\
-1 & 4
\end{array} \\
& =6+14+80-(-20)-(-28)-12=136
\end{aligned}
\)
Okay, since this is not zero we know that they are not in the same plane.

Hint

Take a deep look to the picture, we have to find \(\alpha\) first. The method is below.
\(
\begin{aligned}
\frac{A}{\sin \beta} & =\frac{B}{\sin \alpha} \\
\frac{40}{\sin 66.53^{\circ}} & =\frac{30}{\sin \alpha} \\
\sin \alpha & =0.687 \\
\alpha & =\sin ^{-1} 0.687 \\
& \approx 43.47^{\circ}
\end{aligned}
\)
Once we know \(\alpha\), we can calculate \(\theta\).
\(
\begin{aligned}
\theta & =\alpha+\beta \\
& =43.47^{\circ}+66.53^{\circ} \\
& =110^{\circ}
\end{aligned}
\)
Since we already know \(\theta\) we can finally calculate the magnitude of \(\vec{R}\).
\(
\begin{aligned}
R & =\sqrt{A^2+B^2+2 A B \cos \theta} \\
& =\sqrt{40^2+30^2+2(40)(30) \cos 110^{\circ}} \\
& \approx 40.98 \text { units }
\end{aligned}
\)
Based on the above calculation, we get \(\alpha=43.47^{\circ}\) and \(|R|=40.98\) units(satuan)

Hint

First, we operate using \(\hat{i}\) and \(\hat{j}\) notation.
\(
\begin{aligned}
\vec{R} & =\vec{A}-\vec{B}+\vec{C} \\
& =(4 \hat{i}+5 \hat{j})-(9 \hat{i}-7 \hat{j})+(-3 \hat{i}+2 \hat{j}) \\
& =(4-9-3) \hat{i}+(5+7+2) \hat{j} \\
& =-8 \hat{i}+14 \hat{j}
\end{aligned}
\)
Then we can calculate the resultant magnitude as follows.
\(
\begin{aligned}
R & =\sqrt{-8^2+14^2} \\
& =16.12 \text { units }
\end{aligned}
\)
Based on our calculation, the resultant length of the \(\vec{A}-\vec{B}+\vec{C}\) is \(16.12\) units.

Hint

Take a more deep look at \(\vec{A}\) and \(\vec{B}\). The y-components of the two vectors have the same magnitude but opposite directions, so they cancel each other out. So the resultant of \(\vec{A}+\vec{B}\) is simply an addition of the x-component of the two vectors and points to the left.
\(|A+B|\) is the resultant magnitude of the \(\vec{A}+\vec{B}\) which we can calculate as follows.
\(
\begin{aligned}
|A+B| & =A_x+B_x \\
& =A \sin 20+B \sin 20 \\
& =30 \sin 20+30 \sin 20 \\
& \approx 20.521 \text { units }
\end{aligned}
\)
Next, we can calculate the magnitude of the resultant vector.
\(
\begin{aligned}
R & =\sqrt{|A+B|^2+C^2} \\
& =\sqrt{20.521^2+30^2} \\
& =36.35 \text { units }
\end{aligned}
\)
Based on the calculation we have done, the resultant magnitude of the addition of the three vectors is \(\mathbf{3 6 . 3 5}\) units.

Hint

We can calculate the resultant force first.
\(
\begin{aligned}
& F_R=\sqrt{{F_1}^2+{F_2}^2} \\
& =\sqrt{20^2+5^2} \\
& =\sqrt{425} \\
& =5 \sqrt{17} \mathrm{~N} \\
&
\end{aligned}
\)
Then using Newton’s law formula \(F_R=m a\) we can find the acceleration of the object.
\(
\begin{aligned}
a & =\frac{F}{m} \\
& =\frac{5 \sqrt{17}}{5} \\
& =\sqrt{17} \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
The direction of the object’s acceleration is the same as the direction of the resultant force and can be calculated as follows.
\(
\begin{aligned}
\theta & =\tan ^{-1} \frac{\text { opposite }}{\text { adjacent }} \\
& =\tan ^{-1} \frac{5}{20} \\
& \approx 14^{\circ}
\end{aligned}
\)
So the object’s acceleration is \(\sqrt{17} \mathrm{~m} / \mathrm{s}^2\) with direction about \(-14^{\circ}\) from the negative \(y\)-axis.

Hint

An object is said to be at rest if it satisfies two conditions.
The first condition is that the force to the right is equal to the force to the left. Look at the second slide of the image above. In this case, the force to the right is \(F_1[latex] while the force to the left is [latex]F_{2 x}[latex]. So to fulfill the first condition we must write as follows.
[latex]
F_{2 x}=F_1=20 \text { Newton }
\)
Since we already know \(F_{2 x}\), we can find \(F_2\). The method is below.
\(
\begin{aligned}
F_{2 x} & =F_2 \cos 30^{\circ} \\
F_2 & =\frac{F_{2 x}}{\cos 30^{\circ}} \\
& =\frac{20}{\cos 30^{\circ}} \\
& =\frac{40}{3} \sqrt{3} \text { Newton }
\end{aligned}
\)
The second condition is the upward force must be equal to the downward force. Slide the image above and see. In this case, there are 2 upward forces, namely \(F_{2 y}\) and N(normal force). While there is one downward force, namely \(w\) (weight). To fulfill the condition that the upward force equals the downward force, we must write the following mathematical equation.
\(
N+F_{2 y}=W
\)
We can operate equation above further
\(
\begin{aligned}
N+F_{2 y} & =W \\
N+F_2 \sin 30 & =50 \\
N+\frac{40}{3} \sqrt{3} \times \frac{1}{2} & =50 \\
N & =50-\frac{40}{6} \sqrt{3} \\
& =38.45 \text { Newton }
\end{aligned}
\)
So \(F_2=\frac{40}{3} \sqrt{3}\) Newtons and \(N=38.45\) Newton.

Hint

Slide the image above to see the resultant vector. The resultant vector magnitude calculation is as follows.

\(
\begin{aligned}
R & =\sqrt{A^2+B^2+2 A B \cos \theta} \\
& =\sqrt{75^2+25^2+2(75)(25) \cos 30^{\circ}} \\
& =97.46 \text { meter }
\end{aligned}
\)

Hint

First we have to realize that the angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta=30^{\circ}+\) \(67^{\circ}=97^{\circ}\). Then we can calculate the magnitude.
\(
\begin{aligned}
R & =\sqrt{A^2+B^2-2 A B \cos \theta} \\
& =\sqrt{45^2+30^2-2(45)(30) \cos 97^{\circ}} \\
& \approx 57 \text { unit }
\end{aligned}
\)
So, the magnitude of \(\vec{A}-\vec{B}\) is about 57 units.

Hint

Two forces, \(F_1\) and \(F_2\), are perpendicular to each other, which means that the magnitude of \(F_{\text {total }}\) is equal to the hypotenuse of the triangle formed by these vectors.
\(
\begin{aligned}
& F_{\text {total }}=\sqrt{F_1{ }^2+F_2{ }^2} \\
& F_{\text {total }}=\sqrt{3^2+4^2}=5 \text { Newtons }
\end{aligned}
\)

Hint

Let \(v\) represent the velocity and use the given information to write \(\mathrm{v}\) in unit vector form:
\(
\mathrm{v}=70\left(\cos \left(35^{\circ}\right)\right) i+70\left(\sin \left(35^{\circ}\right)\right) j
\)
Simplify the scalars, we get:
\(
\mathrm{v} \approx 57.34 i+40.15 j
\)
Since the scalars are the horizontal and vertical components of \(v\),
Therefore, the horizontal component is \(57.34\) feet per second and the vertical component is \(40.15\) feet per second.

Hint

First we write \(\mathrm{F}_1\) and \(\mathrm{F}_2\) in component form:
\(
\mathrm{v} \approx 57.34 i+40.15 j
\)
Simplify the scalars, we get:
\(
\begin{aligned}
& \mathrm{F}_1=\left(20 \cos \left(45^{\circ}\right)\right) i+\left(20 \sin \left(45^{\circ}\right)\right) j \\
& =20\left(\frac{\sqrt{2}}{2}\right) i+20\left(\frac{\sqrt{2}}{2}\right) j \\
& =10 \sqrt{2} i+10 \sqrt{2} j \\
& \mathrm{~F}_2=\left(30 \cos \left(150^{\circ}\right)\right) i+\left(30 \sin \left(150^{\circ}\right)\right) j \\
& =30\left(-\frac{\sqrt{3}}{2}\right) i+30\left(\frac{1}{2}\right) j \\
& =-15 \sqrt{3} i+15 j
\end{aligned}
\)
So, the resultant force \(\mathrm{F}\) is
\(
\begin{aligned}
& \mathrm{F}=\mathrm{F}_1+\mathrm{F}_2 \\
& =(10 \sqrt{2} i+10 \sqrt{2} j)+(-15 \sqrt{3} i+15 j) \\
& =(10 \sqrt{2}-15 \sqrt{3}) i+(10 \sqrt{2}+15) j \\
& \approx-12 i+29 j
\end{aligned}
\)

Hint

First we find the Displacement.
The displacement vector is
\(
\mathrm{D}=\langle 5-1,9-3\rangle=\langle 4,6\rangle
\)
By using the formula, the work done is
\(
\mathrm{W}=\mathrm{F} \cdot \mathrm{D}=\langle 2,3\rangle \cdot\langle 4,6\rangle=26
\)
If the unit of force is pounds and the distance is measured in feet, then the work done is \(26 \mathrm{ft}-\mathrm{lb}\).

Hint

For vector problems, we first draw a neat sketch of the vectors and the vector operation of interest. Here we are adding three vectors.

Then to solve the problem numerically, we break the vectors into their components.
\(
\mathbf{A}=i\left[57 \cos \left(47^{\circ}\right)\right]+j\left[57 \sin \left(47^{\circ}\right)\right]=i[38.8739]+j[41.6872]
\)
\(
\begin{aligned}
& \mathrm{B}=i\left[72 \cos \left(15^{\circ}\right)\right]+j\left[-72 \sin \left(15^{\circ}\right)\right]=i[69.5467]+j[-18.6350] \\
& \mathrm{C}=i\left[-24 \sin \left(30^{\circ}\right)\right]+j\left[24 \cos \left(30^{\circ}\right)\right]=i[-12]+j[20.7846]
\end{aligned}
\)
Next we add them to get the components of vector \(\mathbf{D}\).
\(
\mathrm{D}=i[38.8739+69.5467+-12]+j[41.6872+-18.6350+20.7846]=i[96.4206]+j[43.8368]
\)
Then we convert to polar coordinate form. Using Pythagoras’ Theorem, \(D=\left[(96.4206)^2+(43.8368)^2\right]^{1 / 2}=105.92 \mathrm{~m}\) The angle \(\theta=\) \(\arctan \left(\mid \mathrm{D}_{\mathrm{y}} / \mathrm{D}_{\mathrm{x}}\mid\right)=\arctan (43.8368 / 96.4206)=24.45^{\circ}\). Thus the person’s displacement is \(106 \mathrm{~m}\) at \(24.4^{\circ}\) north of east.

Hint

We are looking for \(\mathbf{F}_2=\mathbf{F}_3-\mathbf{F}_1\), a vector subtraction. For vector problems, we first draw a neat sketch of the vectors and the vector operation of interest.
Then to solve the problem numerically, we break the vectors into their components:
\(
\begin{aligned}
& \mathbf{F}_1=i\left[150 \sin \left(15^{\circ}\right)\right]+j\left[150 \cos \left(15^{\circ}\right)\right]=i[38.823]+j[144.889] \\
& \mathbf{F}_3=i\left[-100 \sin \left(10^{\circ}\right)\right]+j\left[100 \cos \left(10^{\circ}\right)\right]=i[-17.365]+j[98.481]
\end{aligned}
\)
We subtract the components to get the components of vector \(\mathbf{F}_2\).
\(
\mathbf{F}_2=i[38.823-(-17.365)]+j[144.889-98.481]=i[56.188]+j[46.408]
\)
Then we convert to polar coordinate form Using Pythagoras’ Theorem, \(F_2=\left[(56.188)^2+(46.408)^2\right]^{1 / 2}=72.875 \mathrm{~N}\). The angle \(\theta=\) \(\arctan \left(\mid \mathrm{F}_{2 \mathrm{y}} / \mathrm{F}_{2 \mathrm{x}}\mid\right)=\arctan (46.408 / 56.188)=39.55^{\circ}\). Thus the other force is \(72.9 \mathrm{~N}\) at \(39.6^{\circ}\) north of east.