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Exercise Set 1E
These questions are considered beginner to intermediate Ordinary Level.
Question 1
Write the vector \(\vec{a}=\langle 5 \,\, \angle \,\, 15^{\circ}\rangle\) in component form, correct to two decimal places.
\(\vec{a}=4.83\vec{i} + 1.29\vec{j}\)
\begin{align}|\vec{a_x}| &= |\vec{a}| \cos A \\&= 5\cos 15^{\circ} \\&\approx 4.83\end{align}
\begin{align}|\vec{a_y}| &= |\vec{a}| \sin A \\&= 5\sin 15^{\circ} \\&\approx 1.29 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}a_x = 4.83 && a_y = 1.29\end{align}
\begin{align}\downarrow\end{align}
\[\vec{a}=4.83\vec{i} + 1.29\vec{j}\]
We are being asked to convert a vector from polar form into component form and should therefore follow the relevant steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) given in polar form is the same angle that is within the triangle, i.e. \(A=15^{\circ}\).
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{a_x}| = |\vec{a}|\cos A} \) and \(\boldsymbol{|\vec{a_y}| = |\vec{a}|\sin A} \).
\begin{align}|\vec{a_x}| &= |\vec{a}| \cos A \\&= 5\cos 15^{\circ} \\&\approx 4.83\end{align}
\begin{align}|\vec{a_y}| &= |\vec{a}| \sin A \\&= 5\sin 15^{\circ} \\&\approx 1.29 \end{align}
Step 4: Calculate \(\boldsymbol{a_x}\) and \(\boldsymbol{a_y}\) using \(\boldsymbol{|\vec{a_x}|}\) and \(\boldsymbol{|\vec{a_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{a_x}\) is in the same direction as \(\vec{i}\) and \(a_y\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}a_x = 4.83 && a_y = 1.29\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{a} =a_x \vec{i} + a_y \vec{j}}\).
\[\vec{a}=4.83\vec{i} + 1.29\vec{j}\]
Question 2
Write the vector \(\vec{a}=\langle 46 \mbox{ km} \,\, \angle \,\, 180^{\circ}\rangle\) in component form.
\(-46\vec{i}\mbox{ km}\)
Vector is pointing purely in the negative \(x\) direction \(\rightarrow \vec{a}=-46\vec{i}\mbox{ km}\).
We could use our usual method to convert this vector from polar form into component form.
However, in this case, we can be a bit smarter!
As the argument is \(180^{\circ}\), this vector is pointing purely in the negative \(x\) direction, i.e. in the opposite direction to \(\vec{i}\).
Therefore, we can immediately write this vector in component form as \(-46\vec{i}\mbox{ km}\)!
Question 3
Write the vector \(\vec{p}=\langle 12\mbox{ m/s} \,\, \angle \,\, 105^{\circ}\rangle\) in component form, correct to two decimal places.
\(\vec{p}=-3.11\vec{i} + 11.59\vec{j}\mbox{ m/s}\)
\begin{align}|\vec{p_x}| &= |\vec{p}| \cos A \\&= 12\cos 75^{\circ} \\&\approx 3.11\end{align}
\begin{align}|\vec{p_y}| &= |\vec{p}| \sin A \\&= 12\sin 75^{\circ} \\&\approx 11.59 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}p_x = -3.11 && p_y = 11.59\end{align}
\begin{align}\downarrow\end{align}
\[\vec{p}=-3.11\vec{i} + 11.59\vec{j}\mbox{ m/s}\]
We are being asked to convert a vector from polar form into component form and should therefore follow the relevant steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(A\) in the triangle is related to the argument \(\theta\) as follows:
\begin{align}A &= 180^{\circ}-105^{\circ}\\&=75^{\circ}\end{align}
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{p_x}| = |\vec{p}|\cos A} \) and \(\boldsymbol{|\vec{p_y}| = |\vec{p}|\sin A} \).
\begin{align}|\vec{p_x}| &= |\vec{p}| \cos A \\&= 12\cos 75^{\circ} \\&\approx 3.11\end{align}
\begin{align}|\vec{p_y}| &= |\vec{p}| \sin A \\&= 12\sin 75^{\circ} \\&\approx 11.59 \end{align}
Step 4: Calculate \(\boldsymbol{p_x}\) and \(\boldsymbol{p_y}\) using \(\boldsymbol{|\vec{p_x}|}\) and \(\boldsymbol{|\vec{p_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{p_x}\) is in the opposite direction to \(\vec{i}\) and \(p_y\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}p_x = -3.11 && p_y = 11.59\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{p} =p_x \vec{i} + p_y \vec{j}}\).
\[\vec{p}=-3.11\vec{i} + 11.59\vec{j}\mbox{ m/s}\]
Question 4
Consider the following vectors, where \(|\vec{a}|=|\vec{b}|=20\mbox{ N}\) and \(\theta = \tan^{-1}\left(\dfrac{4}{3}\right)\).
(a) Write each vector in polar form.
(b) Write each vector in component form.
(c) What is \(\vec{a}\cdot\vec{b}\)?
(a) \(\vec{a}=\langle 20\mbox{ N} \,\, \angle \,\, 90^{\circ}\rangle\) and \(\vec{b}=\langle 20\mbox{ N} \,\, \angle \,\, 53.13^{\circ}\rangle\)
(b) \(\vec{a}=20\vec{j}\mbox{ N}\) and \(\vec{b}=12\vec{i} + 16\vec{j} \mbox{ N}\)
(c) \(320\)
(a)
\begin{align}\vec{a}=\langle 20\mbox{ N} \,\, \angle \,\, 90^{\circ}\rangle\end{align}
and
\begin{align}\theta &= \tan^{-1} \left(\frac{4}{3}\right)\\&\approx 53.13^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\vec{b}=\langle 20\mbox{ N} \,\, \angle \,\, 53.13^{\circ}\rangle\end{align}
(b)
\begin{align}H &= \sqrt{3^2+4^2} \\&= 5\end{align}
\begin{align}\downarrow\end{align}
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{3}{5}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{4}{5}\end{align}\]
\begin{align}\downarrow\end{align}
\[\begin{align}|\vec{b}_x| &= |\vec{b}| \cos A \\& = 20 \left(\frac{3}{5}\right) \\&= 12\end{align}\]
\[\begin{align}|\vec{b}_y| &= |\vec{b}| \sin A \\&= 20 \left(\frac{4}{5}\right) \\&= 16\end{align}\]
\begin{align}b_x = 12 && b_y = 16\end{align}
\begin{align}\downarrow\end{align}
\[\vec{b}=12\vec{i} + 16\vec{j} \mbox{ N}\]
(c)
\begin{align}\vec{a}\cdot\vec{b} &= |\vec{a}||\vec{b}|\cos \alpha \\&=(20)(20)\cos (90^{\circ}-53.13^{\circ})\\&=(20)(20)(0.8)\\&=320\end{align}
(a) As \(\vec{a}\) is pointing in the positive \(y\) direction, it has an argument of \(90^{\circ}\). As the magnitude of \(\vec{a}\) is \(20\mbox{ N}\), we write \(\vec{a}\) in polar form as follows:
\begin{align}\vec{a}=\langle 20\mbox{ N} \,\, \angle \,\, 90^{\circ}\rangle\end{align}
For \(\vec{b}\), we are already given the argument in the form of an inverse tan. Typically, when writing a vector in polar form, the convention is to instead write the angle in degrees which, according to a calculator, is given by
\begin{align}\theta &= \tan^{-1} \left(\frac{4}{3}\right)\\&\approx 53.13^{\circ}\end{align}
Therefore, as \(\vec{b}\) also has a magnitude of \(20\mbox{ N}\), we write \(\vec{b}\) in polar form as follows:
\begin{align}\vec{b}=\langle 20\mbox{ N} \,\, \angle \,\, 53.13^{\circ}\rangle\end{align}
(b) We are being asked to convert vectors from polar form into component form and should therefore follow the relevant steps.
\(\vec{a}\), however, is pointing in the positive \(y\) direction, i.e. in the same direction as \(\vec{j}\). Therefore, as it has a magnitude of \(20\mbox{ N}\), we can immediately write it in component form as follows:
\begin{align}\vec{a}=20\vec{j}\mbox{ N}\end{align}
As \(\vec{b}\) is instead an arbitrary vector, we must instead use our usual steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) given in polar form is the same angle that is within the triangle, i.e.
\[A = \tan^{-1} \left(\frac{4}{3}\right)\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{b_x}| = |\vec{b}|\cos A} \) and \(\boldsymbol{|\vec{b_y}| = |\vec{b}|\sin A} \).
As we are given the angle in terms of the ratio of sides of a triangle, we must first draw the a triangle which correctly includes this angle \(A\).
Next, we calculate the hypotenuse \(H\) of this triangle using Pythagoras’ Theorem.
\begin{align}H &= \sqrt{3^2+4^2} \\&= 5\end{align}
Now, we can easily find \(\cos A\) and \(\sin A\) by using their definitions.
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{3}{5}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{4}{5}\end{align}\]
Using these, we can directly calculate our components as
\[\begin{align}|\vec{b}_x| &= |\vec{b}| \cos A \\& = 20 \left(\frac{3}{5}\right) \\&= 12\end{align}\]
\[\begin{align}|\vec{b}_y| &= |\vec{b}| \sin A \\&= 20 \left(\frac{4}{5}\right) \\&= 16\end{align}\]
\[\,\]
Step 4: Calculate \(\boldsymbol{b_x}\) and \(\boldsymbol{b_y}\) using \(\boldsymbol{|\vec{b_x}|}\) and \(\boldsymbol{|\vec{b_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{b_x}\) is in the same direction as \(\vec{i}\) and \(\vec{b_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}b_x = 12 && b_y = 16\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{b} =b_x \vec{i} + b_y \vec{j}}\).
\[\vec{b}=12\vec{i} + 16\vec{j} \mbox{ N}\]
(c) According to the diagram in the question, the angle between vectors \(\vec{a}\) and \(\vec{b}\) is
\begin{align}\alpha &= 90^{\circ}-\theta\\&\approx 90^{\circ} – 53.13^{\circ}\\&=36.87^{\circ}\end{align}
Therefore, the dot product of these two vectors is given by
\begin{align}\vec{a}\cdot\vec{b} &= |\vec{a}||\vec{b}|\cos \alpha \\&=(20)(20)(\cos 36.87^{\circ})\\&=(20)(20)(0.8)\\&=320\end{align}
Question 5
Write the vector \(\vec{a}=\langle 50\mbox{ N} \,\, \angle \,\, 140^{\circ}\rangle\) in component form, correct to two decimal places.
\(\vec{a}=-38.30\vec{i} + 32.14\vec{j}\mbox{ N}\)
\begin{align}|\vec{a_x}| &= |\vec{a}| \cos A \\&= 50\cos 40^{\circ} \\&\approx 38.30\end{align}
\begin{align}|\vec{a_y}| &= |\vec{a}| \sin A \\&= 50\sin 40^{\circ} \\&\approx 32.14 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}a_x = -38.30 && a_y = 32.14\end{align}
\begin{align}\downarrow\end{align}
\[\vec{a}=-38.30\vec{i} + 32.14\vec{j}\mbox{ N}\]
We are being asked to convert a vector from polar form into component form and should therefore follow the relevant steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(A\) in the triangle is related to the argument \(\theta\) as follows:
\begin{align}A &= 180^{\circ}-140^{\circ}\\&=40^{\circ}\end{align}
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{a_x}| = |\vec{a}|\cos A} \) and \(\boldsymbol{|\vec{a_y}| = |\vec{a}|\sin A} \).
\begin{align}|\vec{a_x}| &= |\vec{a}| \cos A \\&= 50\cos 40^{\circ} \\&\approx 38.30\end{align}
\begin{align}|\vec{a_y}| &= |\vec{a}| \sin A \\&= 50\sin 40^{\circ} \\&\approx 32.14 \end{align}
Step 4: Calculate \(\boldsymbol{a_x}\) and \(\boldsymbol{a_y}\) using \(\boldsymbol{|\vec{a_x}|}\) and \(\boldsymbol{|\vec{a_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{a_x}\) is in the opposite direction to \(\vec{i}\) and \(a_y\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}a_x = -38.30 && a_y = 32.14\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{a} =a_x \vec{i} + a_y \vec{j}}\).
\[\vec{a}=-38.30\vec{i} + 32.14\vec{j}\mbox{ N}\]
Question 6
A vector \(\vec{p}\) has a magnitude of \(13\) units and subtends an angle \(\theta\) relative to the positive \(x\) axis, where \(\theta = \tan^{-1}\left(\displaystyle\frac{5}{12}\right)\).
Write \(\vec{p}\) in component form without using a calculator.
\(\vec{p}=12\vec{i} + 5\vec{j}\)
\begin{align}H &= \sqrt{5^2+12^2} \\&= 13\end{align}
\begin{align}\downarrow\end{align}
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{12}{13}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{5}{13}\end{align}\]
\begin{align}\downarrow\end{align}
\[\begin{align}|\vec{p}_x| &= |\vec{p}| \cos A \\& = 13 \left(\frac{12}{13}\right) \\&= 12\end{align}\]
\[\begin{align}|\vec{p}_y| &= |\vec{p}| \sin A \\&= 13 \left(\frac{5}{13}\right) \\&= 5\end{align}\]
\begin{align}p_x = 12 && p_y = 5\end{align}
\begin{align}\downarrow\end{align}
\[\vec{p}=12\vec{i} + 5\vec{j}\]
We are being asked to convert a vector from polar form into component form and should therefore follow the relevant steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) given in polar form is the same angle that is within the triangle, i.e.
\[A = \tan^{-1} \left(\frac{5}{12}\right)\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{p_x}| = |\vec{p}|\cos A} \) and \(\boldsymbol{|\vec{p_y}| = |\vec{p}|\sin A} \).
As we are given the angle in terms of the ratio of sides of a triangle, we must first draw the a triangle which correctly includes this angle \(A\).
Next, we calculate the hypotenuse \(H\) of this triangle using Pythagoras’ Theorem.
\begin{align}H &= \sqrt{5^2+12^2} \\&= 13\end{align}
Now, we can easily find \(\cos A\) and \(\sin A\) by using their definitions.
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{12}{13}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{5}{13}\end{align}\]
Using these, we can directly calculate our components as
\[\begin{align}|\vec{p}_x| &= |\vec{p}| \cos A \\& = 13 \left(\frac{12}{13}\right) \\&= 12\end{align}\]
\[\begin{align}|\vec{p}_y| &= |\vec{p}| \sin A \\&= 13 \left(\frac{5}{13}\right) \\&= 5\end{align}\]
\[\,\]
Step 4: Calculate \(\boldsymbol{p_x}\) and \(\boldsymbol{p_y}\) using \(\boldsymbol{|\vec{p_x}|}\) and \(\boldsymbol{|\vec{p_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{p_x}\) is in the same direction as \(\vec{i}\) and \(\vec{p_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}p_x = 12 && p_y = 5\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{p} =p_x \vec{i} + p_y \vec{j}}\).
\[\vec{p}=12\vec{i} + 5\vec{j}\]
Question 7
A vector \(\vec{a}\) has a magnitude of \(20\mbox{ m/s}\) and is at an angle \(\theta\) relative to the negative \(x\) axis, where \(\theta = \tan^{-1}\left(\displaystyle\frac{3}{4}\right)\).
Write \(\vec{a}\) in component form without using a calculator.
\(\vec{a}=-16\vec{i}-12\vec{j}\mbox{ m/s}\)
\begin{align}H &= \sqrt{3^2+4^2} \\&= 5\end{align}
\begin{align}\downarrow\end{align}
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{4}{5}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{3}{5}\end{align}\]
\begin{align}\downarrow\end{align}
\[\begin{align}|\vec{a}_x| &= |\vec{a}| \cos A \\& = 20 \left(\frac{4}{5}\right) \\&= 16\end{align}\]
\[\begin{align}|\vec{a}_y| &= |\vec{a}| \sin A \\&= 20 \left(\frac{3}{5}\right) \\&= 12\end{align}\]
\begin{align}a_x = -16 && a_y = -12\end{align}
\begin{align}\downarrow\end{align}
\[\vec{a}=-16\vec{i}-12\vec{j}\mbox{ m/s}\]
We are being asked to write a vector in component form and should therefore follow the relevant steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the given angle \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) that is given is the same angle that is within the triangle, i.e.
\[A = \tan^{-1} \left(\frac{3}{4}\right)\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{a_x}| = |\vec{a}|\cos A} \) and \(\boldsymbol{|\vec{a_y}| = |\vec{a}|\sin A} \).
As we are given the angle in terms of the ratio of sides of a triangle, we must first draw the a triangle which correctly includes this angle \(A\).
Next, we calculate the hypotenuse \(H\) of this triangle using Pythagoras’ Theorem.
\begin{align}H &= \sqrt{3^2+4^2} \\&= 5\end{align}
Now, we can easily find \(\cos A\) and \(\sin A\) by using their definitions.
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{4}{5}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{3}{5}\end{align}\]
Using these, we can directly calculate our components as
\[\begin{align}|\vec{a}_x| &= |\vec{a}| \cos A \\& = 20 \left(\frac{4}{5}\right) \\&= 16\end{align}\]
\[\begin{align}|\vec{a}_y| &= |\vec{a}| \sin A \\&= 20 \left(\frac{3}{5}\right) \\&= 12\end{align}\]
\[\,\]
Step 4: Calculate \(\boldsymbol{a_x}\) and \(\boldsymbol{a_y}\) using \(\boldsymbol{|\vec{a_x}|}\) and \(\boldsymbol{|\vec{a_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{a_x}\) is in the opposite direction to \(\vec{i}\) and \(\vec{a_y}\) is in the opposite direction to \(\vec{j}\). Therefore
\begin{align}a_x = -16 && a_y = -12\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{a} =a_x \vec{i} + a_y \vec{j}}\).
\[\vec{a}=-16\vec{i}-12\vec{j}\mbox{ m/s}\]
Question 8
Consider the following vectors, where \(\theta = \tan^{-1}\left(\dfrac{12}{5}\right)\), \(\alpha = 40^\circ\) and the magnitudes of \(\vec{a}\) and \(\vec{b}\) are \(10\mbox{ m/s}\) and \(15\mbox{ m/s}\) respectively.
(a) Write each vector in polar form.
(b) Write each vector in component form.
(c) What is \(\vec{a}\cdot\vec{b}\)?
(a) \(\vec{a}=\langle 10\mbox{ m/s} \,\, \angle \,\, 140^{\circ}\rangle\) and \(\vec{b}=\langle 15\mbox{ m/s} \,\, \angle \,\, 67.38^{\circ}\rangle\)
(b) \(\vec{a}=-7.66\vec{i} + 6.43\vec{j} \mbox{ m/s}\) and \(\vec{b}=5.77\vec{i} + 13.85\vec{j} \mbox{ N}\)
(c) \(44.81\)
(a)
\begin{align}\beta &= 180^{\circ}-\alpha \\&= 180^{\circ}-40^{\circ}\\&=140^{\circ} \end{align}
\begin{align}\downarrow\end{align}
\begin{align}\vec{a}=\langle 10\mbox{ m/s} \,\, \angle \,\, 140^{\circ}\rangle\end{align}
and
\begin{align}\theta &= \tan^{-1} \left(\frac{12}{5}\right)\\&\approx 67.38^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\vec{b}=\langle 15\mbox{ m/s} \,\, \angle \,\, 67.38^{\circ}\rangle\end{align}
(b)
\(\mathbf{\vec{a}}\):
\[\begin{align}|\vec{a}_x| &= |\vec{a}| \cos A \\& = 10\cos(40^{\circ}) \\&\approx 7.66\end{align}\]
\[\begin{align}|\vec{a}_y| &= |\vec{a}| \sin A \\& = 10\sin(40^{\circ}) \\&\approx 6.43\end{align}\]
\begin{align}a_x = -7.66 && a_y = 6.43\end{align}
\begin{align}\downarrow\end{align}
\[\vec{a}=-7.66\vec{i} + 6.43\vec{j} \mbox{ m/s}\]
\[\,\]
\(\mathbf{\vec{b}}\):
\begin{align}H &= \sqrt{12^2+5^2} \\&= 13\end{align}
\begin{align}\downarrow\end{align}
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{5}{13}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{12}{13}\end{align}\]
\begin{align}\downarrow\end{align}
\[\begin{align}|\vec{b}_x| &= |\vec{b}| \cos A \\& = 15 \left(\frac{5}{13}\right) \\&\approx 5.77\end{align}\]
\[\begin{align}|\vec{b}_y| &= |\vec{b}| \sin A \\&= 15 \left(\frac{12}{13}\right) \\&\approx 13.85\end{align}\]
\begin{align}b_x = 5.77 && b_y = 13.85\end{align}
\begin{align}\downarrow\end{align}
\[\vec{b}=5.77\vec{i} + 13.85\vec{j} \mbox{ N}\]
(c)
\begin{align}\vec{a}\cdot\vec{b} &= |\vec{a}||\vec{b}|\cos \alpha \\&=(10)(15)\cos (180^{\circ}-40^{\circ}-67.38^{\circ}\\&\approx44.81\end{align}
(a) The argument of a vector is defined as the angle subtended counter clockwise from the positive \(x\) axis to that vector.
For \(\vec{a}\), according to the diagram, the argument \(\beta\) is
\begin{align}\beta &= 180^{\circ}-\alpha \\&= 180^{\circ}-40^{\circ}\\&=140^{\circ} \end{align}
(We have used \(\beta\) to represent the argument as \(\theta\) is already used in the diagram.)
Therefore, as the magnitude of \(\vec{a}\) is \(10\mbox{ m/s}\), we write \(\vec{a}\) in polar form as follows:
\begin{align}\vec{a}=\langle 10\mbox{ m/s} \,\, \angle \,\, 140^{\circ}\rangle\end{align}
For \(\vec{b}\), we are already given the argument in the form of an inverse tan. Typically, when writing a vector in polar form, the convention is to instead write the angle in degrees which, according to a calculator, is given by
\begin{align}\theta &= \tan^{-1} \left(\frac{12}{5}\right)\\&\approx 67.38^{\circ}\end{align}
Therefore, as \(\vec{b}\) has a magnitude of \(15\mbox{ m/s}\), we write \(\vec{b}\) in polar form as follows:
\begin{align}\vec{b}=\langle 15\mbox{ m/s} \,\, \angle \,\, 67.38^{\circ}\rangle\end{align}
(b) We are being asked to convert vectors from polar form into component form and should therefore follow the relevant steps.
First, we will apply our usual method to \(\vec{a}\).
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the given angle to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\alpha\) is the same angle that is within the triangle, i.e.
\[A = 40^{\circ}\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{a_x}| = |\vec{a}|\cos A} \) and \(\boldsymbol{|\vec{a_y}| = |\vec{a}|\sin A} \)
\[\begin{align}|\vec{a}_x| &= |\vec{a}| \cos A \\& = 10\cos(40^{\circ}) \\&\approx 7.66\end{align}\]
\[\begin{align}|\vec{a}_y| &= |\vec{a}| \sin A \\& = 10\sin(40^{\circ}) \\&\approx 6.43\end{align}\]
\[\,\]
Step 4: Calculate \(\boldsymbol{a_x}\) and \(\boldsymbol{a_y}\) using \(\boldsymbol{|\vec{a_x}|}\) and \(\boldsymbol{|\vec{a_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{a_x}\) is in the opposite direction to \(\vec{i}\) and \(\vec{a_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}a_x = -7.66 && a_y = 6.43\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{a} =a_x \vec{i} + a_y \vec{j}}\).
\[\vec{a}=-7.66\vec{i} + 6.43\vec{j} \mbox{ m/s}\]
Next, we shall apply the same method to \(\vec{b}\).
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the given angle to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) is the same angle that is within the triangle, i.e.
\[A = \tan^{-1} \left(\frac{12}{5}\right)\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{b_x}| = |\vec{b}|\cos A} \) and \(\boldsymbol{|\vec{b_y}| = |\vec{b}|\sin A} \).
As we are given the angle in terms of the ratio of sides of a triangle, we must first draw the a triangle which correctly includes this angle \(A\).
Next, we calculate the hypotenuse \(H\) of this triangle using Pythagoras’ Theorem.
\begin{align}H &= \sqrt{12^2+5^2} \\&= 13\end{align}
Now, we can easily find \(\cos A\) and \(\sin A\) by using their definitions.
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{5}{13}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{12}{13}\end{align}\]
Using these, we can directly calculate our components as
\[\begin{align}|\vec{b}_x| &= |\vec{b}| \cos A \\& = 15 \left(\frac{5}{13}\right) \\&\approx 5.77\end{align}\]
\[\begin{align}|\vec{b}_y| &= |\vec{b}| \sin A \\&= 15 \left(\frac{12}{13}\right) \\&\approx 13.85\end{align}\]
\[\,\]
Step 4: Calculate \(\boldsymbol{b_x}\) and \(\boldsymbol{b_y}\) using \(\boldsymbol{|\vec{b_x}|}\) and \(\boldsymbol{|\vec{b_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{b_x}\) is in the same direction as \(\vec{i}\) and \(\vec{b_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}b_x = 5.77 && b_y = 13.85\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{b} =b_x \vec{i} + b_y \vec{j}}\).
\[\vec{b}=5.77\vec{i} + 13.85\vec{j} \mbox{ N}\]
(c) According to the diagram in the question, the angle between vectors \(\vec{a}\) and \(\vec{b}\) is
\begin{align}\alpha &= 180^{\circ}-\alpha-\theta\\&\approx 180^{\circ} -40^{\circ}-67.38^{\circ}\\&=72.62^{\circ}\end{align}
Therefore, the dot product of these two vectors is given by
\begin{align}\vec{a}\cdot\vec{b} &= |\vec{a}||\vec{b}|\cos \alpha \\&=(10)(15)(\cos 72.62^{\circ})\\&\approx44.81\end{align}
Question 9
If \(\vec{m}=\displaystyle\frac{1}{2}\vec{i}-k\vec{j}\) is a unit vector, find two possible values of \(k\).
\(k=\pm \sqrt{\dfrac{3}{4}}\)
\begin{align}|\vec{m}|& =\sqrt{\left(\frac{1}{2}\right)^2+k^2}\end{align}
\begin{align}\downarrow\end{align}
\[\sqrt{\frac{1}{4}+k^2}=1\]
\begin{align}\downarrow\end{align}
\[\frac{1}{4}+k^2=1\]
\begin{align}\downarrow\end{align}
\[k=\pm \sqrt{\frac{3}{4}}\]
As \(\vec{m}\) is a unit vector, it must have a magnitude of \(1\). Its magnitude is given by
\begin{align}|\vec{m}|& =\sqrt{\left(\frac{1}{2}\right)^2+k^2}\end{align}
Setting this to \(1\) gives
\[\sqrt{\frac{1}{4}+k^2}=1\]
\begin{align}\downarrow\end{align}
\[\frac{1}{4}+k^2=1\]
\begin{align}\downarrow\end{align}
\[k=\pm \sqrt{\frac{3}{4}}\]
Question 10
Find the dot product of the vectors \(\vec{p}=\langle 2 \,\, \angle \,\, 90^{\circ}\rangle\) and \(\vec{q}=\langle 3 \,\, \angle \,\, 30^{\circ}\rangle\):
(a) without converting the vectors into component form
(b) using the vectors in component form
(a) \(3\)
(b) \(3\)
(a)
\begin{align}\vec{p} \cdot \vec{q} & = |\vec{p}| |\vec{q}| \cos \theta \\&= (2)(3) \cos (90^\circ-30^{\circ})\\&= 3\end{align}
(b)
\[\vec{p} = 2\vec{j}\]
\begin{align}|\vec{q_x}| &= |\vec{q}| \cos A \\&= 3\cos 30^{\circ} \\&\approx 2.60\end{align}
\begin{align}|\vec{q_y}| &= |\vec{q}| \sin A \\&= 3\sin 30^{\circ} \\&=1.5 \end{align}
\begin{align}\downarrow\end{align}
\begin{align}q_x = 2.60 && q_y = 1.5\end{align}
\begin{align}\downarrow\end{align}
\[\vec{q}=2.60\vec{i} + 1.5\vec{j}\]
\[\,\]
\begin{align}\vec{p} \cdot \vec{q} &= p_xq_x + p_yq_y\\&=(0)(2.60)+(2)(1.5)\\&=3\end{align}
(a) \(\vec{p}\) has a magnitude of \(2\) units and is at angle of \(90^{\circ}\) relative to the positive \(x\) axis.
\(\vec{q}\) has a magnitude of \(3\) units and is at angle of \(30^{\circ}\) relative to the positive \(x\) axis.
The angle \(\theta\) between the vectors is therefore:
\begin{align}\theta &= 90^{\circ}-30^{\circ} \\&= 60^{\circ}\end{align}
Hence, their dot product is:
\begin{align}\vec{p} \cdot \vec{q} & = |\vec{p}| |\vec{q}| \cos \theta \\&= (2)(3) \cos 60^{\circ}\\&= 3\end{align}
(b) Let us now convert both \(\vec{p}\) and \(\vec{q}\) into component vectors.
For \(\vec{p}\), while we could certainly use our usual steps, we can be smarter and write it down in component form much quicker!
\(\vec{p}\) is at an angle of \(90^{\circ}\) relative to the positive \(x\) axis. Hence, it points along the positive \(y\) axis, i.e. in the same direction as \(\vec{j}\). Therefore, as it has a magnitude of \(2\), we can immediately write \(\vec{p}\) in component form as
\[\vec{p} = 2\vec{j}\]
Next, let us decompose \(\vec{q}\) by following the relevant steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) given in polar form is the same angle that is within the triangle, i.e. \(A=30^{\circ}\).
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{q_x}| = |\vec{q}|\cos A} \) and \(\boldsymbol{|\vec{q_y}| = |\vec{q}|\sin A} \).
\begin{align}|\vec{q_x}| &= |\vec{q}| \cos A \\&= 3\cos 30^{\circ} \\&\approx 2.60\end{align}
\begin{align}|\vec{q_y}| &= |\vec{q}| \sin A \\&= 3\sin 30^{\circ} \\&=1.5 \end{align}
\[\,\]
Step 4: Calculate \(\boldsymbol{q_x}\) and \(\boldsymbol{q_y}\) using \(\boldsymbol{|\vec{q_x}|}\) and \(\boldsymbol{|\vec{q_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{q_x}\) is in the same direction as \(\vec{i}\) and \(\vec{q_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}q_x = 2.60 && q_y = 1.5\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{q} =q_x \vec{i} + q_y \vec{j}}\).
\[\vec{q}=2.60\vec{i} + 1.5\vec{j}\]
The dot product of these two vectors in component form is therefore
\begin{align}\vec{p} \cdot \vec{q} &= p_xq_x + p_yq_y\\&=(0)(2.60)+(2)(1.5)\\&=3\end{align}
As expected, we get the same value for the dot product independent of the method used.
