Question

\((a)\) Calculate \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\), where \(\overrightarrow{\mathbf{a}}=5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\), \(\overrightarrow{\mathbf{b}}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\), and \(\overrightarrow{\mathbf{c}}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} .\) (b) Calculate the angle between \(\overrightarrow{\mathbf{r}}\) and the \(+z\) axis. ( \(c\) ) Find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\).

Short answer

So, the vector \(\overrightarrow{\mathbf{r}}\) is \(11\hat{\mathbf{i}} + 5\hat{\mathbf{j}} - 7\hat{\mathbf{k}}\). The angle between the vector \(\overrightarrow{\mathbf{r}}\) and the \(+z\) axis is given by \(acos(-7 /sqrt{(11^{2} + 5^{2} + (-7)^{2}})) degrees\), and the angle between the vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is given by \(acos ( (5*-2 + 4*2 - 6*3) / (sqrt{(5^{2} + 4^{2} + (-6)^{2}}) * sqrt{((-2)^{2} + 2^{2} + 3^{2}})) degrees\).

Step-by-step solution

Calculation of vector \(\overrightarrow{\mathbf{r}}\)

Our first task is to calculate the vector \(\overrightarrow{\mathbf{r}}\), which is defined as \(\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}\). We simply subtract the \(i\), \(j\), and \(k\) components of \(\overrightarrow{\mathbf{b}}\) from those of \(\overrightarrow{\mathbf{a}}\), and then add the corresponding components of \(\overrightarrow{\mathbf{c}}\) to the results. Thus, \(\overrightarrow{\mathbf{r}} = (5-(-2)+4)\hat{\mathbf{i}} + (4-2+3)\hat{\mathbf{j}} + (-6-3+2)\hat{\mathbf{k}} = 11\hat{\mathbf{i}} + 5\hat{\mathbf{j}} - 7\hat{\mathbf{k}}\).

Calculation of the angle between \(\overrightarrow{\mathbf{r}}\) and the \(+z\) axis

The angle between a vector and the \(+z\) axis can be calculated using the dot product formula. The dot product of two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is given by \(\overrightarrow{A} . \overrightarrow{B} = |A||B| cos(\theta)\), where \(\theta\) is the angle between the vectors. The \(+z\) axis is represented by the unit vector \(\hat{\mathbf{k}}\). Hence, \(\theta = acos (\overrightarrow{\mathbf{r}} . \hat{\mathbf{k}} / |r|\) = acos ( -7 /sqrt{(11^{2} + 5^{2} + (-7)^{2}}))\).

Calculation of the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\)

This is similar to step 2 and requires the use of the dot product formula. Hence, \(\theta_{ab} = acos (\overrightarrow{\mathbf{a}} . \overrightarrow{\mathbf{b}} / |a||b|\) = acos ( (5*-2 + 4*2 - 6*3) / (sqrt{(5^{2} + 4^{2} + (-6)^{2}} * sqrt{((-2)^{2} + 2^{2} + 3^{2})}))\)

Key concepts

Understanding the Dot Product

The dot product is a fundamental operation in vector mathematics. It provides a way to find the angle between two vectors, or it can simplify complex vector calculations. To compute the dot product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), you multiply their corresponding components and then sum these products:

• \( \overrightarrow{A} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}} \)
• \( \overrightarrow{B} = b_1\hat{\mathbf{i}} + b_2\hat{\mathbf{j}} + b_3\hat{\mathbf{k}} \)

The dot product formula then becomes: \[ \overrightarrow{A} . \overrightarrow{B} = a_1b_1 + a_2b_2 + a_3b_3 \]This product can also be written using the magnitudes of the vectors and the cosine of the angle \( \theta \) between them:\[ \overrightarrow{A} . \overrightarrow{B} = |A||B| \cos(\theta) \]By rearranging, you can solve for \( \theta \), which is useful to find the angle between the vectors. The dot product reflects both the magnitude and direction, making it essential for problems involving angles.

Exploring Vector Addition

Vector addition combines multiple vectors to form a single resultant vector. This is a straightforward operation where each vector component is added separately. For example, if you have vectors \( \overrightarrow{a} \), \( \overrightarrow{b} \), and \( \overrightarrow{c} \), and you want to find \( \overrightarrow{r} = \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c} \), you do the following:

• Subtract \( \overrightarrow{b} \) from \( \overrightarrow{a} \) by separately subtracting each of the \( i \), \( j \), and \( k \) components.
• Add the result to \( \overrightarrow{c} \) by adding each of the components again.

For this exercise: - \( \overrightarrow{r} = (5 - (-2) + 4)\hat{\mathbf{i}} + (4 - 2 + 3)\hat{\mathbf{j}} + (-6 - 3 + 2)\hat{\mathbf{k}} \)- Final result: \( 11\hat{\mathbf{i}} + 5\hat{\mathbf{j}} - 7\hat{\mathbf{k}} \) This method of breaking vectors into components and dealing with each separately makes vector addition manageable and intuitive, emphasizing the directionality inherent in vectors.

Calculating the Angle Between Vectors

To find the angle between two vectors, you use the dot product in conjunction with the magnitudes of the vectors. The formula connects the dot product with the angle \( \theta \):\[ \cos(\theta) = \frac{\overrightarrow{A} . \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \]For example, to calculate the angle \( \theta \) between vector \( \overrightarrow{r} \) and the positive z-axis:

• Find \( \overrightarrow{r} . \hat{\mathbf{k}} \), where \( \hat{\mathbf{k}} \) is the vector of the z-axis, yielding the product of the z component of \( \overrightarrow{r} \) and 1.
• Use the magnitudes: \( |\overrightarrow{r}| = \sqrt{11^2 + 5^2 + (-7)^2} \).

Substitute these into the formula and calculate \( \theta \) using the inverse cosine (\( \cos^{-1} \)). This approach also applies to finding the angle between any two vectors, such as \( \overrightarrow{a} \) and \( \overrightarrow{b} \). Calculate the dot product and divide by the product of their magnitudes, then use inverse cosine. Understanding this concept aids in visualizing how vectors relate spatially, offering insights into physical interpretations like forces or velocities.