Problem 41 - Fundamentals of Physics

Question

Use the definition of scalar product, \(\vec{a} \cdot \vec{b}=a b \cos \theta\), and the fact that \(\vec{a} \cdot \vec{b}=a_{x} b_{x}+a_{y} b_{y}+a_{z} b_{z}\) to cal- culate the angle between the two vectors given by \(\vec{a}=3.0 \hat{\mathrm{i}}+\) \(3.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and \(\vec{b}=2.0 \hat{\mathrm{i}}+1.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\)

Step-by-Step Solution

Answer

The angle between the vectors is approximately 21.52 degrees.

Step 1: Calculate the Scalar Product Using Components

We need to calculate the scalar product of the vectors \( \vec{a} \) and \( \vec{b} \) using their components. The formula is \[ \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \] Plug in the given components: \[ (3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0) = 6.0 + 3.0 + 9.0 = 18.0 \] Therefore, the scalar product \( \vec{a} \cdot \vec{b} = 18.0 \).

Step 2: Calculate the Magnitudes of the Vectors

Next, we calculate the magnitudes of the vectors \( \vec{a} \) and \( \vec{b} \). The magnitude of a vector \( \vec{v} = x \hat{\imath} + y \hat{\jmath} + z \hat{k} \) is given by \[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \] For \( \vec{a} = 3.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}} \): \[ |\vec{a}| = \sqrt{3.0^2 + 3.0^2 + 3.0^2} = \sqrt{27.0} = 3\sqrt{3} \approx 5.20 \] For \( \vec{b} = 2.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}} \): \[ |\vec{b}| = \sqrt{2.0^2 + 1.0^2 + 3.0^2} = \sqrt{14.0} \approx 3.74 \]

Step 3: Use the Scalar Product Definition to Find the Cosine of the Angle

According to the definition of the scalar product: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \] We can rearrange this formula to find \( \cos \theta \): \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \] Plug in the values: \[ \cos \theta = \frac{18.0}{(3\sqrt{3})(\sqrt{14})} \approx \frac{18.0}{5.20 \times 3.74} \approx 0.9285 \]

Step 4: Calculate the Angle

Finally, we use the inverse cosine function to find the angle \( \theta \): \[ \theta = \cos^{-1}(0.9285) \] Using a calculator, we find: \[ \theta \approx 21.52^\circ \]

Key Concepts

Vector Magnitude

Understanding the concept of vector magnitude is vital when dealing with vectors. The magnitude of a vector, often denoted as \(|\vec{v}|\), represents its length and is a scalar value. To calculate the magnitude of a vector given in three-dimensional space, such as \(\vec{v} = x \hat{\imath} + y \hat{\jmath} + z \hat{k}\), you can use the formula:\[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \]For example, if you have a vector \(\vec{a} = 3.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}}\), its magnitude would be calculated as:\[ |\vec{a}| = \sqrt{3.0^2 + 3.0^2 + 3.0^2} = \sqrt{27.0} = 3\sqrt{3} \approx 5.20 \]Magnitude helps in determining how large a vector is, irrespective of its direction. It's akin to determining the length of a line segment in geometry.

Dot Product Formula

The dot product, also known as the scalar product, is a crucial concept in vector algebra. It combines two vectors to yield a scalar (number). The formula for the dot product is:\[ \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \]This formula sums the products of corresponding components of vectors \(\vec{a}\) and \(\vec{b}\). For instance, if \(\vec{a} = 3.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}}\) and \(\vec{b} = 2.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}}\), their dot product would be:\[ \vec{a} \cdot \vec{b} = (3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0) = 6.0 + 3.0 + 9.0 = 18.0 \]The dot product can also be expressed using the cosine of the angle between the vectors, encapsulated in the formula:\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]This not only provides a physical meaning to the dot product but also links it to the angle between the vectors, which is instrumental in further calculations.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors is an important quantity that indicates the directional relationship between the vectors. It can be found using the dot product and the magnitudes of the vectors, with the formula:\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]Using this, you can determine how "aligned" or "opposed" two vectors are in multidimensional space. A cosine value of \(1\) indicates that vectors are parallel, while \(-1\) means they point in opposite directions.For example, with vectors \(\vec{a} = 3.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}}\) and \(\vec{b} = 2.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} + 3.0 \hat{\mathrm{k}}\), the cosine of the angle \(\theta\) can be calculated as:\[ \cos \theta = \frac{18.0}{(3\sqrt{3})(\sqrt{14})} \approx 0.9285 \]To find the angle \(\theta\), apply the inverse cosine function:\[ \theta = \cos^{-1}(0.9285) \approx 21.52^\circ \]This calculation reveals not only the angle between the vectors but also offers valuable insight into their spatial orientation.