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A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) \(A(8,0,0), \quad B(8,18,0)\), \(C(0,18,8),\) and \(D(0,0,8)\) (see the following figure). a. Find vector \(\mathbf{n}=\overrightarrow{A B} \times \overrightarrow{A D}\) perpendicular to the surface of the solar panels. Express the answer using standard unit vectors. b. Assume unit vector \(\mathbf{s}=\frac{1}{\sqrt{3}} \mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\) points toward the Sun at a particular time of the day and the flow of solar energy is \(\mathbf{F}=900 \mathbf{s}\) (in watts per square meter [ \(\mathrm{W} / \mathrm{m}^{2}\) ]). Find the predicted amount of electrical power the panel can produce, which is given by the dot product of vectors \(\mathbf{F}\) and \(\mathbf{n}\) (expressed in watts). c. Determine the angle of elevation of the Sun above the solar panel. Express the answer in degrees rounded to the nearest whole number. (Hint: The angle between vectors \(\mathbf{n}\) and \(\mathbf{s}\) and the angle of elevation are complementary.)

Step by step solution

01

Find Position Vectors

First, we find vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) based on the given coordinates.\[\overrightarrow{AB} = B - A = (8, 18, 0) - (8, 0, 0) = \langle 0, 18, 0 \rangle \\overrightarrow{AD} = D - A = (0, 0, 8) - (8, 0, 0) = \langle -8, 0, 8 \rangle\]

02

Calculate Vector Cross Product

Next, calculate the cross product \( \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AD} \).\[\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 18 & 0 \ -8 & 0 & 8 \end{vmatrix} = \mathbf{i}(18 \times 8 - 0 \times 0) - \mathbf{j}(0 \times 8 - 0 \times -8) + \mathbf{k}(0 \times 0 - (-8) \times 18)\]\[= 144 \mathbf{i} + 0 \mathbf{j} + 144 \mathbf{k} \]Thus, \( \mathbf{n} = 144 \mathbf{i} + 0 \mathbf{j} + 144 \mathbf{k} \).

03

Find Dot Product for Power

Using the given \( \mathbf{F} = 900 \mathbf{s} \), compute the dot product \( \mathbf{F} \cdot \mathbf{n} \):\[\mathbf{F} = 900 \left( \frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k} \right) = 300 \sqrt{3} \mathbf{i} + 300 \sqrt{3} \mathbf{j} + 300 \sqrt{3} \mathbf{k}\]\[\mathbf{F} \cdot \mathbf{n} = (300 \sqrt{3} \mathbf{i} + 300 \sqrt{3} \mathbf{j} + 300 \sqrt{3} \mathbf{k}) \cdot (144 \mathbf{i} + 0 \mathbf{j} + 144 \mathbf{k}) \]\[= 43200 \sqrt{3} + 0 + 43200 \sqrt{3} = 86400 \sqrt{3} \]Thus, the predicted amount of electrical power is \( 86400 \sqrt{3} \) watts.

04

Calculate Angle of Elevation

First, find the angle between \( \mathbf{n} \) and \( \mathbf{s} \) using the dot product formula:\[\cos \theta = \frac{\mathbf{n} \cdot \mathbf{s}}{\|\mathbf{n}\| \|\mathbf{s}\|}\]Compute norms:\[\|\mathbf{n}\| = \sqrt{144^2 + 0^2 + 144^2} = 144\sqrt{2}, \quad \|\mathbf{s}\| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2} = 1\]Calculate\( \mathbf{n} \cdot \mathbf{s} \):\[\mathbf{n} \cdot \mathbf{s} = 144 \times \frac{1}{\sqrt{3}} + 0 \times \frac{1}{\sqrt{3}} + 144 \times \frac{1}{\sqrt{3}} = 96 \sqrt{3}\]\[\cos \theta = \frac{96 \sqrt{3}}{144\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}\left(\frac{2}{3}\sqrt{2} \right)= \frac{1}{\sqrt{2}}\]\[\theta = 45^\circ\]Since \( \theta + \text{elevation angle} = 90^\circ \), the angle of elevation is:\[90^\circ - 45^\circ = 45^\circ\]

05

Conclusion

The vector \( \mathbf{n} \) perpendicular to the solar panel is \( 144\mathbf{i} + 0\mathbf{j} + 144\mathbf{k}\). The power output is \( 86400\sqrt{3} \) watts and the angle of elevation of the Sun is \(45^\circ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product

The cross product is a fundamental operation in vector calculus, essential for finding a vector that is perpendicular to two given vectors. Understanding this concept is critical for tasks like determining the direction normal to a surface, as seen in our exercise involving a solar panel. When you have two vectors in three-dimensional space, defined as \( \overrightarrow{AB}\) and \( \overrightarrow{AD}\), the cross product \( \overrightarrow{AB} \times \overrightarrow{AD}\) results in a new vector, \( \mathbf{n} \), which is perpendicular to both \( \overrightarrow{AB}\) and \( \overrightarrow{AD}\). This can be computed using the determinant formula:

• Arrange the vectors in matrix form.
• Add the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) as the first row.
• Use the coordinates of the vectors as subsequent rows.

Finally, solve the determinant to find the vector components. For our example, the cross product resulted in \( \mathbf{n} = 144 \mathbf{i} + 0 \mathbf{j} + 144 \mathbf{k} \), signifying the direction perpendicular to the solar panel.

Dot Product

The dot product of two vectors is a scalar quantity that measures the extent to which two vectors are aligned. In practical applications, like calculating potential energy output from a solar panel, the dot product determines how much of one vector acts in the direction of another. In this context, it is used to predict the amount of solar energy hitting the panel. The formula involves multiplying corresponding components of the vectors and summing the results:

• Given \( \mathbf{F} = 900 \mathbf{s} \, (900 \text{ watts per meter squared}) \) and \( \mathbf{n} = 144 \mathbf{i} + 0 \mathbf{j} + 144 \mathbf{k} \).
• Calculate the dot product: \( \mathbf{F} \cdot \mathbf{n} = 86400\sqrt{3} \) watts.

This measure illustrates the influence of the solar vector \( \mathbf{s} \) on the panel's surface normal vector \( \mathbf{n} \), showing that the solar panel effectively absorbs energy in this alignment.

Angle of Elevation

The angle of elevation is the angle between the horizontal plane and a line pointing toward an object above the plane. In the context of our solar panel problem, it determines the angle at which sunlight strikes the panel. The relationship between vectors \( \mathbf{n} \) and \( \mathbf{s} \) is essential here. We've established that the angle between \( \mathbf{n} \) and \( \mathbf{s} \) is complementary to the angle of elevation, so:

• Find the cosine of the angle utilizing the dot product ratio: \( \cos \theta = \frac{\mathbf{n} \cdot \mathbf{s}}{\|\mathbf{n}\| \|\mathbf{s}\|} \).
• We computed \( \theta = 45^\circ \), implying that sunlight strikes the panel at an angle equal to the angle of elevation.

Understanding this concept helps dictate the panel's ability to capture solar energy efficiently, emphasizing the importance of correctly aligning solar panels to optimize for the angle of elevation.