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Chapter 5: Problem 111

Find the angle between the diagonal of a cube and the diagonal of one of its sides.

Short Answer

Expert verified
The angle between the diagonal of a cube and the diagonal of one of its sides is found by taking the inverse cosine of \(\frac{3\sqrt{2}}{2}\), which is approximately 35.26 degrees.

Step by step solution

01

Calculate the diagonals

The diagonal of the cube can be found by applying Pythagoras' theorem in three dimensions: \(d_c = \sqrt{l^2 + l^2 + l^2} = l\sqrt{3}\). The diagonal of one of the cube's sides can be calculated with Pythagoras' theorem in two dimensions: \(d_s = \sqrt{l^2 + l^2} = l\sqrt{2}\).
02

Find the angle

The angle \(胃\) between these two diagonals can be found using the dot product formula: \(cos(胃) = \frac{V_1.V_2}{||V_1||.||V_2||}\). In this case, because the diagonals of the cube and its side both originate from the same vertex and extend to the midpoints of the opposite edges and faces, respectively, the vectors \(V_1\) and \(V_2\) representing these diagonals have the same components, making their dot product \(l^2+l^2+l^2 = 3l^2\) and their magnitudes \(||V_1|| = l\sqrt{3}\), \(||V_2|| = l\sqrt{2}\). Thus \(cos(胃) = \frac{3l^2}{l\sqrt{6}} = \frac{3\sqrt{2}}{2}\). Therefore, the angle can be found by taking the inverse cosine of this value.

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Most popular questions from this chapter

Find the area of the triangle with the given vertices. Use the fact that the area of the triangle having \(\mathbf{u}\) and \(\mathbf{v}\) as adjacent sides is given by \(A=\frac{1}{2}\|\mathbf{u} \times \mathbf{v}\|\). $$(2,-3,4),(0,1,2),(-1,2,0)$$

Use a graphing utility with vector capabilities to find \(\mathbf{u} \times \mathbf{v}\) and then show that it is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). $$\mathbf{u}=2 \mathbf{i}-\mathbf{j}+\mathbf{k}, \quad \mathbf{v}=\mathbf{i}-2 \mathbf{j}+\mathbf{k}$$

Find the least squares quadratic polynomial for the data points. $$(-2,6),(-1,5),\left(0, \frac{7}{2}\right),(1,2),(2,-1)$$

Prove that \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\).

Find the area of the parallelogram that has the vectors as adjacent sides. $$\mathbf{u}=\mathbf{j}, \quad \mathbf{v}=\mathbf{j}+\mathbf{k}$$
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