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Chapter 1: Problem 10

By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle between any two diagonals of a cube, where e?ch diagonal connects diametrically opposite corners and passes through the center of the cube.

Short Answer

Expert verified
Answer: The smaller angle between any two diagonals of a cube is approximately \(70.53^\circ\).

Step by step solution

01

Express the diagonals as vectors

Let's first label the vertices of the cube as A, B, C, D, E, F, G, and H, where A is the origin (0,0,0). We can represent the diagonals of the cube with vectors connecting pairs of diametrically opposite points. Let AB = a, BC = b, DA = c. Connecting (0,0,0) to the diagonally opposite vertex E, we obtain the following diagonals: - For diagonal AE: \(\vec{d_1} = a+b+c\) - For diagonal BG: \(\vec{d_2} = -a+b+c\) - For diagonal DH: \(\vec{d_3} = a-b+c\) - For diagonal CF: \(\vec{d_4} = -a-b+c\)
02

Find the dot product of the diagonals

Using the dot product definition, the dot product of \(\vec{d_1}\) and \(\vec{d_2}\) is given by: \(\vec{d_1} \cdot \vec{d_2} = (a+b+c) \cdot (-a+b+c)\) By calculating the products: \(\vec{d_1} \cdot \vec{d_2} = -a^2+b^2+c^2+ab-ac+bc\) We can choose any pair of vectors since all the diagonal pairs will make the same angle with each other.
03

Find the magnitudes of the vectors

Now, we need to find the magnitudes of both vectors. This is given by: \(|\vec{d_1}| = \sqrt{a^2+b^2+c^2}\) \(|\vec{d_2}| = \sqrt{(-a)^2+b^2+c^2}\) As both vectors have the same magnitudes, we can denote them as M: \(M = \sqrt{a^2+b^2+c^2}\)
04

Apply the dot product formula

Applying the dot product formula to find the angle θ between \(\vec{d_1}\) and \(\vec{d_2}\): \(\cos(\theta) = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{-a^2+b^2+c^2+ab-ac+bc}{M^2}\)
05

Solve for the angle θ

Now let's solve for the angle θ: \(\cos(\theta) = \frac{-a^2+b^2+c^2+ab-ac+bc}{(a^2+b^2+c^2)}\) As we know that in a cube, all edges are equal in length, a = b = c. Let a = b = c = s. Substitute s into the equation: \(\cos(\theta) = \frac{-s^2+s^2+s^2+s^2-s^2+s^2}{3s^2} = \frac{s^2}{3s^2}\) Finally, take the inverse cosine: \(\theta = \arccos(\frac{s^2}{3s^2}) = \arccos(\frac{1}{3})\) The smaller angle between any two diagonals of the cube is approximately \(70.53^\circ\).

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Most popular questions from this chapter

Two unit vectors, \(\mathbf{a}_{1}\) and \(\mathbf{a}_{2}\), lie in the \(x y\) plane and pass through the origin. They make angles \(\phi_{1}\) and \(\phi_{2}\), respectively, with the \(x\) axis \((a)\) Express each vector in rectangular components; \((b)\) take the dot product and verify the trigonometric identity, \(\cos \left(\phi_{1}-\phi_{2}\right)=\cos \phi_{1} \cos \phi_{2}+\sin \phi_{1} \sin \phi_{2} ;(c)\) take the cross product and verify the trigonometric identity \(\sin \left(\phi_{2}-\phi_{1}\right)=\sin \phi_{2} \cos \phi_{1}-\cos \phi_{2} \sin \phi_{1}\)

A field is given as \(\mathbf{G}=\left[25 /\left(x^{2}+y^{2}\right)\right]\left(x \mathbf{a}_{x}+y \mathbf{a}_{y}\right) .\) Find \((a)\) a unit vector in the direction of \(\mathbf{G}\) at \(P(3,4,-2) ;(b)\) the angle between \(\mathbf{G}\) and \(\mathbf{a}_{x}\) at \(P\); (c) the value of the following double integral on the plane \(y=7\). $$ \int_{0}^{4} \int_{0}^{2} \mathbf{G} \cdot \mathbf{a}_{y} d z d x $$

(a) Express the field \(\mathbf{D}=\left(x^{2}+y^{2}\right)^{-1}\left(x \mathbf{a}_{x}+y \mathbf{a}_{y}\right)\) in cylindrical components and cylindrical variables. ( \(b\) ) Evaluate \(\mathbf{D}\) at the point where \(\rho=2, \phi=0.2 \pi\), and \(z=5\), expressing the result in cylindrical and rectangular components.

Given the points \(M(0.1,-0.2,-0.1), N(-0.2,0.1,0.3)\), and \(P(0.4,0,0.1)\), find \((a)\) the vector \(\mathbf{R}_{M N} ;(b)\) the dot product \(\mathbf{R}_{M N} \cdot \mathbf{R}_{M P} ;(c)\) the scalar projection of \(\mathbf{R}_{M N}\) on \(\mathbf{R}_{M P} ;(d)\) the angle between \(\mathbf{R}_{M N}\) and \(\mathbf{R}_{M P}\).

Three vectors extending from the origin are given as \(\mathbf{r}_{1}=(7,3,-2)\), \(\mathbf{r}_{2}=(-2,7,-3)\), and \(\mathbf{r}_{3}=(0,2,3)\). Find \((a)\) a unit vector perpendicular to both \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2} ;(b)\) a unit vector perpendicular to the vectors \(\mathbf{r}_{1}-\mathbf{r}_{2}\) and \(\mathbf{r}_{2}-\mathbf{r}_{3} ;\) (c) the area of the triangle defined by \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2} ;(d)\) the area of the triangle defined by the heads of \(\mathbf{r}_{1}, \mathbf{r}_{2}\), and \(\mathbf{r}_{3}\).
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