The cosine of the angle between any two diago | vedclass

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(A) Let the vertices of the cube be represented in a $3D$ coordinate system. Consider a cube with side length $a$. The four diagonals of the cube conn

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(A) Let the vertices of the cube be represented in a $3D$ coordinate system. Consider a cube with side length $a$. The four diagonals of the cube connect opposite vertices. Let us consider two diagonals,for example,the diagonal from $(0,0,0)$ to $(a,a,a)$ and the diagonal from $(a,0,0)$ to $(0,a,a)$.The direction vector of the first diagonal $\vec{d_1}$ is $(a-0, a-0, a-0) = (a, a, a)$.The direction vector of the second diagonal $\vec{d_2}$ is $(0-a, a-0, a-0) = (-a, a, a)$.The cosine of the angle $	heta$ between these two vectors is given by the formula:$\cos 	heta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|}$Calculating the dot product:$\vec{d_1} \cdot \vec{d_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2$Calculating the magnitudes:$|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$$|\vec{d_2}| = \sqrt{(-a)^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$Therefore,$\cos 	heta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.Thus,the cosine of the angle between any two diagonals of a cube is $1/3$.

The cosine of the angle between any two diagonals of a cube is:

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