The cosine of the angle between any two diago | vedclass

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(A) Let the vertices of a cube be $(0,0,0)$ and $(a,a,a)$. The four diagonals of the cube can be represented by the vectors connecting opposite vertic

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(A) Let the vertices of a cube be $(0,0,0)$ and $(a,a,a)$. The four diagonals of the cube can be represented by the vectors connecting opposite vertices: $\vec{d_1} = (a,a,a)$,$\vec{d_2} = (-a,a,a)$,$\vec{d_3} = (a,-a,a)$,and $\vec{d_4} = (a,a,-a)$.Consider two diagonals with direction ratios $(1,1,1)$ and $(-1,1,1)$.The cosine of the angle $	heta$ between two vectors $\vec{u} = (a_1, b_1, c_1)$ and $\vec{v} = (a_2, b_2, c_2)$ is given by $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the values: $\cos 	heta = \frac{|(1)(-1) + (1)(1) + (1)(1)|}{\sqrt{1^2 + 1^2 + 1^2} \sqrt{(-1)^2 + 1^2 + 1^2}} = \frac{|-1 + 1 + 1|}{\sqrt{3} \sqrt{3}} = \frac{1}{3}$.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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