Question

Show that the angle between two diagonals of a cube is cos inverse 1/3

Answer 1

 



Let OABCDEFG be a cube with vertices as below

O(0,0,0), A(a,0,0), B(a,a,0), C(0,a,0),

D(0,a,a), E(0,0,a), F(a,0,a) and G(a,a,a)

There are four diagonals OG,CF,AD and BE for the cube.

Let us consider any two say OG and AD

We know that if A(x1,y1,z1) and B(x2,y2,z2) A(x1,y1,z1) and B(x2,y2,z2) are two points in space then AB−→−=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^AB→=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^

⇒OG−→−=(a−0)i^+(a−0)j^+(a−0)k^=ai^+aj^+ak^⇒OG→=(a−0)i^+(a−0)j^+(a−0)k^=ai^+aj^+ak^ and

AD−→−=(0−a)i^+(a−0)j^+(a−0)k^=−ai^+aj^+ak^AD→=(0−a)i^+(a−0)j^+(a−0)k^=−ai^+aj^+ak^

|OG−→−|=a2+a2+a2−−−−−−−−−−√=3–√a|OG→|=a2+a2+a2=3a

|AD−→−|=(−a)2+a2+a2−−−−−−−−−−−−−√=3–√a|AD→|=(−a)2+a2+a2=3a

OG−→−.AD−→−=−a2+a2+a2=a2OG→.AD→=−a2+a2+a2=a2

We know that angle between any two vectors a→andb→a→andb→ =cos−1(a→.b→|a→||b→|)=cos−1(a→.b→|a→||b→|)

⇒⇒Angle between the two diagonals OG−→−OG→ and AD−→−AD→=

cos−1(OG−→−.AD−→−|OG−→−||AD−→−|)cos−1(OG→.AD→|OG→||AD→|)

=cos−1(a23√a.3√a)=cos−1a23a2=cos−1(a23a.3a)=cos−1a23a2

=cos−113=cos−113

Hence proved.