Math, asked by anamikakr038, 3 months ago

Find the angle between any two diagonals of a
cube​

Answers

Answered by danishjibran
0

Answer: 90

Step-by-step explanation:

Answered by helenfaustina12
1

Answer:

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),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(x

1

,y

1

,z

1

) and B(x

2

,y

2

,z

2

) are two points in space then

AB

=(x

2

−x

1

)i+(y

2

−y

1

)j+(z

2

−z

1

)k

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.

Therefore ∣

OG

∣=

a

2

+a

2

+a

2

=a

3

and ∣

AD

∣=

(−a)

2

+a

2

+a

2

=a

3

OG

AD

=−a

2

+a

2

+a

2

=a

2

We know that angle between two vectors

a

,

b

is given by θ=cos

−1

a

∣∣

b

a

b

Thus angle between two diagonals

OG

and

AD

is

θ=cos

−1

a

3

×a

3

a

2

=cos

−1

(

3

1

)

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