Find the angle between any two diagonals of a
cube
Answers
Answered by
0
Answer: 90
Step-by-step explanation:
Answered by
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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