We need to prove that the scalar triple product of the vectors $a+b$, $b+c$, and $c+a$ is equal to twice the scalar triple product of the vectors $a$, $b$, and $c$. In other words, we need to prove: $[a+b, b+c, c+a] = 2[a, b, c]$.

GeometryVector AlgebraScalar Triple ProductVector Operations3D Geometry
2025/6/15

1. Problem Description

We need to prove that the scalar triple product of the vectors a+ba+b, b+cb+c, and c+ac+a is equal to twice the scalar triple product of the vectors aa, bb, and cc. In other words, we need to prove: [a+b,b+c,c+a]=2[a,b,c][a+b, b+c, c+a] = 2[a, b, c].

2. Solution Steps

We will use the properties of the scalar triple product to expand the left side of the equation:
[a+b,b+c,c+a]=(a+b)((b+c)×(c+a))[a+b, b+c, c+a] = (a+b) \cdot ((b+c) \times (c+a))
First, expand the cross product:
(b+c)×(c+a)=b×c+b×a+c×c+c×a(b+c) \times (c+a) = b \times c + b \times a + c \times c + c \times a
Since c×c=0c \times c = 0, we have:
(b+c)×(c+a)=b×c+b×a+c×a(b+c) \times (c+a) = b \times c + b \times a + c \times a
Now, take the dot product with (a+b)(a+b):
(a+b)(b×c+b×a+c×a)=a(b×c+b×a+c×a)+b(b×c+b×a+c×a)(a+b) \cdot (b \times c + b \times a + c \times a) = a \cdot (b \times c + b \times a + c \times a) + b \cdot (b \times c + b \times a + c \times a)
Distribute the dot product:
a(b×c)+a(b×a)+a(c×a)+b(b×c)+b(b×a)+b(c×a)a \cdot (b \times c) + a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times c) + b \cdot (b \times a) + b \cdot (c \times a)
Recall that if two vectors in a scalar triple product are the same, the scalar triple product is zero.
Therefore, a(b×a)=0a \cdot (b \times a) = 0, a(c×a)=0a \cdot (c \times a) = 0, b(b×c)=0b \cdot (b \times c) = 0, and b(b×a)=0b \cdot (b \times a) = 0.
We are left with:
a(b×c)+b(c×a)a \cdot (b \times c) + b \cdot (c \times a)
We know that a(b×c)=[a,b,c]a \cdot (b \times c) = [a, b, c] and b(c×a)=[b,c,a]b \cdot (c \times a) = [b, c, a].
Also, the scalar triple product is invariant under cyclic permutations: [a,b,c]=[b,c,a]=[c,a,b][a, b, c] = [b, c, a] = [c, a, b].
Therefore, b(c×a)=[b,c,a]=[a,b,c]b \cdot (c \times a) = [b, c, a] = [a, b, c].
So we have:
[a,b,c]+[a,b,c]=2[a,b,c][a, b, c] + [a, b, c] = 2[a, b, c]
Thus, [a+b,b+c,c+a]=2[a,b,c][a+b, b+c, c+a] = 2[a, b, c].

3. Final Answer

[a+b,b+c,c+a]=2[a,b,c][a+b, b+c, c+a] = 2[a, b, c]

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