已知 $a$、$b$ 和 $c$ 是非零向量。 若 $|a+b| = |a-b|$，证明这两个向量 $a$ 和 $b$ 互相垂直。 (i) 解释为什么 $a\times(b+c)+b\times(c+a)+c\times(a+b)$ 的结果是一个向量。 (ii) 化简 $a\times(b+c)+b\times(c+a)+c\times(a+b)$。清晰地写出你的求解过程。 解释为什么 $a\times(b+c)+b\times(c+a)+c\times(a+b)$ 的结果是一个向量。 化简 $a\times(b+c)+b\times(c+a)+c\times(a+b)$。清晰地写出你的求解过程。

1. Square both sides of the given equation to relate magnitude to scalar product.

$|a+b| = |a-b|$

$|a+b|^2 = |a-b|^2$

2. Expand both sides using the scalar product property $|v|^2 = v \cdot v$.

$(a+b)\cdot(a+b) = (a-b)\cdot(a-b)$

$a\cdot a + a\cdot b + b\cdot a + b\cdot b = a\cdot a - a\cdot b - b\cdot a + b\cdot b$

$|a|^2 + 2(a\cdot b) + |b|^2 = |a|^2 - 2(a\cdot b) + |b|^2$

3. Simplify the equation to solve for $a \cdot b$.

$2(a\cdot b) = -2(a\cdot b)$

$4(a\cdot b) = 0$

$a\cdot b = 0$

4. Conclude the geometric relationship between the vectors.

Since $a \cdot b = 0$ and it is given that $a \neq 0$ and $b \neq 0$, the angle $\theta$ between them satisfies $\cos \theta = 0$.

Thus, $\theta = 90^\circ$, which means $a$ and $b$ are perpendicular to each other. (Shown)

1. Explain how the vector operations in the expression guarantee a vector result

The addition of two vectors (e.g., $b+c$) results in a vector. The vector product (cross product) of two vectors (e.g., $a \times (b+c)$) also yields a vector. Since the expression is a sum of these individual vector products, the final result is a vector.

1. Expand the expression using the distributive property of the vector product over vector addition.

$a\times(b+c)+b\times(c+a)+c\times(a+b)$
$= a\times b + a\times c + b\times c + b\times a + c\times a + c\times b$

2. Apply the anti-commutative property of the vector product to match like terms.

Since $u\times v = -v\times u$ for any vectors $u$ and $v$:
$b\times a = -a\times b$
$c\times a = -a\times c$
$c\times b = -b\times c$
Substitute these into the expression:
$= a\times b + a\times c + b\times c - a\times b - a\times c - b\times c$

3. Simplify the expression to find the final resulting vector.

$= (a\times b - a\times b) + (a\times c - a\times c) + (b\times c - b\times c)$
$= \mathbf{0}$