Both sides are antisymmetric in

j, k

and in

p, q

. If either pair has equal indices, both sides vanish. It therefore suffices to take

j \neq = k

and

p \neq = q

. The sum over

i

is nonzero exactly when

(p, q) = (j, k)

(k, j)

; it is respectively

1

- 1

. This is exactly

ϵ_{ijk} ϵ_{i pq} = δ_{j p} δ_{k q} - δ_{j q} δ_{k p},

ii

Solution

Expanding a sum of squares gives the

n

\frac{1}{2} \sum_{i, j} (x_{i} y_{j} - x_{j} y_{i})^{2} = \sum_{i, j} x_{i}^{2} y_{j}^{2} - \sum_{i, j} x_{i} x_{j} y_{i} y_{j} = ∣ x ∣^{2} ∣ y ∣^{2} - (x \cdot y)^{2} .

The left-hand side is nonnegative, so

∣ x ∣^{2} ∣ y ∣^{2} \geq ∣ x \cdot y ∣^{2} .

In components, part (i) gives

[a \times (b \times c)]_{i} = ϵ_{ijk} a_{j} ϵ_{k l m} b_{l} c_{m} = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) a_{j} b_{l} c_{m} = b_{i} (a \cdot c) - c_{i} (a \cdot b) .

a \times (b \times c) = b (a \cdot c) - c (a \cdot b) .

There are no discussions about this article yet.

There are currently no matching articles.