7  Vectors in 3D space

A vector in three-dimensional Euclidean space is specified by three components. We write \[ \mathbf{x}=(x_1,x_2,x_3)\in\mathbb{R}^3. \] Geometrically, \(\mathbf{x}\) can be regarded as a directed displacement in space. Its Euclidean length is \[ \|\mathbf{x}\|=\sqrt{x_1^2+x_2^2+x_3^2}. \] Vectors can be added componentwise and multiplied by scalars, exactly as for vectors in \(\mathbb{R}^n\).

Two particularly useful operations are available for vectors in three dimensions. The dot product of \(\mathbf{x}\) and \(\mathbf{y}\) is the scalar \[ \mathbf{x}\cdot\mathbf{y}=x_1y_1+x_2y_2+x_3y_3=\sum_{i=1}^3x_i y_i. \] It contains geometric information about the relative directions of the vectors: \[ \mathbf{x}\cdot\mathbf{y}=\|\mathbf{x}\|\,\|\mathbf{y}\|\cos\theta, \] where \(\theta\) is the angle between them. In particular, two nonzero vectors are perpendicular if and only if \(\mathbf{x}\cdot\mathbf{y}=0\).

The cross product \(\mathbf{x}\times\mathbf{y}\) is instead a vector. It is perpendicular to both \(\mathbf{x}\) and \(\mathbf{y}\), with direction determined by the right-hand rule, and its length is \[ \|\mathbf{x}\times\mathbf{y}\|=\|\mathbf{x}\|\,\|\mathbf{y}\|\sin\theta. \] In components, \[ \mathbf{x}\times\mathbf{y}=(x_2y_3-x_3y_2,\;x_3y_1-x_1y_3,\;x_1y_2-x_2y_1). \]

For manipulating vector expressions, it is often more convenient to work with components rather than with this explicit formula. The Kronecker delta \[ \delta_{ij}= \begin{cases} 1,&i=j,\\ 0,&i\neq j, \end{cases} \] acts as an identity for indices, while the Levi-Civita symbol \(\varepsilon_{ijk}\) encodes the orientation of three-dimensional space. It is \(+1\) for even permutations of \((1,2,3)\), \(-1\) for odd permutations, and zero whenever two indices are equal. With this notation the cross product has the compact representation \[ (\mathbf{x}\times\mathbf{y})_i=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}x_jy_k. \] This index notation is especially useful for deriving identities involving several dot and cross products: complicated-looking vector expressions can be reduced to systematic manipulations of sums, Kronecker deltas, and Levi-Civita symbols.

Exercises

Exercise 7.1 (The vector triple product and the Levi-Civita symbol) It is recommended that you also look at Exercise 3.12 before doing this exercise, and also Exercise 3.11 on the Kronecker delta.

Recall that the \(i\)th Cartesian component of the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be written as \[ (\mathbf{a}\times\mathbf{b})_i=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}a_jb_k, \] where \(\varepsilon_{ijk}\) is the Levi-Civita symbol.

You may also use the contraction identity \[ \sum_{i=1}^3\varepsilon_{ijk}\varepsilon_{imn}=\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km}, \] where \(\delta_{ij}\) is the Kronecker delta.

  1. Write the \(i\)th component of \(\mathbf{a}\times(\mathbf{b}\times\mathbf{c})\) using two Levi-Civita symbols. Be careful to use different dummy indices for the two cross products.

  2. Use the cyclic property \[ \varepsilon_{ijk}=\varepsilon_{kij} \] and the contraction identity above to show that \[ \sum_{k=1}^3\varepsilon_{ijk}\varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}. \]

  3. Use the Kronecker deltas to carry out the remaining sums and show that \[ [\mathbf{a}\times(\mathbf{b}\times\mathbf{c})]_i=b_i\sum_{j=1}^3a_jc_j-c_i\sum_{j=1}^3a_jb_j. \]

  4. Hence prove the vector triple-product identity \[ \mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b}). \] This identity is sometimes remembered as ``BAC minus CAB’’.