14.7. PLANES 303

and so(u×v)1 = (u2v3 −u3v2) .

From the above formula in the proposition,

ε1 jku jvk ≡ u2v3 −u3v2,

the same thing. The cases for (u×v)2 and (u×v)3 are verified similarly. The last claimfollows directly from the definition.

With this notation, you can easily discover vector identities and simplify expressionswhich involve the cross product.

Example 14.6.5 Discover a formula which simplifies (u×v)×w.

From the above reduction formula,

((u×v)×w)i = ε i jk (u×v) j wk = ε i jkε jrsurvswk

= −ε jikε jrsurvswk =−(δ irδ ks −δ isδ kr)urvswk

= −(uivkwk −ukviwk) = u ·wvi −v ·wui

= ((u ·w)v− (v ·w)u)i .

Since this holds for all i, it follows that

(u×v)×w = (u ·w)v− (v ·w)u.

14.7 PlanesThis section concerns something called a level surface of a function of many variables. Itis a little outside the goals of this book but it seems a shame not to consider it because it isa nice illustration of the geometric significance of the dot product. To find the equation of aplane, you need two things, a point contained in the plane and a vector normal to the plane.Let p0 = (x0,y0,z0) denote the position vector of a point in the plane, let p= (x,y,z) be theposition vector of an arbitrary point in the plane, and let n denote a vector normal to theplane. This means that

n·(p−p0) = 0

whenever p is the position vector of a point in the plane. The following picture illustratesthe geometry of this idea.

p0p

n

Expressed equivalently, the plane is just the set of all points p such that the vectorp−p0 is perpendicular to the given normal vector n.