Expanding Products of Vectors

Both products distribute

The vector product distributes over addition: a × (b + c) = a × b + a × c. The dot product does too: a · (b + c) = a · b + a · c.

Check it on a = (2, 1, 3), b = (1, 0, 2) and c = (0, 4, 1). b + c = (1, 4, 3), so a · (b + c) = 2 + 4 + 9 = 15. Separately, a · b = 2 + 0 + 6 = 8 and a · c = 0 + 4 + 3 = 7, and 8 + 7 = 15.

Expanding $|a+b{|}^2$|a + b|²

The square of a length is a dot product: $|v{|}^2=v\cdot v$|v|² = v · v. So $|a+b{|}^2=\left(a+b\right)\cdot \left(a+b\right)$|a + b|² = (a + b) · (a + b).

Distributivity lets this expand like brackets of numbers: a · a + a · b + b · a + b · b. Two facts tidy it. $a\cdot a=|a{|}^2$a · a = |a|², and the dot product is commutative, b · a = a · b. So $|a+b{|}^2=|a{|}^2+2a\cdot b+|b{|}^2$|a + b|² = |a|² + 2a · b + |b|².

In the same way, $|a-b{|}^2=|a{|}^2-2a\cdot b+|b{|}^2$|a − b|² = |a|² − 2a · b + |b|².

Lengths without components

The expansion needs only three numbers: the two lengths and a · b. With |a| = 3, |b| = 2 and a · b = 1, $|a+b{|}^2=9+2\times 1+4=15$|a + b|² = 9 + 2 × 1 + 4 = 15, so $|a+b|=\sqrt{15}$|a + b| = √15.

A second example: |a| = 4, |b| = 3 and a · b = 5 give $|a-b{|}^2=16-10+9=15$|a − b|² = 16 − 10 + 9 = 15.

The vector product keeps the order

(a + b) × (a − b) expands to a × a − a × b + b × a − b × b. The vector product is not commutative, so each term keeps the order it was written in.

a × a = 0 and b × b = 0. b × a = −(a × b), so b × a adds another −a × b. The result is −2(a × b).

Check it with a = (1, 0, 0) and b = (0, 1, 0). a + b = (1, 1, 0) and a − b = (1, −1, 0), and their vector product is (0, 0, −2). a × b = (0, 0, 1), and −2(a × b) = (0, 0, −2).

A difference of squares

With the dot product, $\left(a+b\right)\cdot \left(a-b\right)=|a{|}^2-a\cdot b+b\cdot a-|b{|}^2=|a{|}^2-|b{|}^2$(a + b) · (a − b) = |a|² − a · b + b · a − |b|² = |a|² − |b|², because the middle terms cancel.

This is 0 exactly when |a| = |b|. In a rhombus the two sides from one corner are equal in length, and the diagonals are a + b and a − b, so the diagonals of a rhombus are perpendicular.

The usual slips

Dropping the middle terms. $|a+b{|}^2$|a + b|² is not $|a{|}^2+|b{|}^2$|a|² + |b|²: the 2a · b is there unless a and b are perpendicular.

Adding the lengths. ${\left(|a|+|b|\right)}^2$(|a| + |b|)² is $|a+b{|}^2$|a + b|² only when a and b point the same way, which is when a · b = |a| |b|.

Swapping the order in a vector product. b × a = −(a × b), so (a + 2b) × (a − b) is −3(a × b), not a × b.