You already know distribution from arithmetic: a(b+c)=ab+ac — multiplication "spreads" over addition. The cross product obeys the same kind of rule, with one important twist: it is not commutative, so the order of the vectors must be respected.
Intuition
Put two vectors u and v at a common point; their sum u+v is the diagonal of the parallelogram they form. Crossing a third vector w with this sum, w×(u+v), gives the same result as crossing w with each piece separately and adding. This works because the cross product is bilinear — its geometry (parallelogram area, right-hand rule) is linear in each argument.
The Precise Statement
For any vectors a,b,c in R3 and any scalar k:
Left distributivity:a×(b+c)=a×b+a×c
Right distributivity:(b+c)×a=b×a+c×a
Scalar multiplication:(ka)×b=k(a×b)=a×(kb)
Watch out
The cross product is anti-commutative: a×b=−(b×a). So left and right distributivity are different statements — you may not swap the order across the × without flipping the sign. In particular,
a×(b+c)=a×b+a×c,
nota×b+c×a. Keep the left vector on the left.
Why It Matters
Distributivity lets you expand a product of sums term by term, exactly like FOIL in algebra:
(p+q)×(r+s)=p×r+p×s+q×r+q×s.
Every term keeps its left–right order intact. This is the routine behind expanding cross products in proofs (areas of triangles, testing collinearity, deriving vector identities). …
Crossing a+b+c=0 with a and with b (using x×x=0 and anti-commutativity) gives a×b=b×c=c×a; geometrically all three equal twice the vector area of the triangle the vectors form.
Tools we use
Two cross-product facts do all the work:
x×x=0 (a vector crossed with itself is zero),
x×y=−(y×x) (anti-commutativity).
Step 1: cross the relation with a
Start from a+b+c=0 and take the cross product of both sides with a on the left:
Method: Cross-multiplying a vector identity to expose equal cross products
Use this reasoning pattern when a relation among vectors (like a+b+c=0) must be converted into statements about cross products.
Steps
Step 1: Recall the two cross-product facts that do the work.
x×x=0andx×y=−(y×x).
The first kills self-terms; the second lets you flip an order at the cost of a sign.
Step 2: Cross the whole relation by one vector at a time.
Taking ×a of a+b+c=0 removes a×a and yields one equality; taking ×b yields another. Use anti-commutativity to rewrite each term in a consistent order. …
Mistake 1: Ignoring anti-commutativity when reordering cross products.
Why it's wrong: a×c=−c×a, so the sign must flip when you rewrite a term; dropping it produces a false equality. Correct approach: apply x×y=−(y×x) every time you swap.
Mistake 2: Forgetting that a×a=0.
Why it's wrong: the whole method relies on the self-cross term vanishing when you cross the relation with a or b. Correct approach: cancel a×a and b×b to zero. …
Expanding via u=aˉ−bˉ and discarding degenerate (repeated-vector) scalar triple products collapses the whole expression to 3[aˉbˉcˉ].
Concept and Intuition
Whenever a scalar triple product [xˉyˉzˉ]=xˉ⋅(yˉ×zˉ) has a repeated vector among x,y,z, it is automatically zero (the vectors can't span a parallelepiped of nonzero volume, or equivalently yˉ×zˉ is perpendicular to both yˉ,zˉ, so dotting with either gives 0). This is the key simplification tool here.
Step-by-Step Solution
Let u=aˉ−bˉ. Then (aˉ−bˉ)×(aˉ−bˉ−cˉ)=u×(u−cˉ)=u×u−u×cˉ=−u×cˉ.
So the bracketed term equals −(aˉ−bˉ)×cˉ=cˉ×(aˉ−bˉ)=cˉ×aˉ−cˉ×bˉ.
Dot with (aˉ+2bˉ−cˉ): (aˉ+2bˉ−cˉ)⋅(cˉ×aˉ−cˉ×bˉ)
Expand into six scalar triple products; four vanish because they repeat a vector: aˉ⋅(cˉ×aˉ)=0, bˉ⋅(cˉ×bˉ)=0, cˉ⋅(cˉ×aˉ)=0, cˉ⋅(cˉ×bˉ)=0.
What remains: −aˉ⋅(cˉ×bˉ)+2bˉ⋅(cˉ×aˉ)=−[aˉcˉbˉ]+2[bˉcˉaˉ]. …
Q.If aˉ=2iˉ−5jˉ+8kˉ, bˉ=7iˉ−5jˉ+3kˉ are two vectors and (2aˉ−3bˉ)×(4aˉ+bˉ)=xiˉ+yjˉ+zkˉ, then x+y+z=
(A) −1000
(B) 1400
(C) 1000
(D) −1400
›Reveal solutionSolution
Expand the cross product bilinearly; the self-cross terms vanish, leaving 14(aˉ×bˉ), which computes to (350,700,350) and sums to 1400.
Concept and Intuition
Cross product distributes over addition and is bilinear, and vˉ×vˉ=0 for any vector. So (2aˉ−3bˉ)×(4aˉ+bˉ) collapses to a single multiple of aˉ×bˉ — no need to compute the full 3×3 determinant with the combined vectors.