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Exercise 6.10 · Q4

Q.If a⃗,b⃗,c⃗\vec a,\vec b,\vec c are three unit vectors such that a⃗\vec a is perpendicular to b⃗\vec b, and is parallel to c⃗\vec c then (a⃗×b⃗)×c⃗(\vec a\times\vec b)\times\vec c is equal to

(1) a⃗\vec a
(2) b⃗\vec b
(3) c⃗\vec c
(4) 0⃗\vec 0
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Concept understanding — Vector Triple Product

The Vector Triple Product

When you cross three vectors together as a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}), the result is again a vector — this is the vector triple product. The remarkable thing is that it can always be rewritten without any cross products at all, using only dot products.

The Key Identity (the "BAC −- CAB" rule)

a⃗×(b⃗×c⃗)=(a⃗⋅c⃗) b⃗−(a⃗⋅b⃗) c⃗\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}

A memorable way to recall it: the answer is B times (A dot C) minus C times (A dot B) — "BAC minus CAB". The two survivors are b⃗\vec{b} and c⃗\vec{c} (the vectors inside the inner bracket); each is scaled by a dot product involving the outside vector a⃗\vec{a}.

Why the Result Lies in the Plane of b⃗\vec{b} and c⃗\vec{c}

The inner product b⃗×c⃗\vec{b}\times\vec{c} is perpendicular to the plane containing b⃗\vec{b} and c⃗\vec{c}. Crossing a⃗\vec{a} with that perpendicular swings the result back into the b⃗\vec{b}–c⃗\vec{c} plane. So the answer must be a combination λb⃗+μc⃗\lambda\vec{b}+\mu\vec{c} — and the identity tells you exactly what λ\lambda and μ\mu are.

Order Matters — the Product Is Not Associative

The brackets are not decoration. In general,

a⃗×(b⃗×c⃗)≠(a⃗×b⃗)×c⃗.\vec{a}\times(\vec{b}\times\vec{c}) \ne (\vec{a}\times\vec{b})\times\vec{c}.

The left side lies in the plane of b⃗,c⃗\vec{b},\vec{c}; the right side lies in the plane of a⃗,b⃗\vec{a},\vec{b}. Its own expansion is

(a⃗×b⃗)×c⃗=(a⃗⋅c⃗) b⃗−(b⃗⋅c⃗) a⃗,(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{b}\cdot\vec{c})\,\vec{a},

which is a different vector. Always keep the parentheses where they are given.

Watch out

A frequent slip is to "cancel" and write a⃗×(b⃗×c⃗)\vec{a}\times(\vec{b}\times\vec{c}) as some multiple of a⃗\vec{a}. It is not — the surviving vectors are b⃗\vec{b} and c⃗\vec{c}, never the outer vector. …

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