Proof of the Vector Triple Product

DefinitionFormulaProofPropertiesSolved Examples

Vector Triple Product is a branch in vector algebra where we deal with the cross product of three vectors. The value of the vector triple product can be found by the cross product of a vector with the cross product of the other two vectors. It gives a vector as a result. When we simplify the vector triple product it gives us an identity name as BAC – CAB identity.

Vector Triple Product Definition

Vector triple product of three vectors a,b,c\vec a, \vec b, \vec c is defined as the cross product of vector a\vec awith the cross product of vectors b and c\vec b\ and\ \vec c, i.e. a×(b×c)\vec a \times (\vec b \times \vec c).

Here a×(b×c)\vec a \times (\vec b \times \vec c) is coplanar with the vectors b and c\vec b\ and\ \vec c and perpendicular to a\vec a .

Hence we can write a×(b×c)\vec a \times (\vec b \times \vec c) as linear combination of vectors b and c\vec b\ and\ \vec c,

That is, a×(b×c)=xb+yc\vec a \times (\vec b \times \vec c) = x \vec b + y \vec c

Vector Triple Product Formula

a×(b×c)=(a.c)b –(a.b)c\vec a \times (\vec b \times \vec c) = (\vec a . \vec c) \vec b\ – (\vec a . \vec b) \vec c (a×b)×c=(a.c)b –(b.c)a(\vec a \times \vec b) \times \vec c = (\vec a . \vec c) \vec b\ – (\vec b . \vec c) \vec a

In general, a×(b×c)(a×b)×c\vec a \times (\vec b \times \vec c) \neq (\vec a \times \vec b) \times \vec c

Vector Triple Product Proof

We can write (a×b)×c(\vec a \times \vec b) \times \vec c as linear combination of vectors a and b\vec a\ and\ \vec b .

So, (a×b)×c=xa+yb(\vec a \times \vec b) \times \vec c = x \vec a + y \vec b

=> c.(a×b)×c=c.(xa+yb)\vec c . (\vec a \times \vec b) \times \vec c = \vec c . (x \vec a + y \vec b)

= x.(c.a+y(c.b)x . (\vec c . \vec a + y (\vec c . \vec b)

=> 0 = x.(a.c+y(b.c)x . (\vec a . \vec c + y (\vec b . \vec c)

=> xb.c=ya.c=λ\frac{x}{\vec b . \vec c} = \frac{-y}{\vec a . \vec c} = \lambda

Substituting value of x and y in = (a×b)×c=xa+yb(\vec a \times \vec b) \times \vec c = x \vec a + y \vec b we have,

(a×b)×c=(λb.c)a+(λa.c)b(\vec a \times \vec b) \times \vec c = (\lambda \vec b . \vec c) \vec a + (-\lambda \vec a . \vec c) \vec b

= (λb.c)a(λa.c)b(\lambda \vec b . \vec c) \vec a – (\lambda \vec a . \vec c) \vec b

It is valid for every value of a,b,c \vec a, \vec b, \vec c because it is an identity

Put a=i^,b=j^,c=i^ \vec a = \hat i , \vec b = \hat j, \vec c = \hat i

=> (i^×j^)×i^=(λj^.i^)i^(λi^.i^)j^(\hat i \times \hat j) \times \hat i = (\lambda \hat j . \hat i) \hat i – (\lambda \hat i . \hat i) \hat j

=> j^=λj^\hat j = – \lambda \hat j

=> λ=1\lambda = -1

Hence, (a×b)×c=(a.c)b(b.c)a(\vec a \times \vec b) \times \vec c = (\vec a . \vec c)\vec b – (\vec b . \vec c)\vec a

Properties

1. Vector triple product is a vector quantity.

2. Unit vector coplanar with a and b\vec a\ and\ \vec b and perpendicular to c\vec c is ±(a×b)×c(a×b)×c\pm \frac{(\vec a \times \vec b)\times \vec c}{|(\vec a \times \vec b)\times \vec c|}.

3. a×(b×c)(a×b)×c\vec a \times (\vec b \times \vec c) \neq (\vec a \times \vec b) \times \vec c.

4.

Vector Triple Product Properties

Note that if a,b,c \vec a, \vec b, \vec c are non coplanar vector then a×b,b×c and c×a \vec a \times \vec b, \vec b \times \vec c\ and\ \vec c \times \vec a are also non coplanar.

 

Some other useful results:

Vector Triple Product Useful Results

Solved Examples

Example 1: Find the value of i^×(j^×k^)+j^×(k^×i^) \hat i \times (\hat j \times \hat k) + \hat j \times (\hat k \times \hat i)

Solution: i^×(j^×k^)+j^×(k^×i^)+k^×(i^×j^)=i^×i^+j^×j^=0 \hat i \times (\hat j \times \hat k) + \hat j \times (\hat k \times \hat i) + \hat k \times (\hat i \times \hat j) = \hat i \times \hat i + \hat j \times \hat j = 0

Example 2: If a,b,c \vec a, \vec b, \vec c are three vectors such that a| \vec a| = 1, b| \vec b| = 2, c| \vec c| = 1 and a×(a×b)+c \vec a \times (\vec a \times \vec b) + \vec c = 0 then find the acute angle between a and b \vec a\ and\ \vec b.

Solution: let angle between a andb \vec a\ and \vec b is A, then

a.b=abcos A \vec a . \vec b = |\vec a||\vec b|cos\ A

= 1.2. cos A

= 2 cos A

But a×(a×b)+c \vec a \times (\vec a \times \vec b) + \vec c = 0

=> (a.b)a(a.a)b+c=0 (\vec a . \vec b)\vec a – (\vec a . \vec a)\vec b + \vec c = 0

=> 2 cos A ab+c=0\vec a – \vec b + \vec c = 0

=> 2 cos A ab=c\vec a – \vec b =- \vec c

Squaring both side we have

=> [2 cos A ab]2=[c]2\vec a – \vec b]^2 = [- \vec c]^2

=> 4cos2Aa24cosAa.b+b2=c24 cos^2 A |\vec a|^2 – 4 cos A \vec a . \vec b + |\vec b|^2 = |\vec c|^2

=> 4 cos2 A – 4 cos A . 2 cos A + 4 = 1

=> 4 cos2 A – 8 cos2 A + 4 = 1

=> 4(1cos2A)=14 (1 – cos^2 A) = 1

=> 4 sin2 A = 1

or sin A = 1/2

or A = π/6

Example 3: If a,b,c \vec a, \vec b, \vec c are coplanar then prove that a×b,b×c,c×a \vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a are also coplanar.

Solution: If a,b,c \vec a, \vec b, \vec c are coplanar then [a    b    c][\vec a\;\; \vec b\;\; \vec c]

=> [a    b    c]2[\vec a\;\; \vec b\;\; \vec c]^2 = 0

=> [a×b    b×c    c×a][\vec a \times \vec b\;\; \vec b \times \vec c\;\; \vec c \times \vec a] = 0

So we can say that a×b,b×c,c×a \vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a are coplanar.

Example 4: Let c=2i^+j^2k^ and b=i^+j^ \vec c = 2 \hat i + \hat j – 2 \hat k\ and\ \vec b = \hat i + \hat j and if vector a \vec a is such that c.a=a,ac \vec c . \vec a = |\vec a|, |\vec a – \vec c| = 2√2 and angle between (c×b)(\vec c \times \vec b) and a \vec a is π/6 then find the value of (c×b)×a|(\vec c \times \vec b) \times \vec a|.

Solution: c=2i^+j^2k^ and b=i^+j^ \vec c = 2 \hat i + \hat j – 2 \hat k\ and\ \vec b = \hat i + \hat j c |\vec c| = 3 and b |\vec b| = √2

ac |\vec a – \vec c| = 2√2 (Given)

Squaring both side we have

Example 5: If a×b=c,b×c=a\mathbf{a}\times \mathbf{b}=\mathbf{c},\,\,\mathbf{b}\times \mathbf{c}=\mathbf{a} and a, b, c be moduli of the vectors a, b, c respectively, then find the values of a and b.

Solution: 

a = b × c and a × b = c

∴ a is perpendicular to both b and c and c is perpendicular to both a and b.

a, b, c are mutually perpendicular

Now, a = b × c = b × (a × b) = (b . b) a − (b . a) b or a=b2a(b.a)b=b2a,{becauseab}1=b2,thereforec=a×b=absin90n^\mathbf{a}={{b}^{2}}\mathbf{a}-(\mathbf{b}\,.\,\mathbf{a})\mathbf{b}={{b}^{2}}\mathbf{a}, \left\{ \text because \,\mathbf{a}\,\bot \,\mathbf{b} \right\} \\\Rightarrow 1={{b}^{2}}, \\\text therefore \,\mathbf{c}=\mathbf{a}\times \mathbf{b}=ab\sin 90{}^\circ \,\mathbf{\hat{n}}

Take moduli of both sides, then c = ab, but b = 1 ⇒ c = a.

Example 6: Given the following simultaneous equations for vectors x and y.

x + y = a …..(i)

x × y = b …..(ii)

x . a = 1 …..(iii)

Then find the values of x and y.

Solution: 

Multiplying (i) scalarly by a, we get

a . x + a . y = a2

∴ a . y = a2 − 1 ..(iv),

{By (iii)} Again a × (x × y) = a × b or (a . y) x − (a . x) y = a × b

(a2 − 1) x − y = a × b ..(v),

Adding and subtracting (i) and (v),

we get x = [a + (a × b)] / [a2] and y = a − x

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