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Proof of the Vector Triple Product

DefinitionFormulaProofPropertiesSolved Examples

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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

aβ†’,bβ†’,cβ†’ is defined as the cross product of vector aβ†’
   
with the cross product of vectors bβ†’ and cβ†’,
i.e., aβ†’Γ—(bβ†’Γ—cβ†’)

Here,

aβ†’Γ—(bβ†’Γ—cβ†’) is coplanar with the vectors bβ†’ and cβ†’ and perpendicular to aβ†’.

Hence we can write

aβ†’Γ—(bβ†’Γ—cβ†’) as linear combination of vectors bβ†’ and cβ†’.

That is,

a→×(b→×c→)=xb→+yc→

Vector Triple Product Formula

aβ†’Γ—(bβ†’Γ—cβ†’)=(aβ†’.cβ†’)bβ†’ β€“(aβ†’.bβ†’)cβ†’

and

(aβ†’Γ—bβ†’)Γ—cβ†’=(aβ†’.cβ†’)bβ†’ β€“(bβ†’.cβ†’)aβ†’

In general,

a→×(b→×c→)≠(a→×b→)×c→

Vector Triple Product Proof

We can write

(aβ†’Γ—bβ†’)Γ—cβ†’ as a linear combination of vectors aβ†’ and bβ†’.

So,

(a→×b→)×c→=xa→+yb→
⇒c→.(a→×b→)×c→=c→.(xa→+yb→)
=x.(c→.a→)+y(c→.b→)
⇒0=x.(a→.c→)+y(b→.c→)
β‡’xbβ†’.cβ†’=βˆ’yaβ†’.cβ†’=Ξ»
Substituting value of x and y in (aβ†’Γ—bβ†’)Γ—cβ†’=xaβ†’+ybβ†’,
we have;

(aβ†’Γ—bβ†’)Γ—cβ†’=(Ξ»bβ†’.cβ†’)aβ†’+(βˆ’Ξ»aβ†’.cβ†’)bβ†’
=(Ξ»bβ†’.cβ†’)a→–(Ξ»aβ†’.cβ†’)bβ†’
It is valid for every value of aβ†’,bβ†’,cβ†’ because it is an identity.

Put

aβ†’=i^,bβ†’=j^ and cβ†’=i^
β‡’(i^Γ—j^)Γ—i^=(Ξ»j^.i^)i^–(Ξ»i^.i^)j^
β‡’j^=–λj^
β‡’Ξ»=βˆ’1

Hence,

(aβ†’Γ—bβ†’)Γ—cβ†’=(aβ†’.cβ†’)b→–(bβ†’.cβ†’)aβ†’

Properties

  • A vector triple product is a vector quantity.
  • Unit vector coplanar with aβ†’ and bβ†’
     
    and perpendicular to cβ†’ is Β±(aβ†’Γ—bβ†’)Γ—cβ†’|(aβ†’Γ—bβ†’)Γ—cβ†’|.
  • aβ†’Γ—(bβ†’Γ—cβ†’)β‰ (aβ†’Γ—bβ†’)Γ—cβ†’
    .
  • Vector Triple Product Properties

Note that,

if aβ†’,bβ†’,cβ†’ are non-coplanar vector then
 
aβ†’Γ—bβ†’,bβ†’Γ—cβ†’ and cβ†’Γ—aβ†’ are also non-coplanar.

 

Some other useful results:

Vector Triple Product Useful Results

Solved Examples

Example 1:

Find the value of i^Γ—(j^Γ—k^)+j^Γ—(k^Γ—i^).

Solution:

i^Γ—(j^Γ—k^)+j^Γ—(k^Γ—i^)=i^Γ—i^+j^Γ—βˆ’j^=1βˆ’1=0

Example 2:

If aβ†’,bβ†’,cβ†’ are three vectors such that
 
|aβ†’|=1, |bβ†’|=2, |cβ†’|=1, and aβ†’Γ—(aβ†’Γ—bβ†’)+cβ†’=0,
 
then find the acute angle between aβ†’ and bβ†’.

Solution:

Let A be the angle between aβ†’ and bβ†’.

Then,

aβ†’.bβ†’=|aβ†’||bβ†’|cos A

= 1.2. cos A

= 2 cos A

But

a→×(a→×b→)+c→=0
β‡’(aβ†’.bβ†’)a→–(aβ†’.aβ†’)bβ†’+cβ†’=0
β‡’2 cos Aa→–bβ†’+cβ†’=0
β‡’2 cos Aa→–bβ†’=βˆ’cβ†’

Squaring both sides, we have;

β‡’[2 cos Aa→–bβ†’]2=[βˆ’cβ†’]2
β‡’4cos2A|aβ†’|2–4cosAaβ†’.bβ†’+|bβ†’|2=|cβ†’|2

β‡’ 4 cos2 A – 4 cos A . 2 cos A + 4 = 1

β‡’ 4 cos2 A – 8 cos2 A + 4 = 1

β‡’4(1–cos2A)=1

β‡’ 4 sin2 A = 1

or sin A = 1/2

or A = Ο€/6

[neglected -ve value]

Example 3:

If aβ†’,bβ†’,cβ†’ are coplanar, then prove that
 
aβ†’Γ—bβ†’, bβ†’Γ—cβ†’, cβ†’Γ—aβ†’ are also coplanar.

Solution:

If aβ†’,bβ†’,cβ†’ are coplanar then
[a→b→c→]
⇒[a→b→c→]2=0
⇒[a→×b→b→×c→c→×a→]=0
So we can say that aβ†’Γ—bβ†’,bβ†’Γ—cβ†’,cβ†’Γ—aβ†’ are coplanar.

Example 4:

Let cβ†’=2i^+j^–2k^ and bβ†’=i^+j^
and if vector aβ†’ is such that cβ†’.aβ†’=|aβ†’|, |a→–cβ†’|=22
 
and angle between (cβ†’Γ—bβ†’) and aβ†’ is Ο€6,
 
then find the value of |(cβ†’Γ—bβ†’)Γ—aβ†’|.

Solution:

cβ†’=2i^+j^–2k^ and bβ†’=i^+j^
so,

|cβ†’|=3 and |bβ†’|=2
|a→–cβ†’|=22 (Given)

Squaring both sides, we have;

Example 5: If 

aΓ—b=c,bΓ—c=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 Γ— c and a Γ— b = c

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

Therefore, a, b, and c are mutually perpendicular.

Now, a = b Γ— c = b Γ— (a Γ— b) = (b . b) a βˆ’ (b . a) b or

a=b2aβˆ’(b.a)b=b2a,{becauseaβŠ₯b}β‡’1=b2,∴c=aΓ—b=absin⁑90∘n^

Take the moduli of both sides, then c = ab, but b = 1 β‡’ c = a.

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

x + y = a …..(i)

x Γ— y = b …..(ii)

x . a = 1 …..(iii)

Then find the values of x and y.

Solution: 

Multiplying (i) by scalar β€œa”, we get;

a . x + a . y = a2

∴ a . y = a2 βˆ’ 1 ..(iv),

{By (iii)} Again a Γ— (x Γ— y) = a Γ— b or (a . y) x βˆ’ (a . x) y = a Γ— b

(a2 βˆ’ 1) x βˆ’ y = a Γ— b ..(v),

Adding and subtracting (i) and (v), we get;

x = [a + (a Γ— b)] / [a2] and y = a βˆ’ x

Vector Triple Point

36,926

Frequently Asked Questions

Q1

What do you mean by vector triple product?

Let a, b, and c be three vectors. The vector product of a, b, and c is the cross product of vector a with the cross product of vector b and vector c.

Q2

Give the vector triple product formula.

If a, b, c are three vectors, then
a Γ— (b Γ— c) = (a.c)b – (a.b)c.
(a Γ— b) Γ— c = (a.c)b – (b.c)a.

Q3

Is the vector triple product associative?

No, the vector triple product is not associative.

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