Scalar Triple Product of Vectors

Scalar triple product is one of the important concepts of the vector algebra in which we take the product of three vectors. This can be performed by taking dot product of one vector with the cross product of the other two vectors and results in some scalar quantity as dot product always gives some particular value.

Scalar triple product is also known as box product, triple scalar product and mixed product. In this section, aspirants will learn about geometrical interpretation of scalar triple product, formula and its expansion, properties and many more.

Definition

If a,b and c\vec{a}, \vec{b}\ and\ \vec{c} are three vectors, then scalar triple product is defined as the dot product of a\vec{a} with the cross product of b\vec{b} and c\vec{c}, i.e. dot product of a\vec{a} with b×c\vec{b} \times \vec{c} .

It is generally denoted by a.(b×c)\vec{a} . ( \vec{b} \times \vec{c}) or [a   b   c][\vec{a}\ \; \vec{b}\ \; \vec{c}]

Scalar triple product is read as box product of a,b,c\vec{a}, \vec{b}, \vec{c}.

Similarly we can define other scalar triple product as:

b.(c×a) or c.(a×b)\vec{b}. (\vec{c} \times \vec{a})\ or\ \vec{c} . (\vec{a} \times \vec{b})

Geometrical interpretation of Scalar Triple Product

We can find the volume of parallelopiped using the scalar triple product.

Scalar Triple Product

Area of parallelogram is cross product of vectors a and b\vec{a}\ and\ \vec{b}, i.e. a×b=absin(θ)\vec{a} \times \vec{b}= |\vec{a}||\vec{b}|\sin(\theta)

Height of parallelopiped is c cos(ϕ)|\vec{c}|\ cos(\phi).

Then volume of parallelopiped is (absin(θ))×(ccos(ϕ))=(aasin(θ)n^.c)=(a×b).c|(|\vec{a}|\vec{b}|sin(\theta)) \times (|\vec{c}|cos(\phi))| = |(|\vec{a}||\vec{a}|sin(\theta)\hat{n} . \vec{c})| = |(\vec{a} \times \vec{b}).\vec{c}|

So we can say that (a×b).c=[a   b   c]|(\vec{a} \times \vec{b}).\vec{c}| = |[\vec{a}\;\ \vec{b}\;\ \vec{c}]| is the volume of parallelopiped with coterminous edges a,b and c\vec{a}, \vec{b}\ and\ \vec{c}.

Formula

Scalar triple product equation is given as

(a×b).c=[a   b   c]=a1a2a3b1b2b3c1c2c3(\vec{a} \times \vec{b}).\vec{c}| = [\vec{a}\;\ \vec{b}\;\ \vec{c}] = \begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 & b_2 &b_3 \\ c_1& c_2 &c_3 \end{vmatrix}

Where a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^c=c1i^+c2j^+c3k^\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\\ \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\\ \vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}

Properties

Below are some of the important properties of scalar triple product:

1. If we interchange the position of (.) and (x), the result will be same i.e. a.(b×c)=(a×b).c\vec{a}. (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b} ). \vec{c}

2. Value of scalar triple product doesn’t change if we don’t change the cyclic order of a,b and c\vec{a}, \vec{b}\ and\ \vec{c} [a   b   c]=[b   c   a]=[c   a   b][\vec{a}\; \ \vec{b}\; \ \vec{c}] = [\vec{b}\; \ \vec{c}\; \ \vec{a}] = [\vec{c}\; \ \vec{a}\; \ \vec{b}]

3. If we change the cyclic order of the vectors then sign of scalar triple product is changed.

i.e. [a   b   c]=[a   c   b] or a.(b×c)=a.(c×b)[\vec{a}\; \ \vec{b}\; \ \vec{c}] = -[\vec{a}\; \ \vec{c}\; \ \vec{b}]\ or\ \vec{a}. (\vec{b} \times \vec{c}) = -\vec{a}. (\vec{c} \times \vec{b})

4. If any two of three vectors are equal or parallel, then scalar triple product is zero.

[a   b   a]=[a   c   c]=0[\vec{a}\; \ \vec{b}\; \ \vec{a}] = [\vec{a}\; \ \vec{c}\; \ \vec{c}]=0

5. If three vectors are mutually perpendicular, then scalar triple product is ±1.

[i^   j^   k^]=1,[j^   i^   k^]=1[\hat{i}\; \ \hat{j}\; \ \hat{k}] = 1, [\hat{j}\; \ \hat{i}\; \ \hat{k}] = -1

6. If three vectors are coplanar then [a   b   c][\vec{a}\; \ \vec{b}\; \ \vec{c}] = 0, this is the necessary and sufficient condition for three non zero and non collinear vectors to be coplanar.

7. For any four vectors a,b c and d\vec{a}, \vec{b}\ \vec{c}\ and\ \vec{d},

[a+d   b   c]=[a   b   c]+[d   b   c][\vec{a}+ \vec{d}\; \ \vec{b}\; \ \vec{c}] = [\vec{a}\; \ \vec{b}\; \ \vec{c}] + [\vec{d}\; \ \vec{b}\; \ \vec{c}]

8. [a+b   b+c   c+a]=2[a   b   c][\vec{a}+ \vec{b}\; \ \vec{b}+ \vec{c}\; \ \vec{c}+ \vec{a}] = 2[\vec{a}\; \ \vec{b}\; \ \vec{c}]

9. [ab   bc   ca][\vec{a}- \vec{b}\; \ \vec{b}- \vec{c}\; \ \vec{c}- \vec{a}] is always zero.

10. [l   m   n][a   b   a]=l.al.bl.cm.am.bm.cn.an.bn.c[\vec{l}\; \ \vec{m}\; \ \vec{n}][\vec{a}\; \ \vec{b}\; \ \vec{a}] = \begin{vmatrix} \vec{l}.\vec{a} & \vec{l}.\vec{b} & \vec{l}.\vec{c}\\ \vec{m}.\vec{a} & \vec{m}.\vec{b} &\vec{m}.\vec{c} \\ \vec{n}.\vec{a}&\vec{n}.\vec{b} & \vec{n}.\vec{c} \end{vmatrix}

Where l,m,n and a,b and c\vec{l}, \vec{m}, \vec{n}\ and\ \vec{a}, \vec{b}\ and\ \vec{c} are non-coplanar vectors.

11. [l   m   n](a×b)=l.al.blm.am.bmn.an.bn[\vec{l}\; \ \vec{m}\; \ \vec{n}](\vec{a} \times \vec{b}) = \begin{vmatrix} \vec{l}.\vec{a} & \vec{l}.\vec{b} & \vec{l}\\ \vec{m}.\vec{a} & \vec{m}.\vec{b} &\vec{m} \\ \vec{n}.\vec{a}&\vec{n}.\vec{b} & \vec{n} \end{vmatrix}

Scalar Triple Product Proof

If a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^c=c1i^+c2j^+c3k^\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\\ \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\\ \vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}

 

a.(b×c)=(a1i^+a2j^+a3k^).i^j^k^b1b2b3c1c2c3\vec{a}.(\vec{b} \times \vec{c}) = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) . \begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ b_1& b_2& b_3\\ c_1&c_2 &c_3 \end{vmatrix}

= a1i^.i^a2j^.j^a3k^.k^b1b2b3c1c2c3\begin{vmatrix} a_1\hat{i} . \hat{i}& a_2\hat{j} .\hat{j}& a_3 \hat{k} .\hat{k}\\ b_1& b_2& b_3\\ c_1&c_2 &c_3 \end{vmatrix}

= [a   b   c] [\vec{a}\; \ \vec{b}\; \ \vec{c}]

Expansion:

a.(b×c)=[a   b   c]=a1a2a3b1b2b3c1c2c3\vec{a}.(\vec{b} \times \vec{c}) = [\vec{a}\; \ \vec{b}\; \ \vec{c}] = \begin{vmatrix} a_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1&c_2 &c_3 \end{vmatrix}

= a1(b2c3b3c2)a2(b1c3b3c1)+a3(b1c2b2c1)a_1(b_2c_3 – b_3c_2) – a_2(b_1c_3 – b_3c_1) + a_3(b_1c_2 – b_2c_1)

Finding Volume of Tetrahedron

Let a,b and c\vec{a}, \vec{b}\ and\ \vec{c} are the position vectors of vertices A, B, C with respect to O, then volume of tetrahedron OABC is given by:

Volume of tetrahedron

Volume = ⅓ (area of base) x height

 

Area of base = 12(a×b)+(b×c)+(c×a)\frac{1}{2} |(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|

Let (a×b)+(b×c)+(c×a)=n(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) = \vec{n}

Area of base = ½ n|\vec{n}|

Height = projection of a on n\vec{a}\ on\ \vec{n}

Find Volume of tetrahedron

Volume = 1/3 x ½ n×[a   b   c]n |\vec{n}| \times \frac{|[\vec{a}\;\ \vec{b}\;\ \vec{c} ]|}{|\vec{n}|}

Volume = 1/6 x [a   b   c] |[\vec{a}\;\ \vec{b}\;\ \vec{c}]|

 

Note: If a,b,c and d\vec{a}, \vec{b}, \vec{c}\ and\ \vec{d}, are position vectors of vertices A, B ,C, D of tetrahedron ABCD, then

Volume of tetrahedron = 16×[AB   AC   AD]\frac{1}{6} \times |[\vec{AB}\; \ \vec{AC}\; \ \vec{AD} ]|

= 16×[ba   ca   da]\frac{1}{6} \times |[\vec{b} – \vec{a}\; \ \vec{c} – \vec{a}\; \ \vec{d}-\vec{a}]|

Scalar Triple Product Examples

Example 1: If a=(i^+2j^+k^),b=(4i^j^+2k^) and c=(3i^+j^\vec{a} = (\hat{i} + 2 \hat{j} + \hat{k}), \vec{b} = (4\hat{i} – \hat{j} + 2\hat{k})\ and\ \vec{c} = (3\hat{i} + \hat{j} represent three coterminous edges of parallelepiped, then find it’s volume.

Solution:

Using Volume formula, we get

Volume = 121412310\begin{vmatrix} 1& 2& 1\\ 4& -1& 2\\ 3&1 &0 \end{vmatrix}

= |3(4 + 1) – 1(2 – 4)| = |15 + 2| = 17

Example 2: If vertices of a tetrahedron are a=(i^+2j^+k^),b=(4i^2j^+3k^),c=2i^+2j^ and d=j^+2k^\vec{a} = (\hat{i} + 2 \hat{j} + \hat{k}), \vec{b} = (4\hat{i} – 2 \hat{j} + 3\hat{k}), \vec{c} = 2\hat{i} + 2\hat{j}\ and\ \vec{d} = \hat{j} + 2\hat{k}. Find the volume of tetrahedron.

Solution:

Let A, B, C, D are the vertices and their position vectors with respect to O are a,b,c and d\vec{a}, \vec{b}, \vec{c}\ and\ \vec{d}, then

AB=ba=4i^2j^+3k^(i^+2j^+k^)=3i^4j^+2k^\vec{AB} = \vec{b} – \vec{a} = 4 \hat{i} – 2\hat{j} + 3\hat{k} – (\hat{i} + 2\hat{j} + \hat{k}) = 3\hat{i} – 4\hat{j} + 2\hat{k} AC=ca=2i^+2j^(i^+2j^+k^)=i^k^\vec{AC} = \vec{c} – \vec{a} = 2 \hat{i} + 2\hat{j} – (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} – \hat{k}

 

AD=da=j^+2k^(i^+2j^+k^)=i^j^+k^\vec{AD} = \vec{d} – \vec{a} = \hat{j} + 2\hat{k} – (\hat{i} + 2\hat{j} + \hat{k}) = -\hat{i} – \hat{j} + \hat{k}

Now,

Example on Scalar Triple Product

Example 3: For any three vectors, a,b and c\vec{a}, \vec{b}\ and\ \vec{c}, prove that [a+b   b+c   c+a]=2[a   b   a][\vec{a}+ \vec{b}\; \ \vec{b}+ \vec{c}\; \ \vec{c}+ \vec{a}] = 2[\vec{a}\; \ \vec{b}\; \ \vec{a}]

Solution: [a+b   b+c   c+a]=(a+b).[(b+c)×(c+a)][\vec{a}+ \vec{b}\; \ \vec{b}+ \vec{c}\; \ \vec{c}+ \vec{a}] = (\vec{a}+ \vec{b}) . [(\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})]

=> (a+b).[(b×c)+(b×a)+(c×c)+(c×a)](\vec{a} + \vec{b}) . [(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{c}) + (\vec{c} \times \vec{a})]

=> (a+b).[(b×c)+(b×a)+(c×a)](\vec{a} + \vec{b}) . [(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a})] [As, (c×c)=0(\vec{c} \times \vec{c}) = 0 ]

=> [a   b   c]+[a   b   a]+[a   c   a]+[b   b   c]+[b   b   a]+[b   c   a][\vec{a}\; \ \vec{b}\; \ \vec{c}] + [\vec{a}\; \ \vec{b}\; \ \vec{a}] + [\vec{a}\; \ \vec{c}\; \ \vec{a}] + [\vec{b}\; \ \vec{b}\; \ \vec{c}] + [\vec{b}\; \ \vec{b}\; \ \vec{a}] + [\vec{b}\; \ \vec{c}\; \ \vec{a}]

=> [a   b   c]+[b   c   a]=2[a   b   c][\vec{a}\; \ \vec{b}\; \ \vec{c}] + [\vec{b}\; \ \vec{c}\; \ \vec{a}] = 2[\vec{a}\; \ \vec{b}\; \ \vec{c}] . Hence Proved.

Example 4:  If a, b, c are three non-coplanar vectors, then a.b×cc×a.b+b.a×cc.a×b=\frac{\mathbf{a}\,.\,\mathbf{b}\times \mathbf{c}}{\mathbf{c}\times \mathbf{a}\,.\,\mathbf{b}}+\frac{\mathbf{b}\,.\,\mathbf{a}\times \mathbf{c}}{\mathbf{c}\,.\,\mathbf{a}\times \mathbf{b}}=

Solution:

a.b×cc×a.b+b.a×cc.a×b=[abc][cab]+[bac][cab]=[abc][cab][abc][cab]=0\frac{\mathbf{a}.\mathbf{b}\times \mathbf{c}}{\mathbf{c}\times \mathbf{a}.\mathbf{b}}+\frac{\mathbf{b}.\mathbf{a}\times \mathbf{c}}{\mathbf{c}.\mathbf{a}\times \mathbf{b}}=\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}+\frac{[\mathbf{b}\,\mathbf{a}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}=\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}-\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}=0\\

Example 5: If a, b, c be any three non-coplanar vectors, then [a+bb+cc+a]=[\mathbf{a}+\mathbf{b}\,\,\,\mathbf{b}+\mathbf{c}\,\,\,\mathbf{c}+\mathbf{a}]=

Solution:

[a+bb+cc+a]=(a+b).{(b+c)×(c+a)}=(a+b).(b×c+b×a+c×c+c×a)=(a+b).(b×c+b×a+c×a),{becausec×c=0}=a.b×c+a.b×a+a.c×a+b.b×c+b.b×a+b.c×a=[abc]+[bca]=2[abc][\mathbf{a}+\mathbf{b}\,\,\mathbf{b}+\mathbf{c}\,\,\mathbf{c}+\mathbf{a}]=(\mathbf{a}+\mathbf{b}).\{(\mathbf{b}+\mathbf{c})\times (\mathbf{c}+\mathbf{a})\} =(\mathbf{a}+\mathbf{b}).(\mathbf{b}\times \mathbf{c}+\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}) =(\mathbf{a}+\mathbf{b}).(\mathbf{b}\times \mathbf{c}+\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{a}), \left\{ \text because \,\mathbf{c}\times \mathbf{c}=0 \right\} =\mathbf{a}.\mathbf{b}\times \mathbf{c}+\mathbf{a}.\mathbf{b}\times \mathbf{a}+\mathbf{a}.\mathbf{c}\times \mathbf{a}+\mathbf{b}.\mathbf{b}\times \mathbf{c}+\mathbf{b}.\mathbf{b}\times \mathbf{a}+\mathbf{b}.\mathbf{c}\times \mathbf{a} =[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]+[\mathbf{b}\,\mathbf{c}\,\mathbf{a}]=2[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\\

Example 6: If the points whose position, vectors are 3i2jk,2i+3j4k,i+j+2k and 4i+5j+λk3\mathbf{i}-2\mathbf{j}-\mathbf{k}, 2\mathbf{i}+3\mathbf{j}-4\mathbf{k}, -\mathbf{i}+\mathbf{j}+2\mathbf{k}\ \text and \ 4\mathbf{i}+5\mathbf{j}+\lambda \mathbf{k} then find the value of λ.

Solution:

Let a=3i2jk,b=2i+3j4k,c=i+j+2k and d=4i+5j+λk\mathbf{a}=3\mathbf{i}-2\mathbf{j}-\mathbf{k},\mathbf{b}=2\mathbf{i}+3\mathbf{j}-4\mathbf{k},\mathbf{c}=-\mathbf{i}+\mathbf{j}+2\mathbf{k} \text \ and \ \mathbf{d}=4\mathbf{i}+5\mathbf{j}+\lambda \mathbf{k}

Since the points are coplanar, [dbc]+[dca]+[dab]=[abc]45λ234112+45λ112321+45λ321234=32123411240+5λ+37λ+94+13λ=25λ=14617[\mathbf{d}\,\mathbf{b}\,\mathbf{c}]+[\mathbf{d}\,\mathbf{c}\,\mathbf{a}]+[\mathbf{d}\,\mathbf{a}\,\mathbf{b}]=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] \Rightarrow \left| \begin{matrix} 4 & 5 & \lambda \\ 2 & 3 & -4 \\ -1 & 1 & 2 \\ \end{matrix} \right|+\left| \begin{matrix} 4 & 5 & \lambda \\ -1 & 1 & 2 \\ 3 & -2 & -1 \\ \end{matrix} \right|+\left| \,\begin{matrix} 4 & 5 & \lambda \\ 3 & -2 & -1 \\ 2 & 3 & -4 \\ \end{matrix}\, \right| =\left| \begin{matrix} 3 & -2 & -1 \\ 2 & 3 & -4 \\ -1 & 1 & 2 \\ \end{matrix} \right| \Rightarrow 40+5\lambda +37-\lambda +94+13\lambda =25\Rightarrow \lambda =\frac{-146}{17}

 

Example 7: Find the altitude of a parallelopiped determined by the vectors a,bandc\vec{a},\vec{b} \: and \: \vec{c} if the base is taken as a parallelogram determined by a  and  b\vec{a}\; and \; \vec{b} and if a=i^+j^+k^\vec{a} = \hat{i}+\hat{j}+\hat{k} , b=2i^+4j^k^\vec{b} = 2\hat{i}+4\hat{j}-\hat{k} and c=i^+j^+3k^\vec{c} = \hat{i}+\hat{j}+3\hat{k}.

Solution:

Let V be the volume of the parallelopiped determined by the vectors a,b  and  c.\vec{a} , \vec{b}\; and \; \vec{c}. then V = [abc]\begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \end{bmatrix}

V = 111241113\begin{vmatrix} 1 & 1& 1\\ 2 & 4 & -1\\ 1 &1 & 3 \end{vmatrix} 

= 1(12–1)-(6–1)+1(2-4)

= (12+1) – (6+1) + (2-4)

= 13-7-2

= 4 cubic units.

Let A be the area of the base of the parallelopiped. Then A = a×b\left | \vec{a}\times \vec{b} \right | = i^j^k^111241\begin{vmatrix} \hat{i}& \hat{j} & \hat{k}\\ 1 & 1 & 1\\ 2& 4 & -1 \end{vmatrix}

= 5i^+3j^+2k^-5\hat{i}+3\hat{j}+2\hat{k}

A = a×b\left | \vec{a} \times \vec{b}\right |

= 52+32+22\sqrt{5^{2}+3^{2}+2^{2}}

= 25+9+4\sqrt{25+9+4}

= 38\sqrt{38}

Volume of parallelopiped = Area of base × altitude

So altitude = V/A = 4/√38 units. 

Example 8: Find [a  b  c][\vec{a}\; \vec{b}\; \vec{c}] when a=2i^3j^+4k\vec{a}= 2\hat{i}-3\hat{j}+4k , b=i^+2j^k\vec{b}= \hat{i}+2\hat{j}-k and c=3i^j^+2k\vec{c}= 3\hat{i}-\hat{j}+2k

 

Solution:

Given