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Scalar Triple Product of Three Vectors a, b and c is the dot product of a single vector with the cross product of the two remaining vectors. The result is not a scalar but a pseudoscalar because reversing the direction of one of the vectors changes the sign of the scalar. The properties of the Scalar triple product of vectors include being the volume of the parallelepiped, being cyclic, etc. It is geometrically written as (a x b).c.
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Key Terms: Scalar, Triple, Product, Vectors, Formula, Dot, Cross, Resultant
What is Scalar Triple Product of Vector?
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Scalar Triple Product of Vectors a, b, c is the scalar product with cross product of the vectors b and c, i.e., a · (b × c). Symbolically, scalar triple product can also be written as [a b c] = [a, b, c] = a · (b × c).
Scalar Triple Product [a b c] gives the volume of a parallelepiped with adjacent sides a, b, and c. The scalar triple products of three vectos a, b, c, will be as follows:
- a · (b × c)
- a · (c × b)
- b · (a × c)
- b · (c × a)
- c · (b × a)
- c · (a × b)
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Scalar Triple Product of Vector: Formula
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The scalar triple product of three vectors a, b, and c is (a×b)⋅c. When Vectors are multiplied and get a scalar triple product, then it’s simply known as the product of three vectors, or, the dot product of a vector with the cross product of the other two vectors is known as the Scalar triple product formula.
It can be written as abc = (a x b).c, In this case, (a x b).c is the resultant vector and a scalar quantity. This formula can also be written by exchanging the cross and dot in it, as (a x b).c = a.(b xc).
A scalar triple product shows the absolute value |(a×b)⋅c| of a parallelepiped’s volume spanned by a, b, and c (i.e., vectors a,b and c are adjacent sides of a parallelepiped.)
Scalar Triple Product of Vector: Proof
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Key points to consider while computing Scalar Triple Product of Vector Proof:
- The Scalar triple product formula shows the volume of a parallelepiped whose three adjacent sides are the three vectors, a, b and c.
- The cross product of two vectors (let a and b) among these three, provides the base’s area. Both of the vectors are perpendicular to the direction of the resultant.
- The third vector component ‘c’, gives the height along the direction of the resultant cross product.
- |a x b| calculates the parallelopiped’s area, while the direction of the resultant vector is perpendicular to the base.
- The height is of the parallelepiped is denoted by |c| cos \(\phi\), where \(\phi\) denotes the angle between (a x b) and c.
Scalar Triple Product of Vector: Proof
It is shown in the above diagram that |c|cos \(\phi\) denotes the height and the direction of |a x b| The vector is perpendicular to the base.
Now, let’s use Let’s calculate the scalar triple product
Let a = a1i^ + a2j^ + a3k^, b = b1i^ + b2j^ + b3k^, c = c1i^ + c2j^ + c3k^
Now, (a x b).c = \(\begin {vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end {vmatrix}\) .(c1i + c2j + c3k)
Scalars and vectors can be calculated by using the determinant property.
(a x b).c = \(\begin {vmatrix} \hat{i}.(c_1\hat{i} +c_2\hat{j}+c_3\hat{k}) & \hat{j}.(c_1\hat{i} +c_2\hat{j}+c_3\hat{k}) & \hat{k}.(c_1\hat{i} +c_2\hat{j}+c_3\hat{k})\\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end {vmatrix}\)
From the dot product vector’s property:
i.i = j.j = k.k =1 (since, cos 0 = 1)
This implies that i.(c1i + c2j + c3k) = c1
j.(c1i + c2j + c3k) = c2
k.(c1i + c2j + c3k) = c3
(a x b).c = \(\begin {vmatrix} c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end {vmatrix}\)
abc = \(\begin {vmatrix} c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end {vmatrix}\)
Key Observations:
We can determine the following factors after observing the formula of scalar triple product:
- The scalar triple product’s resultant is always scalar.
- The cross product of the first two vectors is calculated first and then the dot product of the remaining vector with the resultant vector is calculated, to calculate the formula.
- The triple product will become zero if one of the three vectors taken is of zero magnitudes.
- We can find the volume of a parallelepiped by using the formula.
Properties of Scalar Triple Product of Vector
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- The scalar triple product is cyclic, which means that;
abc = bca = cab = -bac = -cba = -acb
- If the vectors taken for calculation in the scalar triple product formula, say a, b, and c are cyclically permuted, then:
(a x b).c = a.(b x c)
- The given vectors will be coplanar if the scalar triple product of them is zero.
The triple product shows the volume of a parallelepiped. However, If it is zero, then any vector is of zero magnitudes. The cross product of a and b’s direction is perpendicular to their plane. If the vector c lies in the same plane as a x b, then the dot product of the resultant with c will be zero. This is due to the angle between the resultant and c being 90° and cos 90°
- For any k that belongs to a Real number,
kakbkc = kabc
(a + b)cd = (a + b).(c + d)
= a. (cxd) + b.(cxd)
= acd + bcd
Things to Remember
- When Vectors are multiplied and get a scalar triple product, then it’s simply known as the product of three vectors, or, the dot product of a vector with the cross product of the other two vectors is known as the Scalar triple product formula.
- It can be written as abc = (a x b).c, In this case, (a x b).c is the resultant vector and a scalar quantity.
- This formula can also be written by exchanging the cross and dot in it, as (a x b).c = a.(b xc).
- A scalar triple product shows the absolute value |(a×b)⋅c| of a parallelepiped’s volume spanned by a, b and c (i.e., vectors a,b and c are adjacent sides a parallelepiped.)
- The Scalar triple product formula shows the volume of a parallelepiped whose three adjacent sides are the three vectors, a, b and c.
- The cross product of two vectors (let a and b) among these three, provides the base’s area. Both of the vectors are perpendicular to the direction of the resultant.
- The third vector component ‘c’, gives the height along the direction of the resultant cross product.
- |a x b| calculates the parallelopiped’s area, while the direction of the resultant vector is perpendicular to the base.
- The height is of the parallelopiped is denoted by |c| cos \(\phi\), where \(\phi\) denotes the angle between (a x b) and c.
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Previous Year Questions
- If a|(b+c) and a|(b−c) where a,b,c ∈ N then [KCET 2006]
- The vector equation of the symmetrical form of equation of straight line… [VITEEE 2018]
- If a is a unit vector, then… [TS EAMCET 2017]
- Let \(\overrightarrow{a}, \overrightarrow{b}, \& \overrightarrow{c}\) be non-coplanar unit vectors equally inclined to one… [BITSAT 2017]
- From a point A with position vector… [JEE Main 2018]
- If \(\overrightarrow{a} and \overrightarrow{b}\) are unit vectors and… [KCET 2008]
- f a and b are vectors such that… [KCET 2007]
- If \(\overrightarrow{a}= 2\hat{i}+3\hat{j}-\hat{k}, \overrightarrow{b}= \hat{i}+2\hat{j}-5\hat{k}, \overrightarrow{c}= 3\hat{i}+5\hat{j}-\hat{k} \)… [KCET 2007]
- OA and BO are two vectors of magnitudes 5 and 6 respectively… [KCET 2007]
- If a,b and c are three non-coplanar vectors and p,q and r are vectors defined by... [KCET 2012]
Sample Questions
Ques: Can Dot Product and Cross Product be Interchanged in the Scalar Triple Product? (3 marks)
Ans: The scalar triple product is denoted through (a x b).c. Here, dot product and move product may be used interchangeably, without converting the order of vector occurrences. Using these interchangeability assets, diverse residences of the scalar triple product can be derived:
- Associative assets: (a x b).C = a. (b x c)
- Commutative assets: (a x b).C = (b x c).A = (c x a).B
Moreover, if any two vectors taken in scalar and vector triple product are interchanged with recognize to their role, then the fee comes out to be (-1) of the unique end result:
abc = bca = cab = -acb = -cba = -bac
Ques: Three vectors are a = i - j + k, b = 2i + j + k, c= i + j -2k. Calculate the scalar triple product and verify if abc = bca. (4 marks)
Ans: According to the scalar triple product:
abc = \(\begin {vmatrix} c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end {vmatrix}\)
abc = (a x b).c = \(\begin {vmatrix}1 & 1 & -2 \\ 1 & -1 & 1 \\ 2 & 1 & 1 \end {vmatrix}\)= -7
So, the answer is 7.
To verify abc = bca, calculate bca as
bca = \(\begin {vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end {vmatrix}\)
Ques: What is the formula of a vector triple product? (1 marks)
Ans: The formula of vector triple product is given as:
a × (b × c) = (a . c) b – (b . c)a
Ques: What is the scalar product of two vectors? (2 marks)
Ans: In mathematics, the dot product or scalar product is an algebraic operation that takes same-duration sequences of numbers (commonly coordinate vectors) and returns a single quantity.
Thus, the scalar fabricated from vectors a = [a1, a2, a3,…an] and b = [b1, b2, b3,…, bn] is given by way of:
a.B = a1b1 + a2b2 + a3b3 + …. + anbn
Ques: If a,b and c are coplanar, then what is (2a×3b)⋅4c+(5b×3c)⋅6a equal to? (4 marks)
Ans: Given:
a,b and c are coplanar i.e [abc]=0
⇒ (2a×3b)⋅4c=[2a 3b 4c] and (5b×3c)⋅6a=[5b 3c 6a]
⇒ (2a×3b)⋅4c+(5b×3c)⋅6a=[2a 3b 4c]+[5b 3c 6a]
As we know that, [a b c] = [b c a] = [c a b] ⇒ (2a×3b)⋅4c+(5b×3c)⋅6a
=[2a 3b4c]+[6a 5b 3c]
As we know that, [λa, b c] = λ [a b c] ⇒ [2a 3b 4c]+[6a 5b 3c]
=24[abc]+90[a b c]
As we know that, vectors a,b and c are coplanar if and only if [a b c] = 0
⇒ [2a 3b 4c]+[6a 5b 3c]=0
⇒ (2a×3b)⋅4c+(5b×3c)⋅6a=0
Hence, the correct option is 3.
Ques: If the vectors αi^+αj^+γk^,i^+k^ and γi^+γj^+βk^ lie on a plane, where α, β and γ are distinct non-negative numbers, then γ is? (3 marks)
Ans: Given:
vectors αi^+αj^+γk^,i^+k^ and γi^+γj^+βk^ lie on a plane
Vectors lie on the same plane so vectors are coplanar.
Therefore, \(\begin {vmatrix} \alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta \end {vmatrix}\)=0
Expanding R1, we get ⇒ α[0 − γ] − α[β − γ] + γ[γ−0] = 0
⇒ −αγ – αβ + αγ + γ2 = 0
⇒ γ2 = αβ
∴ α, β, γ are in G.P.
Ques: If vector a=i,b→=j^+k^ and c=i^+k^, then find the value of a.(b×c)=? (2 marks)
Ans: Given:
a=i, b=j^+k^ and c→=i^+k^
As we know that scalar triple product a.(b×c)=
⇒ a.(b×c)= ⇒ a.(b×c)=1(1−0)−0+0=1
Hence, option 1 is correct.
Ques: If the origin and the points P(2, 3, 4), Q(1, 2, 3) and R(x, y, z) are coplanar then? (4 marks)
Ans: Given:
the origin (0, 0, 0) and the points P(2, 3, 4), Q(1, 2, 3) and R(x, y, z) are coplanar
⇒ a=OR=(x,y,z)
⇒ b=OP=(2,3,4)
⇒ c=OQ=(1,2,3)
Here, a, b and c are coplanar
The three vectors are coplanar if their scalar triple product is zero..
⇒ a.(b×c)=0
⇒ =0
⇒ x(9 - 8) - y(6 - 4) + z (4 - 3) = 0
⇒ x - 2y + z = 0
Hence, if the origin and the points P(2, 3, 4), Q(1, 2, 3) and R(x, y, z) are co-planar then x - 2y + z = 0
Ques: If the vectors 2i^− j^ + k^,3i^ + pk^ and 4j^−5k^ are coplanar then the value of p is? (2 marks)
Ans: Given:
The vectors 2i^ − j^ + k^, 3i^ + pk^ and 4j^−5k^ are coplanar.
As we know that, if a, b and c vectors are coplanar then [abc]=0
⇒ |230−1041p−5|=0
⇒ 2(0 - 4p) - 3(5 - 4) + 0 = 0 ⇒ 8p = -3
⇒ p = -3/8
Hence, the correct option is 3
Ques: If a, b, c are three non-coplanar vector, then calculate \(\frac{a.b \times c}{c \times a.b}+\frac{b.a \times c}{c.a \times b}|\) [IIT 1985, 86; UPSEAT 2003] (2 Marks)
Ans: \([\frac{\mathbf{a}.\mathbf{b}\times \mathbf{c}}{\mathbf{c}\times \mathbf{a}.\mathbf{b}}+\frac{\mathbf{b}.\mathbf{a}\times \mathbf{c}}{\mathbf{c}.\mathbf{a}\times \mathbf{b}}=\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}+\frac{[\mathbf{b}\,\mathbf{a}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}=\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}-\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}=0.\)
Ques: If a, b, c be any three non-coplanar vectors, then find [a+b b+c c+a]. (3 Marks)
Ans: \([\mathbf{a}+\mathbf{b}\,\,\mathbf{b}+\mathbf{c}\,\,\mathbf{c}+\mathbf{a}]=(\mathbf{a}+\mathbf{b}).\{(\mathbf{b}+\mathbf{c})\times (\mathbf{c}+\mathbf{a})\}\)
\(=(\mathbf{a}+\mathbf{b}).(\mathbf{b}\times \mathbf{c}+\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{c}+\mathbf{c}\times \mathbf{a})\)
\(=(\mathbf{a}+\mathbf{b}).(\mathbf{b}\times \mathbf{c}+\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{a}), \left\{ \because \,\mathbf{c}\times \mathbf{c}=0 \right\}\)
\(=\mathbf{a}.\mathbf{b}\times \mathbf{c}+\mathbf{a}.\mathbf{b}\times \mathbf{a}+\mathbf{a}.\mathbf{c}\times \mathbf{a}+\mathbf{b}.\mathbf{b}\times \mathbf{c}\) \(+\mathbf{b}.\mathbf{b}\times \mathbf{a}+\mathbf{b}.\mathbf{c}\times \mathbf{a} \)
\(=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]+[\mathbf{b}\,\mathbf{c}\,\mathbf{a}]=2[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\)
Ques: If a, b, c are any three coplanar unit vectors, then
A) a.(b×c)=1
B) a.(b×c)=3
C) (a×b).c=0
D) (c×a).b=1 (1 Mark)
Ans: The correct answer is C) (a×b).c=0
a.(b×c)=0
or
(a×b).c=0.
Ques: If a and b be parallel vectors, then find [a c b]. (2 Marks)
Ans: a.(c×b)=c.(b×a)=0, (Since a| and b| are parallel)
Ques: The volume of the parallelopiped whose edges are represented by \(-12\mathbf{i}+\alpha \mathbf{k}\), \(3\mathbf{j}-\mathbf{k}\) and \(2\mathbf{i}+\mathbf{j}-15\mathbf{k}\) is 546. Then find \(\alpha\). (2 Marks)
Ans: Since:
Ques: Three concurrent edges OA, OB, and OC of a parallelopiped are represented by three vectors 2i+j−k,i+2j+3k, and −3i−j+k. Calculate the volume of the solid so formed in cubic unit. (3 Marks)
Ans: Volume can be calculated as follows:
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