Scalar triple product of vectors: Derivation & Important Questions

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Jasmine Grover

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Scalar triple product of three vectors a, b, and c refers to the dot product of a single vector with the cross product of the two remaining vectors is the. Because changing the direction of one of the vectors changes the sign of the scalar, the result is a pseudoscalar. The scalar triple product of vectors is referred to as the product of three vectors in mathematics. This formula yields scalar quantities, which are written as (a x b). c. In this formula, the dot and cross can be swapped out (a x b). c is the same as a. (b x c). It's a scalar product since it evaluates to a single number, exactly like the dot product. In this article, we will learn more about the scalar triple product of vectors.

Read More: Symmetric Matrix

Key TakeawaysScalar Triple Product, dot product, cross product, Parallelepiped, Coterminus Edges, Vector, Coplanar, scalar quantity


Scalar triple product of vectors

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The scalar triple product of vectors is the product of three vectors. It's a scalar product because, like the dot product, it evaluates to a single number. It entails multiplying one of the vectors' dot products by the cross product of the other two. Its mathematical expression like: (a x b). c. The volume of a parallelepiped is represented by the scalar triple product. The scalar triple product (also known as the mixed product, box product, or triple scalar product) is the dot product of one vector with the cross product of the other two vectors. 

Read More: Inverse Trigonometric Formulas


Formula of Scalar triple product of vectors

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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 when vectors are multiplied and a scalar triple product is obtained. It's written like this:

\([\text{a b c}] =(a\times b).c\)

In this instance (a x b). c is a scalar quantity and the resultant vector. This formula can alternatively be expressed by swapping the cross and dot in the middle, as seen below (a x b). c is the same as a. (b x c). The scalar triple product of vectors is obvious from its name: it is the product of three vectors. It requires multiplying one of the vectors by the cross product of the other two.


Geometric Derivation

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The three vectors are represented in this diagram by the coterminous edges, as illustrated. The area of the base is given by the cross product of vectors a and b, and the direction of the cross product of vectors is perpendicular to both vectors. The height in this situation is given by the component of vector c along the direction of the cross product of a and b because volume equals the product of area and height. c cos is the component's name.

Geometric Derivation of Scalar Triple Product of Vector

Geometric Derivation of Scalar Triple Product of Vector

As a result, if the coterminous edges of a parallelepiped are designated by three vectors and a, b, and c, we can deduce that

Volume of parallelepiped = (a × b) c cos α = (a × b) . c 

as α denotes the angle between (a × b) and c. 

The extension of vector cross products is something we're known with. Bearing in mind, if it is given a = a1\(\widehat{i}\) +a2\(\widehat{j}\) + a3\(\widehat{k}\), b = b1\(\widehat{i}\) + b2\(\widehat{j}\) + b3\(\widehat{k}\) and c = c1\(\widehat{i}\) +c2\(\widehat{j}\) + c3\(\widehat{k}\), then we can write the following equation-  

(a x b) . c =\( \begin{matrix} | \widehat{i} & \widehat{j} & \widehat{k} |\\ | a1 & a2 & a3|\\ | b1 & b2 & ib3| \end{matrix} \) . (c1\(\widehat{i}\) +c2 \(\widehat{j}\) + c3\(\widehat{k}\))

The dot product of two vectors is indicated by this symbol. We can expand the previous equation using determinant properties-
(a x b) . c\( \begin{matrix} | \widehat{i}(c1\widehat{i} +c2 \widehat{j} + c\widehat{k}) & \widehat{j}(c1\widehat{i} +c2\widehat{j} + c3\widehat{k}) & \widehat{k}(c1\widehat{i} +c2\widehat{j}+ c3\widehat{k}) |\\ | a1 & a2 & a3|\\ | b1 & b2 & ib3| \end{matrix} \)

As per the dot product of vector properties, 

\(\widehat{i }\).\(\widehat{i }\) = \(\widehat{j}\).\(\widehat{j}\) = \(\widehat{k}\).\(\widehat{k}\) = 1 as cos 0 = 1

\(\widehat{i}\) (c1\(\widehat{i}\) + c2\(\widehat{j}\) + c3\(\widehat{k}\)) = c1

\(\widehat{j}\) (c1\(\widehat{i}\) + c2\(\widehat{j}\) + c3\(\widehat{k}\)) = c2

\(\widehat{k}\) (c1\(\widehat{i}\) + c2\(\widehat{j}\) + c3\(\widehat{k}\)) = c3

⇒(a x b) . c = \( \begin{matrix} | c1 & c2 & c3|\\ | a1 & a2 & a3|\\ | b1 & b2 & ib3| \end{matrix} \)

\(\begin{matrix} [ a & b & c ]\\ \end{matrix} \)\( \begin{matrix} | c1 & c2 & c3|\\ | a1 & a2 & a3|\\ | b1 & b2 & ib3| \end{matrix} \)

Read More: Calculus Formula


Properties of Scalar triple product of vectors

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The properties of scalar triple product of vectors are:

  • If the vectors are permuted cyclically, then (a × b) . c = a.( b × c)
  • If the vectors are permuted cyclically, then a.(b × c) = b.(c × a) = c.(a × b)
  • If the triple product of vectors is zero, it can be assumed that the vectors are coplanar.

The volume of a parallelepiped is represented by the triple product. If it's zero, then such a situation could only occur if one of the three vectors has a magnitude of zero. The cross product of a and b has a direction that is perpendicular to the plane that contains a and b. Only if the vector c is also in the same plane will the dot product of the resultant with c be zero. Because the angle between the resultant and C will be 90° and cos 90°, this is the case.


Outcomes drawn

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The outcomes drawn from scalar triple product of vectors are:

  • The consequence of scalar triple products is always scalar quantities.
  • Calculating the cross-products of two vectors yields scalar triple product formulas. After that, compute the dot product of the remaining vector with the resultant vector.
  • If the triple product is 0, one of the three vectors taken is equivalent to zero magnitudes.
  • Using this method, a parallelepiped can be simply calculated.

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Things to Remember

  • 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 when vectors are multiplied and a scalar triple product is obtained.
  • The mathematical expression for the scalar triple product for vectors is (a x b).c. I.n this instance (a x b). c is a scalar quantity and the resultant vector.
  • The absolute value |(a×b)⋅c| of a parallelepiped's volume spanned by a, b, and c is shown by a scalar triple product (i.e., vectors a,b and c are adjacent sides a parallelepiped.)
  • The volume of a parallelepiped whose three neighboring sides are the three vectors a, b, and c is calculated using the scalar triple product formula.
  • The area of the base is determined by the cross-product of two vectors (let a and b) among these three. The direction of the resultant is perpendicular to both vectors.
  • The height of the resultant cross product is determined by the third vector component, 'c'.
  • The area of the parallelepiped is calculated by |a x b|, with the direction of the resultant vector perpendicular to the base, and the height is calculated by |c| cos \(\Phi\), where \(\Phi\) represents the angle between (a x b) and c.

Sample Questions

Ques. Is it possible to use the Dot Product and Cross Product interchangeably in the Scalar Triple Product? (3 marks)

Ans: (a x b).c denotes the scalar triple product. Without modifying the sequence of vector occurrences, the dot product and cross product can be used interchangeably here. Various features of the scalar triple product can be deduced using this interchangeability property:

Associative property: (a x b).c = a. (b x c)

Commutative property: (a x b).c = (b x c).a = (c x a).b

Ques. The volume of a parallelepiped 7\(\widehat{i}\) + λ\(\widehat{j}\) − 3\(\widehat{k}\), \(\widehat{i}\) + 2\(\widehat{j}\)\(\widehat{k}\), −3\(\widehat{i}\) + 7\(\widehat{j}\) + 5\(\widehat{k}\) with 90 cubic units at their coterminous edges. Calculate the value of λ. (3 marks)

Ans:

volume of a parallelepiped

Ques. Find out the altitude of the parallelepiped as determined by the vectors \(\overrightarrow{a}\) = −2i + 5 j + 3k, \(\overrightarrow{b}\) = i + 3 j − 2k and \(\overrightarrow{c}\) = -3i + j − 2k if the base taken as the parallelogram determined by \(\overrightarrow{b}\) and \(\overrightarrow{c}\)(5 marks)

Ans:

altitude of the parallelepiped

Ques. Consider \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) be three non-zero vectors so that \(\overrightarrow{c}\) is a unit vector perpendicular to both \(\overrightarrow{a}\) and \(\overrightarrow{b}\). If the angle formed between \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is π/6. Show that \([\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]^{2}=\frac{1}{4}|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}\). (3 marks)

Ans:

To Prove

Ques. If given \(\overrightarrow{a}=\widehat{i}-\widehat{k}, \overrightarrow{b}=x\widehat{i}+\widehat{j}+(1-x)\widehat{k},\overrightarrow{c}=y\widehat{i}+x\widehat{j}+(1+x-y)\widehat{k}\), prove that \([\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]\) depends on neither x nor y . (5 marks)

Ans:

to prove the dependency on neither x nor y

Ques. If the given vectors are a\(\widehat{i}\) + a\(\widehat{j}\) + a\(\widehat{k}\) , \(\widehat{i}\) + \(\widehat{k}\) and c\(\widehat{i}\) + c\(\widehat{j}\) + b\(\widehat{k}\) such are coplanar, prove that c is the geometric mean of a and b. (3 marks)

Ans:

c is the geometric mean of a and b

Ques. If \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) are given three non-coplanar vectors denoted by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of \((\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{b}\times\overrightarrow{c})+(\overrightarrow{b}+\overrightarrow{c}).(\overrightarrow{c}\times\overrightarrow{a})+(\overrightarrow{c}+\overrightarrow{a}).(\overrightarrow{a}\times\overrightarrow{b})\)(3 marks)

Ans:

Find the Values

Ques. Find out the volume of the parallelepiped whose coterminous edges are denoted by the vectors -6\(\widehat{i}\) +14\(\widehat{j}\) +10\(\widehat{k}\), 14\(\widehat{i}\) - 10\(\widehat{j}\) - 6\(\widehat{k}\) and 2\(\widehat{i}\) + 4\(\widehat{j}\) - 2\(\widehat{k}\)(3 marks)

Ans:

volume of the parallelepiped whose coterminous edges are denoted by the vectors

Ques. If \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) are xi\(\widehat{i}\) - 12\(\widehat{j}\) - \(\widehat{k}\), 2\(\widehat{i}\) + 2x\(\widehat{j}\) + \(\widehat{k}\) and \(\widehat{i}\) + \(\widehat{k}\) respectively and given vectors \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) create a left-handed system, then find out the range of the system. (3 marks)

Ans: As it’s a right handed system, so

  1. (b × c )= 0

Now

\(\overrightarrow{b}\)x\(\overrightarrow{c}\) = (2\(\widehat{i}\) + 2x\(\widehat{j}\) + \(\widehat{k}\)) x (\(\widehat{i}\) + \(\widehat{k}\) ) = 2x\(\widehat{i}\) – \(\widehat{j}\) – 2x\(\widehat{k}\)

  1. (b × c ) = (x\(\widehat{i}\) - 12\(\widehat{j}\) - \(\widehat{k}\)) . (2x\(\widehat{i}\) – \(\widehat{j}\) – 2x\(\widehat{k}\) ) = 12x2 + 2x - 12 = 0

2 (x2 + x - 6) = 0 

(x-2) (x+3) = 0

So, x will be 2 or -3.

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CBSE CLASS XII Related Questions

  • 1.

    At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


    Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
    On the basis of the above information, answer the following questions :


      • 2.
        Find:

        If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

          • \(0\)
          • \(-2\)
          • \(-1\)
          • \(2\)

        • 3.
          Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


            • 4.
              Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


                • 5.
                  Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).


                    • 6.
                      Find:

                      The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

                        • \(-\frac{\pi}{2}\)
                        • \(-\frac{\pi}{4}\)
                        • \(\frac{\pi}{4}\)
                        • \(\frac{\pi}{2}\)
                      CBSE CLASS XII Previous Year Papers

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