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Chapter 14: Problem 15

Given three vectors \(\boldsymbol{a}, \boldsymbol{b}\) and \(\boldsymbol{c}\), their triple scalar product is defined to be \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\). It can be shown that the modulus of this is the volume of the parallelepiped formed by the three vectors. Find the volume of the parallelepiped formed by the three vectors \(\boldsymbol{a}=3 \boldsymbol{i}+\boldsymbol{j}-2 \boldsymbol{k}, \boldsymbol{b}=\boldsymbol{i}+2 \boldsymbol{j}-2 \boldsymbol{k}\) and \(\boldsymbol{c}=2 \boldsymbol{i}+5 \boldsymbol{j}+\boldsymbol{k}\)

Short Answer

Expert verified
Answer: The volume of the parallelepiped is 24 cubic units.

Step by step solution

01

Compute the cross product of \(\boldsymbol{a}\) and \(\boldsymbol{b}\)

First, we need to calculate the cross product \(\boldsymbol{a} \times \boldsymbol{b}\). The cross product of two vectors can be found using the determinant of a 3x3 matrix: $$ \boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 3 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} $$ Compute the determinant using the standard cofactor expansion method. The result is: $$ \boldsymbol{a} \times \boldsymbol{b} = (4-4)\boldsymbol{i} - (-6+2)\boldsymbol{j} + (6-2)\boldsymbol{k} = 0\boldsymbol{i}+4\boldsymbol{j}+4\boldsymbol{k} $$
02

Compute the dot product of \((\boldsymbol{a} \times \boldsymbol{b})\) and \(\boldsymbol{c}\)

Next, we need to compute the dot product of the cross product we found in the previous step and \(\boldsymbol{c}\). The dot product is calculated as \(\sum_{i=1}^3 (A_i*B_i)\), where \(A_i\) and \(B_i\) are the corresponding components of the vectors. So, we have: $$ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = 0(2) + 4(5) + 4(1) = 20 + 4 = 24 $$
03

Find the modulus of the triple scalar product

The volume of the parallelepiped formed by the three vectors is the modulus of the triple scalar product, i.e., \(|(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}|\). In this case, the triple scalar product is already a positive value, so the modulus remains unchanged: $$ |24| = 24 $$
04

State the final answer

The volume of the parallelepiped formed by the vectors \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) is 24 cubic units.

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Most popular questions from this chapter

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.

State, or find out, which of the following are scalars and which are vectors: (a) the volume of a petrol tank, (b) a length measured in metres, (c) a length measured in miles, (d) the angular velocity of a flywheel, (e) the relative velocity of two aircraft, (f) the work done by a force, (g) electrostatic potential, (h) the momentum of an atomic particle.

Point P has coordinates \((7,8)\). Point \(Q\) has coordinates \((-2,4)\) (a) Draw a sketch showing \(\mathrm{P}\) and \(\mathrm{Q}\). (b) State the position vectors of \(\mathrm{P}\) and \(\mathrm{Q}\). (c) Find an expression for \(\overrightarrow{P Q}\). (d) Find \(|\overrightarrow{\mathrm{PQ}}|\).

Find the volume of the parallelepiped whose edges are represented by the vectors \(12 i+j+k, 2 i\) and \(-2 \boldsymbol{j}+\boldsymbol{k}\)

Points A, B and C have position vectors \((9,1,1),(8,1,1)\) and \((9,0,2)\). Find (a) the equation of the plane containing A, B and \(\mathrm{C}\) (b) the area of the triangle \(\mathrm{ABC}\).
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