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Chapter 9: Problem 1

Which of the following statements about general vectors \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are true? (a) \(\mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})=(\mathbf{b} \times \mathbf{a}) \cdot \mathbf{c} ;\) (b) \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\) (c) \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}\) (d) \(\mathbf{d}=\lambda \mathbf{a}+\mu \mathbf{b}\) implies \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}=0\) (e) \(\mathbf{a} \times \mathbf{c}=\mathbf{b} \times \mathbf{c}\) implies \(\mathbf{c} \cdot \mathbf{a}-\mathbf{c} \cdot \mathbf{b}=c|\mathbf{a}-\mathbf{b}|\) (f) \((\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{b})=\mathbf{b}[\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})]\)

Short Answer

Expert verified
The correct statements are: (a) true, (b) false, (c) true, (d) true, (e) false and (f) false.

Step by step solution

01

Solve statement (a)

This step involves understanding the property of dot product and cross product. The property says that for any three vectors a, b and c, \(a \cdot (b \times c) = (a \times b) \cdot c\), it follows directly from the properties of the scalar triple product. Therefore, the statement (a) is true.
02

Solve statement (b)

In this step, one needs to know that vector cross product is not associative. Meaning, \(a \times (b \times c)\) is not equal to \((a \times b) \times c\). Therefore, the statement (b) is false.
03

Solve statement (c)

This step involves applying the BAC-CAB rule, which says, \(a \times (b \times c) = (a \cdot c) b - (a \cdot b) c\). Therefore, the statement (c) is true.
04

Solve statement (d)

We have the vector d as a linear combination of vectors a and b. The vector \(a \times b\) is orthogonal to the plane formed by a and b. Thus, any linear combination of a and b will lie in the plane formed by a and b and so, it will be orthogonal to \(a \times b\). Hence, \((a \times b) \cdot d = 0\). Thus, statement (d) is true.
05

Solve statement (e)

Statement (e) does not hold true for all vectors a, b and c. Hence, statement (e) is false.
06

Solve statement (f)

The statement (f) does not hold true for all vectors a, b and c. Hence, statement (f) is false.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product
The cross product is a vector operation that results in a third vector. This third vector is perpendicular to the original two vectors. If you have two vectors \(\mathbf{a}\) and \(\mathbf{b}\), the cross product is written as \(\mathbf{a} \times \mathbf{b}\). It is often used in physics to find a vector perpendicular to a plane.
  • The magnitude of the cross product can be calculated as \( |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta\), where \(\theta\) is the angle between the two vectors.
  • The direction follows the right-hand rule, meaning your thumb points in the direction of \(\mathbf{a} \times \mathbf{b}\) if your fingers curl from \(\mathbf{a}\) to \(\mathbf{b}\).
  • Cross product is not commutative, that is, \(\mathbf{a} \times \mathbf{b} eq \mathbf{b} \times \mathbf{a}\).
Understanding these properties helps when solving vector calculus problems involving cross products.
Dot Product
The dot product, also known as the scalar product, is a way to multiply two vectors to obtain a scalar (a real number, not a vector). When you compute the dot product \(\mathbf{a} \cdot \mathbf{b}\), you are finding how much one vector goes in the direction of another.
  • Its formula is \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
  • The dot product is commutative, meaning \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\).
  • If the dot product is zero, \(\mathbf{a} \cdot \mathbf{b} = 0\), it indicates that the vectors are perpendicular to each other.
Using the dot product can help determine angles between vectors or project a vector onto another.
Triple Product
The triple product involves three vectors and can be either a scalar or vector product. The scalar triple product, \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\), measures the volume of the parallelepiped defined by the three vectors. This is because it gives the product of the area of the base and the height of the parallelepiped.
  • The value is calculated as \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\) or equivalently, \((\mathbf{b} \times \mathbf{a}) \cdot \mathbf{c}\).
  • The geometric interpretation is essential in physics and geometry for understanding volumes.
  • Notably, if the scalar triple product is zero, the vectors are coplanar, meaning they lie on the same plane.
This concept is used in understanding the orientation and relationships between three-dimensional vectors.
BAC-CAB Rule
The BAC-CAB rule is a mnemonic that helps to remember the vector triple product identity. It states that \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}\). This rule is helpful for simplifying expressions involving nested cross products.
  • This identity is particularly useful in vector calculus and physics when dealing with forces and moments in three dimensions.
  • It helps to break down complex vector products into more manageable components.
  • Understanding this rule allows for the simplification without the need for directly calculating cross products. Instead, it leverages dot products, which are simpler to compute.
The BAC-CAB rule is crucial for efficiently solving vector calculus problems that involve multiple vector operations.

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Most popular questions from this chapter

By proceeding as indicated below, prove the parallel axis theorem, which states that, for a body of mass \(M\), the moment of inertia \(I\) about any axis is related to the corresponding moment of inertia \(I_{0}\) about a parallel axis that passes through the centre of mass of the body by $$ I=I_{0}+M a_{\perp}^{2} $$ where \(a_{\perp}\) is the perpendicular distance between the two axes. Note that \(I_{0}\) can be written as $$ \int(\mathbf{n} \times \mathbf{r}) \cdot(\hat{\mathbf{n}} \times \mathbf{r}) d m $$ where \(r\) is the vector position, relative to the centre of mass, of the infinitesimal mass \(d m\) and \(\hat{\mathrm{n}}\) is a unit vector in the direction of the axis of rotation. Write a similar expression for \(I\) in which \(r\) is replaced by \(\mathbf{r}^{\prime}=\mathbf{r}-\mathbf{a}\), where \(\mathbf{a}\) is the vector position of any point on the axis to which \(I\) refers. Use Lagrange's identity and the fact that \(\int \mathbf{r} d m=0\) (by the definition of the centre of mass) to establish the result.

Identify the following surfaces: (a) \(|\mathbf{r}|=k ;(\mathrm{b}) \mathbf{r} \cdot \mathbf{u}=l ;(\mathrm{c}) \mathbf{r} \cdot \mathbf{u}=m|\mathbf{r}|\) for \(-1 \leq m \leq+1\) (d) \(|\mathbf{r}-(\mathbf{r} \cdot \mathbf{u}) \mathbf{u}|=n\) Here \(k, l, m\) and \(n\) are fixed scalars and \(\mathbf{u}\) is a fixed unit vector.

Use vector methods to prove that the lines joining the mid-points of the opposite edges of a tetrahedron \(O A B C\) meet at a point and that this point bisects each of the lines.

Prove Lagrange's identity, i.e. $$ (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}) $$

Let \(O, A, B\) and \(C\) be four points with position vectors \(\mathbf{0}, \mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\), and denote by \(\mathbf{g}=\lambda \mathbf{a}+\mu \mathbf{b}+v \mathbf{c}\) the position of the centre of the sphere on which they all lie. (a) Prove that \(\lambda, \mu\) and \(v\) simultaneously satisfy $$ (\mathbf{a} \cdot \mathbf{a}) \lambda+(\mathbf{a} \cdot \mathbf{b}) \mu+(\mathbf{a} \cdot \mathbf{c}) v=\frac{1}{2} a^{2} $$ and two other similar equations. (b) By making a change of origin, find the centre and radius of the sphere on which the points \(\mathbf{p}=3 \mathbf{i}+\mathbf{j}-2 \mathbf{k}, \mathbf{q}=4 \mathbf{i}+3 \mathbf{j}-3 \mathbf{k}, \mathbf{r}=7 \mathbf{i}-3 \mathbf{k}\) and \(\mathbf{s}=6 \mathbf{i}+\mathbf{j}-\mathbf{k}\) all lie.
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