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Chapter 2: Problem 91

If \(\mathbf{a}\) and \(\mathbf{b}\) are two unit vectors perpendicular to each other and \(\mathbf{c}=\lambda_{1} \mathbf{a}+\lambda_{2} \mathbf{b}+\lambda_{3}(\mathbf{a} \times \mathbf{b})\), then the following is (are) true (a) \(\lambda_{1}=\mathbf{a} \cdot \mathbf{c}\) (b) \(\lambda_{2}=|\overrightarrow{\mathrm{b}} \times \overline{\mathrm{a}}|\) (c) \(\lambda_{3}=\mid(\mathrm{a} \times \mathbf{b}) \times \mathbf{c}\) (d) \(\lambda_{1}+\lambda_{2}+\lambda_{3}=(a+b+a \times b) \cdot c\)

Short Answer

Expert verified
Statements (a), (c), and (d) are true.

Step by step solution

01

Understand the Given Vectors

We have two unit vectors, \(\mathbf{a}\) and \(\mathbf{b}\), that are perpendicular to each other. By definition of unit vectors, \(|\mathbf{a}|=1\) and \(|\mathbf{b}|=1\), and since they are perpendicular, \(\mathbf{a} \cdot \mathbf{b} = 0\). \(\mathbf{c}\) is expressed in terms of these vectors as \(\mathbf{c} = \lambda_1 \mathbf{a} + \lambda_2 \mathbf{b} + \lambda_3 (\mathbf{a} \times \mathbf{b})\).
02

Evaluate Option (a)

To test if \(\lambda_1 = \mathbf{a} \cdot \mathbf{c}\), compute \(\mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot (\lambda_1 \mathbf{a} + \lambda_2 \mathbf{b} + \lambda_3 (\mathbf{a} \times \mathbf{b}))\). Since \(\mathbf{a} \cdot \mathbf{b} = 0\) and \(\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0\), this simplifies to \(\lambda_1 \cdot 1 = \lambda_1\). Therefore, option (a) is true.
03

Evaluate Option (b)

Option (b) states that \(\lambda_2 = |\mathbf{b} \times \mathbf{a}|\). Since \(\mathbf{b} \times \mathbf{a} = - (\mathbf{a} \times \mathbf{b})\), we have \(|\mathbf{b} \times \mathbf{a}| = |\mathbf{a} \times \mathbf{b}| = 1\) since both vectors are unit vectors and perpendicular, making the magnitude of their cross product equal to 1. Therefore, \(\lambda_2\) is not intrinsically equal to \(|\mathbf{b} \times \mathbf{a}|\) unless other given conditions specify so. As is, this option is false.
04

Evaluate Option (c)

Check if \(\lambda_3 = |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|\). Using vector identity, \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{c} \cdot \mathbf{a}) \mathbf{b}\), and substitute \(\mathbf{c} = \lambda_1 \mathbf{a} + \lambda_2 \mathbf{b} + \lambda_3 (\mathbf{a} \times \mathbf{b})\). This yields \(\lambda_2 \mathbf{a} - \lambda_1 \mathbf{b}\) giving \(|\mathbf{a} \times \mathbf{b}|\cdot \mathbf{c} = \lambda_3\), making option (c) true.
05

Evaluate Option (d)

For the expression, \(\lambda_1 + \lambda_2 + \lambda_3 = (\mathbf{a} + \mathbf{b} + \mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\), compute the dot product: \((\mathbf{a} + \mathbf{b} + \mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (\lambda_1 \mathbf{a} + \lambda_2 \mathbf{b} + \lambda_3 (\mathbf{a} \times \mathbf{b}))\) gives \(\lambda_1 + \lambda_2 + \lambda_3\). Therefore, option (d) is verified as true.

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Key Concepts

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

Unit Vectors
Unit vectors play an essential role in vector algebra, as they help express direction without accounting for magnitude. A unit vector is a vector with a length of one. You can obtain it by dividing a vector by its magnitude. Key characteristics include:
  • Magnitude: Always equal to 1.
  • Notation: Commonly represented with a lowercase letter and a hat, like \( \hat{i}, \hat{j}, \hat{k} \).
  • Direction: Useful for defining directions in space.
For example, consider vectors \( \mathbf{a} \) and \( \mathbf{b} \) as unit vectors. We know they are already normalized, hence, \( |\mathbf{a}| = 1 \) and \( |\mathbf{b}| = 1 \). If these are perpendicular, their dot product becomes zero, \( \mathbf{a} \cdot \mathbf{b} = 0 \). This property is crucial for calculations in problems involving orthogonality in vectors.
Cross Product
The cross product is a way to multiply two vectors, resulting in a third vector that is perpendicular to both of them. It is especially useful in physics and engineering to find a vector normal to a plane formed by two vectors.
  • Perpendicularity: The result vector is perpendicular to the original vectors.
  • Magintude: Given by \( |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}||\sin \theta| \), where \( \theta \) is the angle between them.
  • Direction: Determined using the right-hand rule.
When both vectors \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors and perpendicular, the magnitude of the cross product becomes 1 (since \( \sin 90^\circ = 1 \)). Thus, \( |\mathbf{a} \times \mathbf{b}| = 1 \) as noted in options of the problem.
Dot Product
Dot products multiply two vectors to give a scalar, providing an idea of how much one vector extends in the direction of another.
  • Scalar Result: Unlike the cross product, the dot product results in a scalar value.
  • Equation: \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta \).
  • Application: Used to find angles between vectors, projection, and check perpendicularity.
If two vectors are perpendicular, their dot product is zero since \( \cos 90^\circ = 0 \). In the solution for Option (a), using the dot product helps to establish that \( \lambda_1 = \mathbf{a} \cdot \mathbf{c} \), leading to the simplification \( \lambda_1 = \lambda_1 \), verifying the condition.
Vector Equations
Vector equations are equations where the unknowns are vectors, and they can describe a line or plane in space. These are crucial in applications such as physics and computer graphics. Consider the vector equation \( \mathbf{c} = \lambda_1 \mathbf{a} + \lambda_2 \mathbf{b} + \lambda_3 (\mathbf{a} \times \mathbf{b}) \) in our exercise, which expresses vector \( \mathbf{c} \) in terms of components along vectors \( \mathbf{a}, \mathbf{b} \), and their cross product.
  • Component Form: Useful to break down vectors along different reference directions.
  • Understanding Coefficients: The \( \lambda \) terms act as scalars that adjust the proportion each vector contributes to \( \mathbf{c} \).
This decomposition is particularly helpful when analyzing vector relationships, making complex vector problems easier to solve by breaking them into simpler components.

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Most popular questions from this chapter

If \(\mathbf{a}, \mathbf{b}\), c are three non-zero vectors, then which of the following statement (s) is/are true? (a) \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c}), \mathbf{b} \times(\mathbf{c} \times \mathbf{a}), \mathbf{c} \times(\mathbf{a} \times \mathbf{b})\) form a right handed system (b) \(c_{(}(a \times b) \times c, a \times b\) form a right handed system (c) \(a \cdot b+b \cdot c+c \cdot a<0\), if \(a+b+c=0\) (d) \(\frac{(a \times b) \cdot(b \times c)}{(b \times c) \cdot(a \times c)}=-1\), if \(a+b+c=0\)

If the angle between the vectors \(\mathbf{a}=\hat{\mathbf{i}}+(\cos x) \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{b}=\left(\sin ^{2} x-\sin x\right) \hat{\mathbf{i}}-(\cos x) \hat{j}+(3-4 \sin x) \hat{k}\) is obtuse and \(x \in\left(0, \frac{\pi}{2}\right)\), then the exhaustive set of values of \(^{\prime} x^{\prime}\) is equal to (a) \(x \in\left(0, \frac{\pi}{6}\right)\) (b) \(x \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right)\) (c) \(x \in\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) (d) \(x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)

Let \(\hat{u}\) and \(\hat{\mathbf{v}}\) are unit vectors and \(\mathbf{w}\) is a vector such that \(\hat{\mathbf{u}} \times \hat{\mathbf{v}}+\hat{\mathbf{u}}=\mathbf{w}\) and \(\mathbf{w} \times \hat{\mathbf{u}}=\hat{\mathbf{v}}\), then find the value of \([\hat{\mathbf{u}} \hat{\mathbf{v}} \mathbf{w}]\).

If the vectors \(\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}\) are mutually orthogonal, then \((\lambda, \mu)\) is equal to [AJEEE 2010] (a) \((-3,2)\) (b) \((2,-3)\) (c) \((-2,3)\) (d) \((3,-2)\)

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are non-coplanar unit vectors also \(\mathrm{b}, \boldsymbol{c}\) are non-collinear and \(2 \mathrm{a} \times(\mathrm{b} \times \mathbf{c})=\mathrm{b}+\mathrm{c}\), then (a) angle between a and \(\mathrm{c}\) is \(60^{\circ}\) (b) angle between \(\mathrm{b}\) and \(\boldsymbol{c}\) is \(30^{*}\) (c) angle between a and \(\mathbf{b}\) is \(120^{\circ}\) (d) \(\mathrm{b}\) is perpendicular to \(\mathrm{c}\)
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