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Chapter 2: Problem 96

Let the unit vectors a and \(\mathrm{b}\) be perpendicular and unit vector \(\mathrm{c}\) is inclined at angle \(\alpha\) to \(\mathrm{a}\) and \(\mathrm{b}\). If \(\mathrm{c}=l \mathrm{a}+m \mathbf{b}+n(\mathrm{a} \times \mathbf{b})\), then (a) \(l=m\) (b) \(n^{2}=1-2 l^{2}\) (c) \(n^{2}=-\cos 2 \alpha\) (d) \(m^{2}=\frac{1+\cos 2 \alpha}{2}\)

Short Answer

Expert verified
Options (a), (b), and (d) are correct.

Step by step solution

01

Expressing Vector Magnitude

First, note that vectors \(\mathbf{a}\) and \(\mathbf{b}\) are unit vectors, which means \(|\mathbf{a}| = 1\) and \(|\mathbf{b}| = 1\). Also, they are perpendicular, so \(\mathbf{a} \cdot \mathbf{b} = 0\). Vector \(\mathbf{c}\) is expressed as \(\mathbf{c} = l \mathbf{a} + m \mathbf{b} + n(\mathbf{a} \times \mathbf{b})\). Since \(\mathbf{c}\) is also a unit vector, \(|\mathbf{c}| = 1\). Therefore, the magnitude can be expressed as \(l^2 + m^2 + n^2 = 1\).
02

Angle Relationship

Since vector \(\mathbf{c}\) is inclined at angle \(\alpha\) from vectors \(\mathbf{a}\) and \(\mathbf{b}\), it means the dot products \(\mathbf{c} \cdot \mathbf{a} = \cos \alpha\) and \(\mathbf{c} \cdot \mathbf{b} = \cos \alpha\). Therefore, \(l = \mathbf{c} \cdot \mathbf{a} = \cos \alpha\) and \(m = \mathbf{c} \cdot \mathbf{b} = \cos \alpha\). This implies \(l = m\).
03

Simplifying the Magnitude Equation

Using \(l = m\) in the magnitude equation from Step 1, \(2l^2 + n^2 = 1\). From here, solving for \(n^2\), we get \(n^2 = 1 - 2l^2\).
04

Evaluating Trigonometric Identities

Given the vector \(\mathbf{c}\) is inclined at angle \(\alpha\) with vectors \(\mathbf{a}\) and \(\mathbf{b}\), use trigonometric identities. We know that double angle formula gives \(\cos 2\alpha = 2\cos^2 \alpha - 1\). Thus, rearranging, we find that \(\cos^2 \alpha = \frac{1 + \cos 2\alpha}{2}\). Thus \(l^2 = m^2 = \frac{1 + \cos 2\alpha}{2}\).
05

Identifying the Solution

Comparing the derived equations with the options given: option (a) \(l = m\) is correct from Step 2. The equation from Step 3 matches with option (b): \(n^2 = 1 - 2l^2\). Also from the trigonometric identity in Step 4, option (d) matches: \(m^2 = \frac{1 + \cos 2\alpha}{2}\). However, option (c) \(n^2 = -\cos 2\alpha\) does not match with any equations.

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Key Concepts

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

Unit Vectors
Unit vectors are vectors with a magnitude of 1. They are used to indicate direction without considering magnitude. Basically, a unit vector is any vector divided by its magnitude, which makes its length equal to 1.
In the given exercise, vectors \(\mathbf{a}\) and \(\mathbf{b}\) are unit vectors. This implies that their magnitudes are both equal to 1: \(|\mathbf{a}| = 1\) and \(|\mathbf{b}| = 1\). Moreover, unit vectors are often used to define high-level concepts like axes in the Cartesian coordinate system, represented by \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\).
  • Because unit vectors have a magnitude of 1, they simplify calculations involving direction.
  • They are essential when performing operations like dot and cross products as they provide a base measurement.
Dot Product
The dot product (also known as the scalar product) of two vectors is a measure of how closely they are aligned. Mathematically, the dot product of two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), is given by \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta\), where \(\theta\) is the angle between the two vectors.
In the context of the exercise, the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular, so \(\mathbf{a} \cdot \mathbf{b} = 0\).
  • The dot product results in a scalar value (a single number), not another vector.
  • When two vectors are perpendicular, their dot product is zero, reflecting their perpendicular nature.
  • In the exercise, the relationship \(\mathbf{c} \cdot \mathbf{a} = \cos \alpha\) and \(\mathbf{c} \cdot \mathbf{b} = \cos \alpha\) helped solve for the components \(l\) and \(m\).
Vector Magnitude
The magnitude of a vector is its length or size calculated using the Euclidean distance formula. For a vector \(\mathbf{v} = \langle x, y, z \rangle\), its magnitude \(|\mathbf{v}|\) is calculated as \(\sqrt{x^2 + y^2 + z^2}\).
In the problem, the magnitude of vector \(\mathbf{c}\) was given as 1 since it is a unit vector: \(|\mathbf{c}| = 1\). This leads to the equation \(l^2 + m^2 + n^2 = 1\) since \(\mathbf{c} = l \mathbf{a} + m \mathbf{b} + n(\mathbf{a} \times \mathbf{b})\).
  • Knowing a vector鈥檚 magnitude is crucial for normalization鈥攃onverting a vector to a unit vector.
  • Magnitude helps in understanding a vector鈥檚 scale relative to other vectors.
  • In this exercise, equating the components of \(\mathbf{c}\) to 1 helped determine relationships between \(l, m,\) and \(n\).
Cross Product
The cross product (or vector product) of two vectors is another vector that is perpendicular to the plane of the two input vectors. It is given by \(\mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle \).
In this exercise, \(\mathbf{a} \times \mathbf{b}\) generates a vector that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
  • The magnitude of the cross product gives the area of the parallelogram formed by the two vectors.
  • The direction of the cross product vector is determined by the right-hand rule.
  • In the exercise, adding \(n(\mathbf{a} \times \mathbf{b})\) to the expression of \(\mathbf{c}\) contributes to the angle \(\alpha\) the vector makes with \(\mathbf{a}\) and \(\mathbf{b}\).

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Most popular questions from this chapter

Let \(\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and \(\mathbf{c}=\boldsymbol{\mathbf { x }}+(x-2) \hat{\mathbf{j}}-\hat{\mathbf{k}}\). If the vector c lies in the plane of \(a\) and \(b\), then \(x\) equal to (a) 0 (b) 1 [AIEEE 2007] (c) \(-4\) (d) \(-2\)

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 \(a+b+c=0\)

Let \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) be three non-zero vectors such that no two of them are collinear and \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\frac{1}{3}|\mathbf{b} \| \mathbf{c}| \mathbf{a}\). If \(\theta\) is the angle between vectors \(\mathbf{b}\) and \(c\), then a value of \(\sin \theta\) [Single Correct Type, 2015 Adv.] is \(\begin{array}{llll}\text { (a) } \frac{2 \sqrt{2}}{3} & \text { (b) } \frac{-\sqrt{2}}{3} & \text { (c) } \frac{2}{3} & \text { (d) } \frac{-2 \sqrt{3}}{3}\end{array}\)

\(\mathbf{A}, \mathbf{B}\) and \(\mathbf{C}\) are three vectors given by \(2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\) Then, find \(\mathbf{R}\), which satisfies the relation \(\mathbf{R} \times \mathbf{B}=\mathbf{C} \times \mathbf{B}\) and \(\mathbf{R} \cdot \mathbf{A}=\mathbf{0}\).

Statement I \(\left(\mathrm{S}_{1}\right):\) If \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)\) are non-collinear points. Then, every point \((x, y)\) in the plane of \(\triangle A B C\), can be expressed in the form \(\left(\frac{k x_{1}+l x_{2}+m x_{3}}{k+l+m}, \frac{k y_{1}+l y_{2}+m y_{3}}{k+l+m}\right)\) Statement II \(\left(\mathrm{S}_{2}\right)\) The condition for coplanarity of four points \(A(\mathbf{a}), B(\mathbf{b}), C(c), D(d)\) is that there exists scalars, \(l, m, n, p\) not all zeros such that $$ l a+m \mathbf{b}+n c+p d=0 $$ where \(l+m+n+p=0\)
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