Problem 30
Question
For two vectors \(\mathbf{A}\) and \(\mathbf{B},|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|\) is always true when (a) \(|A|=|B| \neq 0\) (b) \(\mathrm{A} \perp \mathrm{B}\) (c) \(|\mathrm{A}|=|\mathrm{B}| \neq 0\) and \(\mathrm{A}\) and \(\mathrm{B}\) are parallel or anti parallel (d) when either |A|or |B| is zero.
Step-by-Step Solution
Verified Answer
(b) \(\mathrm{A} \perp \mathrm{B}\) is always true.
1Step 1: Understanding the Condition
We are given \(|\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}|\). This implies the square of the magnitudes are equal, i.e., \(\mathbf{A} + \mathbf{B}\)\cdot(\mathbf{A} + \mathbf{B}) = (\mathbf{A} - \mathbf{B})\cdot(\mathbf{A} - \mathbf{B})."
2Step 2: Expanding Both Sides
Expand both sides of the equation using dot products: \\[\begin{align*} (\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} + \mathbf{B}) &= \mathbf{A}\cdot\mathbf{A} + 2\mathbf{A}\cdot\mathbf{B} + \mathbf{B}\cdot\mathbf{B},\ (\mathbf{A} - \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) &= \mathbf{A}\cdot\mathbf{A} - 2\mathbf{A}\cdot\mathbf{B} + \mathbf{B}\cdot\mathbf{B}.\end{align*}\]
3Step 3: Simplifying the Equation
Set these two expansions equal \\[\mathbf{A}\cdot\mathbf{A} + 2\mathbf{A}\cdot\mathbf{B} + \mathbf{B}\cdot\mathbf{B} = \mathbf{A}\cdot\mathbf{A} - 2\mathbf{A}\cdot\mathbf{B} + \mathbf{B}\cdot\mathbf{B}.\]\Simplifying by canceling \(\mathbf{A}\cdot\mathbf{A}\) and \(\mathbf{B}\cdot\mathbf{B}\), it follows that: \\[2\mathbf{A}\cdot\mathbf{B} = -2\mathbf{A}\cdot\mathbf{B} \Rightarrow 4\mathbf{A}\cdot\mathbf{B} = 0\].
4Step 4: Determining the Conditions
The condition \(\mathbf{A}\cdot\mathbf{B} = 0\) implies that vectors \(\mathbf{A}\) and \(\mathbf{B}\) are perpendicular. Thus, statement (b) is always true. None of the other conditions guarantee perpendicularity under all circumstances.
Key Concepts
vector magnitudesperpendicular vectorsparallel vectors
vector magnitudes
When it comes to understanding vectors, one of the fundamental concepts is the magnitude. Think of the magnitude as the "size" or "length" of a vector. For a vector \(\mathbf{A}\), its magnitude is represented by \(|\mathbf{A}|\). This is analogous to the distance from the origin to the point where the vector points in a coordinate system. So, if you are given a vector \(\mathbf{A} = \langle x, y, z \rangle\), its magnitude can be calculated using the formula:\[|\mathbf{A}| = \sqrt{x^2 + y^2 + z^2}\] A key aspect is that the magnitude is always a non-negative number. It's important when comparing vectors or calculating with them. For instance, if two vectors have the same magnitude, \(|\mathbf{A}| = |\mathbf{B}|\), they can potentially be parallel, perpendicular, or have other interesting geometric relationships. This is part of what was explored in the original exercise, namely underlining conditions like when the magnitudes are equal.
- The magnitude provides a scalar measure of the vector's size.
- Equal magnitudes do not imply the vectors point in the same direction unless more information is known.
- Magnitude is invariant to direction – it only considers the length.
perpendicular vectors
Perpendicular vectors hold a unique position in vector mathematics. When two vectors \(\mathbf{A}\) and \(\mathbf{B}\) are said to be perpendicular, it means that they form a 90-degree angle with each other. Mathematically, this relationship is expressed by the dot product: \[ \mathbf{A} \cdot \mathbf{B} = 0 \]This condition is very significant because it tells us that there is no component of one vector in the direction of the other. In our context of the exercise, the equation \( 4\mathbf{A}\cdot\mathbf{B} = 0 \) perfectly embodies the idea of perpendicularity — if the vectors sum to zero in their products, they are perpendicular.
- The dot product is the key operator to check for perpendicularity.
- Perpendicular vectors imply no overlap in directional influence.
- This concept supports applications like projection, where only the perpendicular component is utilized.
parallel vectors
Parallel vectors not only point in the same or completely opposite directions but also maintain a distinct mathematical relationship. For two vectors \(\mathbf{A}\) and \(\mathbf{B}\) to be considered parallel, there must exist a scalar \(k\) such that:\[ \mathbf{A} = k\mathbf{B} \]This implies that one vector is a scaled version of the other. Parallelism is quite powerful because it simplifies many computations, such as in solving systems of linear equations or geometry problems involving vectors. In the exercise context, the scenario of equal magnitudes and parallel vectors can create interesting conditions.
- Parallel vectors have no concept of perpendicularity unless one of them is a zero vector.
- The directionality is preserved through scalar multiplication.
- They can also be anti-parallel if \(k\) is negative, showing they point in opposite directions.
Other exercises in this chapter
Problem 29
What vector must be added to the sum of two vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and \(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \ha
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Which of the following statements is/are correct? (a) The magnitude of the vector \(3 \hat{i}+4 \hat{j}\) is 5 (b) A force \((3 \hat{i}+4 \hat{j}) \mathrm{N}\)
View solution Problem 31
Given, vector, \(\mathbf{A}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and vector \(\mathbf{B}=3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{
View solution Problem 32
Given, \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B} .\) Then, \(|\mathbf{A} \times \mathbf{B}|\) is equal to (a) \(\sin \theta\) (b) \(\cos \
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