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If the vectors \(\mathbf{A}\) and \(\mathbf{B}\) are perpendicular, what is the component of \(\mathbf{A}\) along the direction of \(\mathbf{B} ?\) What is the component of \(\mathbf{B}\) along the direction of \(\mathbf{A}\) ?

Short Answer

Expert verified
The component of \(\mathbf{A}\) along the direction of \(\mathbf{B}\) and the component of \(\mathbf{B}\) along the direction of \(\mathbf{A}\) are both zero.

Step by step solution

01

Understanding Vector Components and Dot Product

The component of a vector along the direction of another vector can be found using the dot product formula: \(\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(\theta)\), where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\). The component of \(\mathbf{A}\) along \(\mathbf{B}\) is given by the projection formula: \(\text{comp}_{\mathbf{B}}\mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}\). Similarly, the component of \(\mathbf{B}\) along \(\mathbf{A}\) is given by \(\text{comp}_{\mathbf{A}}\mathbf{B} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}|}\).
02

Applying the Perpendicularity Condition

Since \(\mathbf{A}\) and \(\mathbf{B}\) are perpendicular, the angle between them is \(90^\circ\), which means \(\cos(\theta) = \cos(90^\circ) = 0\). Thus, the dot product \(\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(90^\circ) = 0\).
03

Calculating the Components

Since \(\mathbf{A} \cdot \mathbf{B} = 0\), the component of \(\mathbf{A}\) along the direction of \(\mathbf{B}\) is \(\text{comp}_{\mathbf{B}}\mathbf{A} = \frac{0}{|\mathbf{B}|} = 0\), and the component of \(\mathbf{B}\) along the direction of \(\mathbf{A}\) is \(\text{comp}_{\mathbf{A}}\mathbf{B} = \frac{0}{|\mathbf{A}|} = 0\). Therefore, both components are zero.

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Key Concepts

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

Dot Product
The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is fundamental in physics and mathematics, particularly when dealing with vectors.

In physics, the dot product is used to determine the amount of one vector that points in the same direction as another vector. The general formula for the dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(\theta) \), where \( \theta \) represents the angle between the vectors and \( |\mathbf{A}| \) and \( |\mathbf{B}| \) are the magnitudes (or lengths) of vectors \( \mathbf{A} \) and \( \mathbf{B} \) respectively.

When the angle \( \theta = 90^\circ \) (meaning the vectors are perpendicular), the cosine of the angle is zero. Therefore, \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(90^\circ) = 0 \). This property is especially significant because it simplifies many problems in vector mathematics and physics, serving as the basis for orthogonality in vector spaces.
Vectors in Physics
Vectors are essential in physics as they represent quantities having both magnitude and direction, which is crucial for describing physical phenomena. Common examples include velocity, acceleration, force, and displacement. A vector is often denoted by an arrow, where the length represents its magnitude and the direction of the arrow indicates its direction.

Understanding vector components is incredibly important in physics. These components represent the influence of a vector in a particular direction. For instance, the force applied at an angle to an object can be broken down into horizontal and vertical components. Each component can then be analyzed independently to understand the object's motion.

To find vector components, one can use trigonometric functions or the dot product when dealing with non-right angles. Vector components play a significant role in simplifying complex problems by allowing us to consider one direction at a time.
Perpendicular Vectors
Perpendicular vectors are two vectors that meet or intersect at a right angle (\( 90^\circ \)). This special relationship between vectors has important implications in vector mathematics and many practical applications. When two vectors are perpendicular, their dot product is zero, as shown in the exercise with vectors \( \mathbf{A} \) and \( \mathbf{B} \).

The significance of perpendicular vectors can be appreciated in coordinate systems, where axes are defined by mutually perpendicular unit vectors. This orthogonality allows for the separation of space into distinct directions and dimensions, which simplifies calculations and analyses. In physics, perpendicular vectors often signify independence of effects, such as forces acting in different directions having separate impacts on an object's motion.

Knowing that the components of a vector in the direction of another vector are zero when they are perpendicular can significantly ease the resolution of problems involving motion, forces, and other vector-based phenomena.

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Most popular questions from this chapter

A projectile is launched at ground level with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. It strikes a target above the ground \(3.00\) seconds later. What are the \(x\) and \(y\) distances from where the projectile was launched to where it lands?

Solve the following problem using analytical techniques: Suppose you walk \(18.0 \mathrm{~m}\) straight west and then \(25.0 \mathrm{~m}\) straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements \(\mathbf{A}\) and \(\mathbf{B}\), as in Figure, then this problem asks you to find their sum \(\mathbf{R}=\mathbf{A}+\mathbf{B}\).)

You drive \(7.50 \mathrm{~km}\) in a straight line in a direction \(15^{\circ}\) east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

Explain why it is not possible to add a scalar to a vector.

An arrow is shot from a height of \(1.5 \mathrm{~m}\) toward a cliff of height \(H\). It is shot with a velocity of \(30 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal. It lands on the top edge of the cliff \(4.0\) s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow's impact speed just before hitting the cliff?

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