/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 95 You are given vectors \(\overrig... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given vectors \(\overrightarrow{\boldsymbol{A}}=5.0 \hat{\mathfrak{x}}-6.5 \hat{\mathbf{j}}\) and \(\vec{B}=-3.5 \hat{\imath}+\) 7.0\(\hat{\mathrm{J}} .\) A third vector \(\overrightarrow{\boldsymbol{C}}\) lies in the \(x y\) -plane. Vector \(\overrightarrow{\boldsymbol{C}}\) is perpendicular to vector \(\overrightarrow{\boldsymbol{A}},\) and the scalar product of \(\overrightarrow{\boldsymbol{C}}\) with \(\overrightarrow{\boldsymbol{B}}\) is \(15.0 .\) From this information, find the components of vector \(\overrightarrow{\boldsymbol{C}}\).

Short Answer

Expert verified
The components of vector \( \overrightarrow{\boldsymbol{C}} \) are approximately \( 7.948 \) in the \( x \)-direction and \( 6.12 \) in the \( y \)-direction.

Step by step solution

01

Understand Given Vectors

We are given two vectors \( \overrightarrow{\boldsymbol{A}} = 5.0 \hat{\mathfrak{x}} - 6.5 \hat{\mathbf{j}} \) and \( \overrightarrow{\boldsymbol{B}} = -3.5 \hat{\imath} + 7.0 \hat{\mathrm{J}} \). Vector \( \overrightarrow{\boldsymbol{C}} = x \hat{\mathfrak{x}} + y \hat{\mathbf{j}} \) is unknown. \( \overrightarrow{\boldsymbol{C}} \) is perpendicular to \( \overrightarrow{\boldsymbol{A}} \) and its scalar product with \( \overrightarrow{\boldsymbol{B}} \) is 15.
02

Use Perpendicular Condition

Since \( \overrightarrow{\boldsymbol{C}} \) is perpendicular to \( \overrightarrow{\boldsymbol{A}} \), their dot product is zero: \[ \overrightarrow{\boldsymbol{A}} \cdot \overrightarrow{\boldsymbol{C}} = (5.0)x + (-6.5)y = 0 \]. This equation simplifies to: \( 5x - 6.5y = 0 \).
03

Use Given Scalar Product

The dot product of \( \overrightarrow{\boldsymbol{C}} \) with \( \overrightarrow{\boldsymbol{B}} \) is: \[ \overrightarrow{\boldsymbol{C}} \cdot \overrightarrow{\boldsymbol{B}} = (-3.5)x + 7.0y = 15.0 \]. This gives us another equation: \( -3.5x + 7y = 15 \).
04

Solve the System of Equations

Now solve the system of equations: 1. \(5x - 6.5y = 0\) 2. \(-3.5x + 7y = 15\).From Equation 1, express \(x\) in terms of \(y\): \[ x = \frac{6.5}{5}y \] Substitute \(x\) in Equation 2: \[ -3.5 \left( \frac{6.5}{5}y \right) + 7y = 15 \]\[ - \left( \frac{22.75}{5}y \right) + 7y = 15 \]Combine terms and solve for \(y\): \[ \left( \frac{35}{5}y - \frac{22.75}{5}y \right) = 15 \]\[ \frac{12.25}{5}y = 15 \]\[ y = \frac{15 \times 5}{12.25} = 6.12 \].
05

Find x-component

Using \( y = 6.12 \), find \( x \) using \( x = \frac{6.5}{5}y \):\[ x = \frac{6.5}{5} \times 6.12 \approx 7.948 \]
06

Components of Vector C

The components of vector \( \overrightarrow{\boldsymbol{C}} \) are approximately \( x \approx 7.948 \) and \( y \approx 6.12 \). Thus, \( \overrightarrow{\boldsymbol{C}} = 7.948 \hat{\mathfrak{x}} + 6.12 \hat{\mathbf{j}} \).

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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 a fundamental concept in vector calculus. It is a way to multiply two vectors to arrive at a scalar quantity. When working with vectors in a Cartesian coordinate system, the dot product of two vectors \( \overrightarrow{A} = a_1 \hat{i} + a_2 \hat{j} \) and \( \overrightarrow{B} = b_1 \hat{i} + b_2 \hat{j} \) is calculated as:\[ \overrightarrow{A} \cdot \overrightarrow{B} = a_1b_1 + a_2b_2 \] This equation sums the products of their respective components.
- **Key features of the dot product:** - It yields a scalar rather than a vector. - The result quantifies the extent to which one vector extends in the direction of another. - If the dot product is zero, the vectors are perpendicular.
In the original exercise, we used the property of a zero dot product to determine that vectors \( \overrightarrow{A} \) and \( \overrightarrow{C} \) are perpendicular. By equating their dot product to zero, we found the first equation for solving the problem. The exercise also utilized another dot product condition, \( \overrightarrow{C} \cdot \overrightarrow{B} = 15.0 \), to help solve the system of equations.
Solution of Systems of Equations
Solving systems of equations is a crucial aspect of many mathematical problems. In the context of vectors, it often involves finding unknown components that satisfy multiple conditions.
- **Steps to solve the system:** - First, set up equations based on given conditions. Here, the perpendicular vector condition led us to the equation \( 5x - 6.5y = 0 \). - Next, use the condition about the scalar product with vector \( \overrightarrow{B} \) to form another equation: \( -3.5x + 7y = 15 \). - With these two equations, you have a linear system.
- **Solving the system:** - Express one variable in terms of the other from one of the equations. For instance, from \( 5x - 6.5y = 0 \), we expressed \( x \) in terms of \( y \): \( x = \frac{6.5}{5}y \). - Substitute this expression into the other equation to find one of the variables, in this case, \( y \). Once \( y \) is found, substitute back to find \( x \).
This approach ensures that we find a consistent and accurate solution for the unknown vector components.
Perpendicular Vectors
Two vectors are considered perpendicular if they intersect at a right angle. A fundamental property of perpendicular vectors is that their dot product equals zero. This is because if \( \overrightarrow{A} \) and \( \overrightarrow{B} \) are perpendicular, then:\[ \overrightarrow{A} \cdot \overrightarrow{B} = 0 \]
In the exercise, this property is employed to establish the first equation for \( \overrightarrow{C} \)'s components.
- **Why the dot product is zero for perpendicular vectors:** - The cosine of the angle between the two vectors is zero (cosine of 90 degrees is zero). - Thus, any vector projected onto a line perpendicular to it contributes nothing to the dot product.
This relationship makes it easy to check whether vectors are orthogonal in calculations, as demonstrated in the original exercise, where the orthogonality condition helped in forming the system of equations.

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