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Magazine Chapter 13: Problem 31
Find the value(s) of \(a\) making \(\vec{v}=5 a \vec{i}-3 \vec{j}\) parallel to \(\vec{w}=a^{2} \vec{i}+6 \vec{j}\)
Short Answer
Expert verified
The values of \(a\) are 0 and -10.
Step by step solution
01
Understand Vector Parallelism
Two vectors are parallel if one is a scalar multiple of the other. For vectors \(\vec{v} = 5a\vec{i} - 3\vec{j}\) and \(\vec{w} = a^2\vec{i} + 6\vec{j}\), they are parallel if there exists a constant \(k\) such that \(5a = ka^2\) and \(-3 = 6k\).
02
Solve for Scalar from Second Equation
From \(-3 = 6k\), solve for \(k\):\[ k = -\frac{1}{2} \]
03
Substitute Scalar into First Equation
Substitute \(k = -\frac{1}{2}\) into the first equation to find \(a\):\[ 5a = -\frac{1}{2}a^2 \] Rearrange to form the quadratic equation:\[ a^2 + 10a = 0 \]
04
Factor the Quadratic Equation
Factor \(a^2 + 10a = 0\):\[ a(a + 10) = 0 \] This gives solutions \(a = 0\) or \(a = -10\).
05
Verify the Solutions
Verify by substituting \(a = 0\) and \(a = -10\) into the vectors and ensuring they satisfy parallelism.- If \(a = 0\): \(\vec{v} = 0\), \(\vec{w} = 6\vec{j}\) (trivial zero vector case)- If \(a = -10\): \(\vec{v} = -50\vec{i} - 3\vec{j}\), \(\vec{w} = 100\vec{i} + 6\vec{j}\), both are scalar multiples confirming parallelism.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar, which is simply a numerical factor. The result is a new vector that points in the same or exactly opposite direction, depending on the sign of the scalar, and has a magnitude altered by the scalar amount.
In vector parallelism, two vectors are parallel if one vector can be expressed as a scalar multiple of the other. This means multiplying all components of one vector by the same scalar yields the components of the other vector. Consider the vectors \( \vec{v} = 5a\vec{i} - 3\vec{j} \) and \( \vec{w} = a^2\vec{i} + 6\vec{j} \).
In vector parallelism, two vectors are parallel if one vector can be expressed as a scalar multiple of the other. This means multiplying all components of one vector by the same scalar yields the components of the other vector. Consider the vectors \( \vec{v} = 5a\vec{i} - 3\vec{j} \) and \( \vec{w} = a^2\vec{i} + 6\vec{j} \).
- To find if these vectors are parallel, we look for a constant \( k \) such that each component of \( \vec{v} \) is \( k \) times the respective component of \( \vec{w} \).
- This results in the equations \( 5a = ka^2 \) and \( -3 = 6k \).
- These equations allow us to solve for \( k \) and consequently determine the value of \( a \) that makes them parallel.
Quadratic Equation
A quadratic equation is a type of polynomial equation of the second degree, generally in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations involves finding the values of \( x \) that satisfy the equation.
In this exercise, the quadratic equation emerges from substituting the scalar \( k \) into the equation relating \( \vec{v} \) and \( \vec{w} \).
In this exercise, the quadratic equation emerges from substituting the scalar \( k \) into the equation relating \( \vec{v} \) and \( \vec{w} \).
- We arrive at \( 5a = -\frac{1}{2}a^2 \), which is simplified and rearranged to the quadratic equation \( a^2 + 10a = 0 \).
- This equation can be solved either by using the quadratic formula, completing the square, or through factorization, depending on which method seems most straightforward.
Factorization
Factorization is a technique used to simplify expressions or solve equations by expressing a polynomial as a product of its factors. When we factorize, we break it down into simpler components that can be more easily analyzed or solved.
For our quadratic equation \( a^2 + 10a = 0 \), we factorize as follows:
For our quadratic equation \( a^2 + 10a = 0 \), we factorize as follows:
- Identify common factors in the terms of the equation. Here, \( a \) is common, leading us to \( a(a + 10) = 0 \).
- The factored form tells us that for the product to be zero, one or both of the factors must be zero.
- Thus, \( a = 0 \) or \( a + 10 = 0 \), giving solutions \( a = 0 \) and \( a = -10 \).
Constant of Proportionality
The constant of proportionality is a constant value that relates two quantities that are directly proportional. In vector applications, this constant determines how one vector scales to become parallel to another.
- In our problem, the value \( k \) represents the constant of proportionality. It satisfies the condition \( -3 = 6k \), giving \( k = -\frac{1}{2} \).
- This constant helps us determine if a linear relationship exists between two vectors. Essentially, it is the scaling factor needed to convert one vector into another while maintaining direction.
- By solving for \( a \), and verifying values like \( a = -10 \), we confirm that the vectors \( \vec{v} \) and \( \vec{w} \) can indeed become parallel.
