Chapter 14: Problem 43
Find scalars \(p\) and \(q\) such that $$ p\langle 3,-1\rangle+q\langle 4,3\rangle=\langle-6,-11\rangle $$
Short Answer
Expert verified
The scalars are \(p = 2\) and \(q = -3\).
Step by step solution
01
Set Up the Vector Equation
Write the given vectors in the equation form: \( p \langle 3, -1 \rangle + q \langle 4, 3 \rangle = \langle -6, -11 \rangle \). Distribute the scalars \( p \) and \( q \) into the vectors resulting in two equations: \( 3p + 4q = -6 \) and \(-p + 3q = -11 \).
02
Solve for q in Terms of p
Take one equation, for example, \(-p + 3q = -11\), and solve for \(q\). Rearranging gives \(3q = p - 11\), so \(q = \frac{p - 11}{3}\).
03
Substitute q in the Other Equation
Substitute \( q = \frac{p - 11}{3} \) from Step 2 into the other equation \(3p + 4q = -6\). This gives \(3p + 4(\frac{p - 11}{3}) = -6\).
04
Solve for p
Clear the fraction by multiplying through by 3: \(9p + 4(p - 11) = -18\). Distribute the 4: \(9p + 4p - 44 = -18\). Combine like terms to get \(13p - 44 = -18\). Add 44 to both sides to obtain \(13p = 26\). Finally, divide by 13 to find \(p = 2\).
05
Find q Using p
With the value \(p = 2\), substitute back into \(q = \frac{p - 11}{3}\) to solve for \(q\). This gives \(q = \frac{2 - 11}{3} = \frac{-9}{3} = -3\).
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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, a simple but important operation in vector algebra. Here, multiplying a vector like \( \langle 3, -1 \rangle \) by a scalar \( p \) means stretching or compressing the vector by the factor \( p \).
- When you multiply a vector by a positive scalar, the direction of the vector remains unchanged. However, its length will increase by the size of the scalar.
- If you multiply a vector by a negative scalar, the vector's direction reverses, and its length changes according to the scalar's absolute value.
- A scalar of zero reduces any vector to the zero vector \( \langle 0, 0 \rangle \).
Vector Equations
A vector equation sets two vectors equal, representing a geometrical equivalence. The goal is to find unknown values that make the equation true. This often involves breaking it into component form to facilitate the solution process. To solve \( p \langle 3, -1 \rangle + q \langle 4, 3 \rangle = \langle -6, -11 \rangle \), we write down the vector components:
- The x-components equation is \( 3p + 4q = -6 \).
- The y-components equation is \( -p + 3q = -11 \).
Linear Combinations
In vector algebra, a linear combination of vectors involves adding together two or more vectors, each multiplied by a scalar, to form another vector. This concept shows the flexibility and power of linear systems in expressing new directions and magnitudes. In our exercise, \( p \langle 3, -1 \rangle + q \langle 4, 3 \rangle \), the vectors \( \langle 3, -1 \rangle \) and \( \langle 4, 3 \rangle \) are being combined using the scalars \( p \) and \( q \). The goal is to determine the combination of \( p \) and \( q \) that forms the vector \( \langle -6, -11 \rangle \).
- Finding such combinations helps in tasks like solving systems of equations, optimizing solutions, or transforming coordinates.
- They form the backbone for various principles in linear algebra, including vector spaces and span.