Chapter 1: Problem 6
Find a vector \(\mathbf{c}\) which is orthogonal to \((1,3,1)=\mathbf{a}\) and to \((2,1,1)=\mathbf{b}\), and verify that \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) is a basis for \(\mathbb{R}^{3}\).
Short Answer
Expert verified
\(\mathbf{c} = (2, 1, -5)\) is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\), and \(\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}\) is a basis for \(\mathbb{R}^3\).
Step by step solution
01
Identify Orthogonal Vector using Cross Product
To find a vector \(\mathbf{c}\) that is orthogonal to both \(\mathbf{a} = (1,3,1)\) and \(\mathbf{b} = (2,1,1)\), we use the cross product, \(\mathbf{a} \times \mathbf{b}\). Calculate \(\mathbf{a} \times \mathbf{b}\) using the determinant method of a \(3 \times 3\) matrix:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 3 & 1 \2 & 1 & 1\end{vmatrix}.\]Expand the determinant to find the components of \(\mathbf{c}\):\[\mathbf{c} = (3 - 1)\mathbf{i} - (1 - 2)\mathbf{j} + (1 \cdot 1 - 3 \cdot 2)\mathbf{k} = (2, 1, -5).\]
02
Verify Orthogonality of the Result
To verify that \(\mathbf{c} = (2, 1, -5)\) is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\), compute the dot products:1. \(\mathbf{a} \cdot \mathbf{c} = 1\times2 + 3\times1 + 1\times(-5) = 2 + 3 - 5 = 0\), confirming \(\mathbf{a}\) and \(\mathbf{c}\) are orthogonal.2. \(\mathbf{b} \cdot \mathbf{c} = 2\times2 + 1\times1 + 1\times(-5) = 4 + 1 - 5 = 0\), confirming \(\mathbf{b}\) and \(\mathbf{c}\) are orthogonal.
03
Verify Linear Independence
To confirm that \(\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}\) is a basis for \(\mathbb{R}^3\), check that the vectors are linearly independent. Set up the matrix with these vectors as rows and compute the determinant:\[\begin{vmatrix}1 & 3 & 1 \2 & 1 & 1 \2 & 1 & -5\end{vmatrix}.\]Calculate the determinant:\[= 1(1 \cdot (-5) - 1 \cdot 1) - 3(2 \cdot (-5) - 1 \cdot 2) + 1(2 \cdot 1 - 1 \cdot 2).\]\[= 1(-5 - 1) - 3(-10 - 2) + 1(2 - 2) = -6 + 36 = 30.\]The determinant is non-zero, showing that the vectors are linearly independent.
04
Conclusion: Basis Verification
Since we have found \(\mathbf{c}\) and verified orthogonality and linear independence, \(\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}\) forms a basis for \(\mathbb{R}^3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product
To find a vector that is orthogonal to two given vectors, we use the cross product. The cross product of two vectors in three-dimensional space results in another vector that is perpendicular to both of the original vectors. This is especially useful in determining a vector that can complete a set of orthogonal vectors in \(\mathbb{R}^3\).
Given vectors \(\mathbf{a} = (1, 3, 1)\) and \(\mathbf{b} = (2, 1, 1)\), we find \(\mathbf{c}\) by computing \(\mathbf{a} \times \mathbf{b}\).
This simplifies to \(\mathbf{c} = (2, 1, -5)\), which is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\).
Given vectors \(\mathbf{a} = (1, 3, 1)\) and \(\mathbf{b} = (2, 1, 1)\), we find \(\mathbf{c}\) by computing \(\mathbf{a} \times \mathbf{b}\).
How to calculate the cross product:
- Create a 3x3 matrix with the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) in the first row, the components of \(\mathbf{a}\) in the second row, and the components of \(\mathbf{b}\) in the third row.
- Compute the determinant to get the cross product:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 3 & 1 \2 & 1 & 1 \end{vmatrix}.\]
- Expand this determinant:
This simplifies to \(\mathbf{c} = (2, 1, -5)\), which is orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\).
Linear Independence
In the context of vectors, linear independence is a key concept. Three vectors in \(\mathbb{R}^3\) are considered linearly independent if no vector can be expressed as a linear combination of the others.
For our vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), this means that there are no scalar values \(x\), \(y\), and \(z\), not all zero, such that:
If the determinant is non-zero, the vectors are linearly independent.
This result confirms our set is linearly independent and qualifies as a basis for \(\mathbb{R}^3\).
For our vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), this means that there are no scalar values \(x\), \(y\), and \(z\), not all zero, such that:
- \(x\mathbf{a} + y\mathbf{b} + z\mathbf{c} = \mathbf{0}\).
Determining linear independence:
- Arrange \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) as a matrix and calculate its determinant:
If the determinant is non-zero, the vectors are linearly independent.
This result confirms our set is linearly independent and qualifies as a basis for \(\mathbb{R}^3\).
Basis for \(\mathbb{R}^3\)
A basis for \(\mathbb{R}^3\) is a set of three vectors that are linearly independent and span the entire \(\mathbb{R}^3\) space.
This implies any vector in \(\mathbb{R}^3\) can be expressed as a linear combination of the basis vectors. For a set of vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) to form a basis, they must be linearly independent and span \(\mathbb{R}^3\).
This implies any vector in \(\mathbb{R}^3\) can be expressed as a linear combination of the basis vectors. For a set of vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) to form a basis, they must be linearly independent and span \(\mathbb{R}^3\).
Characteristics of a basis:
- Three vectors are needed in three-dimensional space.
- The vectors must be linearly independent.
- The vectors must span the whole space, meaning any vector in the space can be formed by combining the basis vectors.