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Determine whether the points lie on a straight line. (a) $$A(2,4,2), \quad B(3,7,-2), \quad C(1,3,3)$$ (b) $$D(0,-5,5), \quad E(1,-2,4), \quad F(3,4,2)$$

Short Answer

Expert verified
(a) No, (b) Yes

Step by step solution

01

Understanding Vector Collinearity

To determine if points are collinear in 3D space, we calculate the vectors between pairs of points. If the vectors are scalar multiples of one another, the points lie on the same straight line.
02

Calculate Vectors for (a)

For points \(A(2,4,2)\), \(B(3,7,-2)\), and \(C(1,3,3)\), calculate vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):\[\overrightarrow{AB} = (3-2, 7-4, -2-2) = (1, 3, -4)\]\[\overrightarrow{AC} = (1-2, 3-4, 3-2) = (-1, -1, 1)\].
03

Check Collinearity for (a)

To check collinearity, compare \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):The relationships \(1:k = 3:-1 = -4:1\) need to hold for some scalar \(k\). This is not possible because no single scalar \(k\) satisfies all ratios, so the points are not collinear.
04

Calculate Vectors for (b)

For points \(D(0, -5, 5)\), \(E(1, -2, 4)\), and \(F(3, 4, 2)\), calculate vectors \(\overrightarrow{DE}\) and \(\overrightarrow{DF}\):\[\overrightarrow{DE} = (1-0, -2+5, 4-5) = (1, 3, -1)\]\[\overrightarrow{DF} = (3-0, 4+5, 2-5) = (3, 9, -3)\].
05

Check Collinearity for (b)

To check collinearity, compare \(\overrightarrow{DE}\) and \(\overrightarrow{DF}\):Find scalar \(k\) such that \(1:k = 3:k = -1:-3\). Here, \(\overrightarrow{DF} = 3 \cdot \overrightarrow{DE}\), indicating the vectors are scalar multiples of each other. Hence, points D, E, and F are collinear.

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Key Concepts

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

Vector Calculations
In 3D geometry, vectors help us understand the positions and directions of points in space. A vector, in simple terms, is a quantity that has both magnitude (size) and direction. When dealing with the collinearity of points, we calculate vectors between points. Vectors are represented as arrows with a specific direction and length.
To check if points are collinear, we find vectors between pairs of points using their coordinates. For example, for points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), the vector is \((x_2 - x_1, y_2 - y_1, z_2 - z_1)\). This vector represents the direction and distance from point A to point B.
  • For points A and B with coordinates given, calculate \(\overrightarrow{AB}\).
  • Identify the differences in x, y, and z to form the vector components.
Understanding vectors is crucial for analyzing geometric problems, as it allows us to determine relationships between points or shapes efficiently.
3D Geometry
Three-dimensional (3D) geometry is the study of shapes and structures in a space with three dimensions: length, width, and height. In 3D geometry, every point is defined by three coordinates.
The space allows us to define complex shapes and understand the relationships between different points or objects in a way that can represent real-world situations. One essential concept in 3D geometry is collinearity—points lying on a single straight line.
To determine if points are collinear in 3D, we examine vectors, as seen in vector calculations. If two vectors between three points are scalar multiples of each other, the points are collinear. This occurs because the vectors share the same direction and differ only by a scale factor.
  • Visualize the positions using coordinates (x, y, z).
  • Compare vectors to check for collinearity or alignment on a single line.
This fundamental understanding supports various applications in physics, engineering, and computer graphics, where 3D modeling is essential.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar value. A scalar is a single number, unlike vectors, which have multiple components. When we multiply a vector by a scalar, we are stretching or compressing the vector's length without changing its direction.
In the context of collinearity, scalar multiplication helps us check if vectors are aligned in the same direction by being scaled versions of each other.
For instance, given vectors \(\overrightarrow{v}\) and \(\overrightarrow{w}\), if \(\overrightarrow{w} = k \cdot \overrightarrow{v}\), where \(k\) is a scalar, then \(\overrightarrow{w}\) is a scalar multiple of \(\overrightarrow{v}\).
  • Check proportionality of vector components.
  • Identify if \(k\) keeps the same ratio across all components (x, y, z).
If vectors between points are scalar multiples, those points are collinear. This concept streamlines calculations, ensuring we can predict alignment efficiently.

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Most popular questions from this chapter

Suppose that \(A\) is a nonsingular matrix with an eigenvalue\(\lambda .\) Show that 1\(/ \lambda\) is then an eigenvalue of \(A^{-1}\) .

Express the solution to the recursion \(\mathbf{n}_{t+1}=A \mathbf{n}_{t}\) in terms of the eigenvectors and eigenvalues of \(A,\) assuming that \(\mathbf{n}_{0}=\left[ \begin{array}{l}{1} \\ {1}\end{array}\right]\). \(A=\left[ \begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]\)

Suppose that \(A^{2}=0\) for some matrix \(A .\) Prove that the only possible eigenvalues of \(A\) are then \(0 .\)

\(\begin{array}{c}{\text { Mutation Suppose that, as a result of mutation, the }} \\ {\text { number of individuals in a population carrying allele } A \text { and }} \\ {\text { allele } B \text { changes in a way described by the equation }} \\ {\mathbf{y}_{t+1}=\left[ \begin{array}{cc}{0.95} & {0} \\\ {0.05} & {1}\end{array}\right] \mathbf{y}_{t}}\end{array}\) \(\begin{array}{c}{\text { where } \mathbf{y}_{t} \text { is the vector whose components are the numbers }} \\ {\text { of individuals carrying each allele at time } t . \text { An equilibrium }} \\ {\text { is a value of the vector for which no change occurs (that is, }} \\ {\mathbf{y}_{t+1}=\mathbf{y}_{t} ) . \text { Denoting such values by } \hat{\mathbf{y}}, \text { they must therefore }} \\ {\text { satisfy the equation }} \\ {\hat{\mathbf{y}}=\left[ \begin{array}{cc}{0.95} & {0} \\ {0.05} & {1}\end{array}\right] \hat{\mathbf{y}}}\end{array}\) Find all possible equilibrium values.

\(26-29\) Like scalars, vectors with complex components can be broken into real and imaginary parts. For example, \(v=a+b i\) where a and b are the real and imaginary parts of the vector, respectively. The complex conjugate of \(v\) is \(\overline{v}=a-\) bi. The analogous notation holds for matrices. The following rules from complex analysis, familiar from the scalar case, carry over to vectors and matrices: \(\vec{r \mathbf{x}}=\overline{r} \overline{\mathbf{x}}, \overline{B \mathbf{x}}=\overline{B} \overline{\mathbf{x}}, \overline{B C}=\overline{B} \overline{C},\) and \(\overline{r B}=\overline{r} \overline{B},\) where \(r\) is a scalar, \(\mathbf{x}\) is a vector, and \(B\) and \(C\) are matrices. (Note: Complex numbers are discussed in Appendix G.) Suppose that \(A\) is a matrix with real entries. Show that if \(\lambda\) is a complex eigenvalue of \(A,\) then its associated eigen vector will also be complex.

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