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Chapter 3: Problem 20

Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Plane containing the three points \((0,1,1),(2,1,3),\) and (4,2,1).

Short Answer

Expert verified
Plane equation: \[-2x + 8y + 2z - 10 = 0\] or \[x - 4y - z + 5 = 0\]

Step by step solution

01

- Find two vectors in the plane

To determine the plane's equation, first find two vectors that lie in the plane. Use the given points to create vectors: 1. Vector \(\textbf{v}_1\)\from (0,1,1) to (2,1,3): \[\textbf{v}_1 = (2-0, 1-1, 3-1) = (2, 0, 2)\]2. Vector \(\textbf{v}_2\)\from (0,1,1) to (4,2,1): \[\textbf{v}_2 = (4-0, 2-1, 1-1) = (4, 1, 0)\]
02

- Find the normal vector to the plane

To find the normal vector \(\textbf{n}\), calculate the cross product of \(\textbf{v}_1\) and \(\textbf{v}_2\): \[\textbf{n} = \textbf{v}_1 \times \textbf{v}_2 = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 2 & 0 & 2 \ 4 & 1 & 0 \ \ \end{vmatrix} = (-2, 8, 2)\]
03

- Formulate the plane's equation

Use the normal vector \( \textbf{n} = (-2, 8, 2) \)and the point (0,1,1) to write the plane equation: \[-2(x-0) + 8(y-1) + 2(z-1) = 0\] Simplify to get: \[-2x + 8y + 2z - 10 = 0\] Finally: \[x - 4y - z + 5 = 0\]

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Key Concepts

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

Vector Cross Product
The vector cross product is a vector operation that outputs a vector perpendicular to two input vectors. If we have vectors \textbf{a} and \textbf{b}, then their cross product \textbf{a} \times \textbf{b} gives a resultant vector. This is especially useful for finding the normal vector to a plane.
The cross product of two vectors \textbf{a} = \(a_1, a_2, a_3\) and \textbf{b} = \(b_1, b_2, b_3\) is given by the determinant: \textbf{a} \times \textbf{b} = \(a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1\).
This operation helps in finding a vector that is orthogonal to both \textbf{a} and \textbf{b}, which is essential in setting up plane equations.
Normal Vector
The normal vector of a plane is a vector that is perpendicular to the plane. It plays a crucial role in determining the equation of the plane.
In our example, we derived the normal vector by performing the cross product of two vectors that lie within the plane.
These vectors were obtained by subtracting the coordinates of the given points: \textbf{v}_1 = \(2, 0, 2\) and \textbf{v}_2 = \(4, 1, 0\).
The cross product of these vectors yielded the normal vector \textbf{n} = \(-2, 8, 2\), which is used in the plane equation.
Three Points in a Plane
Determining the equation of a plane can be achieved if three non-collinear points are known. This is because these points uniquely define one plane.
For example, given points \textbf{A} = \(0, 1, 1\), \textbf{B} = \(2, 1, 3\) and \textbf{C} = \(4, 2, 1\), two vectors within the plane, \textbf{AB} and \textbf{AC}, are formed by subtracting coordinates: \(2, 0, 2\) and \(4, 1, 0\) respectively.
The cross product of these vectors provided the normal vector, which facilitates the derivation of the plane's equation through \textbf{Ax} + \textbf{By} + \textbf{Cz} = D, where \(A,B,C\) are the components of the normal vector.

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Most popular questions from this chapter

Solve the sets of homogeneous equations by row reducing the matrix. $$\left\\{\begin{aligned}3 x+y+3 z+6 w &=0 \\\4 x-7 y-3 z+5 w &=0 \\\x+3 y+4 z-3 w &=0 \\\3 x+2 z+7 w &=0\end{aligned}\right.$$

Find the eigenvalues and eigenvectors of the matrices in the following problems. $$\left(\begin{array}{lll} 3 & 0 & 1 \\ 0 & 3 & 1 \\ 1 & 1 & 2 \end{array}\right)$$

Find the rank of each of the following matrices. $$\left(\begin{array}{rrrr} 1 & 0 & 1 & 0 \\ -1 & -2 & -1 & 0 \\ 2 & 2 & 5 & 3 \\ 2 & 4 & 8 & 6 \end{array}\right)$$

Problem \(17(\mathrm{b})\) is a special case of the general theorem that the inverse of a product of matrices is the product of the inverses in reverse order. Prove this. Hint: Multiply ABCD times \(\mathrm{D}^{-1} \mathrm{C}^{-1} \mathrm{B}^{-1} \mathrm{A}^{-1}\) to show that you get a unit matrix.

The characteristic equation for a second-order matrix \(M\) is a quadratic equation. We have considered in detail the case in which M is a real symmetric matrix and the roots of the characteristic equation (eigenvalues) are real, positive, and unequal. Discuss some other possibilities as follows: (a) \(\quad \mathrm{M}\) real and symmetric, eigenvalues real, one positive and one negative. Show that the plane is reflected in one of the eigenvector lines (as well as stretched or shrunk). Consider as a simple special case $$M=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right)$$ (b) \(\quad \mathrm{M}\) real and symmetric, eigenvalues equal (and therefore real). Show that \(\mathrm{M}\) must be a multiple of the unit matrix. Thus show that the deformation consists of dilation or shrinkage in the radial direction (the same in all directions) with no rotation (and reflection in the origin if the root is negative). (c) \(\quad M\) real, not symmetric, eigenvalues real and not equal. Show that in this case the eigenvectors are not orthogonal. Hint: Find their dot product. (d) \(\quad \mathrm{M}\) real, not symmetric, eigenvalues complex. Show that all vectors are rotated, that is, there are no (real) eigenvectors which are unchanged in direction by the transformation. Consider the characteristic equation of a rotation matrix as a special case.
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