The equation \(y = -2\ in\ \mathbb{R^{3}}\) represenrs the vertical plane that is parallel to the xz - plane and it locates 2 units left to the xz - plane.
\(RR^{3}\) is the three dimensional coordinate system which contains x, y and z - coordinates.
The equation \(y = -2\ in\ \mathbb{R^{3}}\)
represents the set \({(x, y, z)|y = -2}\),
which is the set of all points in \(\mathbb{R^{3}}\) whose y - coordinate is -2 and x, z - coordinates are any values.
The equation \(y = -2\ in\ \mathbb{R^{3}}\) is sketched as shown in Figure 1
From Figure 1, the equation \(y = -2\ in\ \mathbb{R^{3}}\) represents the vertical plane that is parallel to the xz - plane and it locates 2 units left to the xz - plane.
Thus, the equation \(y = -2\ in\ \mathbb{R^{3}}\) is described.
Descibe in words the region of RR^{3} represented by the equation or inequality.y = -2
Descibe in words the region of RR^{3} represented by the equation or inequality.y = -2
Answers (1)
Relevant Questions
A line L through the origin in \(\displaystyle\mathbb{R}^{{3}}\) can be represented by parametric equations of the form x = at, y = bt, and z = ct. Use these equations to show that L is a subspase of \(RR^3\) by showing that if \(v_1=(x_1,y_1,z_1)\ and\ v_2=(x_2,y_2,z_2)\) are points on L and k is any real number, then \(kv_1\ and\ v_1+v_2\) are also points on L.
(10%) In \(R^2\), there are two sets of coordinate systems, represented by two distinct bases: \((x_1, y_1)\) and \((x_2, y_2)\). If the equations of the same ellipse represented by the two distinct bases are described as follows, respectively: \(2(x_1)^2 -4(x_1)(y_1) + 5(y_1)^2 - 36 = 0\) and \((x_2)^2 + 6(y_2)^2 - 36 = 0.\) Find the transformation matrix between these two coordinate systems: \((x_1, y_1)\) and \((x_2, y_2)\).
Solve the given Alternate Coordinate Systems and give a correct answer 10) Convert the equation from Cartesian to polar coordinates solving for \(r^2\):
\(\frac{x^2}{9} - \frac{y^2}{16} = 25\)
Assume that T is a linear transformation. Find the standard matrix of T.
\(\displaystyle{T}=\mathbb{R}^{{2}}\rightarrow\mathbb{R}^{{4}}\ {s}{u}{c}{h} \ {t}\hat \ {{T}}{\left({e}_{{1}}\right)}={\left({7},{1},{7},{1}\right)},{\quad\text{and}\quad}{T}{\left({e}_{{2}}\right)}={\left(-{8},{5},{0},{0}\right)},{w}{h}{e}{r}{e}{\ e}_{{1}}={\left({1},{0}\right)},{\quad\text{and}\quad}{e}_{{2}}={\left({0},{1}\right)}\).
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
\(\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}\)
To graph: The sketch of the solution set of system of nonlinear inequality
\(\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}\geq\ {9}\right.}]\)
\(\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}\leq\ {25}\)
\(\displaystyle{y}\geq\ {\left|{x}\right|}\)
The given system of inequality:
\(\displaystyle{\left\lbrace\begin{matrix}{y}<{9}-{x}^{2}\\{y}\ge{x}+{3}\end{matrix}\right.}\)
Also find the coordinates of all vertices, and check whether the solution set is bounded.
The quadratic function \(\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}\) whose graph passes through the points (1, 4), (2, 1) and (3, 4).
To solve:
\(\displaystyle{\left(\begin{matrix}{x}-{2}{y}={2}\\{2}{x}+{3}{y}={11}\\{y}-{4}{z}=-{7}\end{matrix}\right)}\)
The reason ehy the point \((-1, \frac{3\pi}{2})\) lies on the polar graph \(r=1+\cos \theta\) even though it does not satisfy the equation.
