Chapter 1: Problem 24
Find the points of intersection of the line \(x=3+2 t, y=7+8 t, z=-2+t,\) that \(\mathrm{is}, \mathrm{I}(t)=\) \((3+2 t, 7+8 t,-2+t),\) with the coordinate planes.
Short Answer
Expert verified
The intersection points are \((7, 23, 0)\), \(\left(\frac{5}{4}, 0, -\frac{23}{8}\right)\), and \((0, -5, -\frac{7}{2})\).
Step by step solution
01
Identify the Parametric Equations
The line given in parametric form is \((x, y, z) = (3+2t, 7+8t, -2+t)\). Here, \(x=3+2t\), \(y=7+8t\), and \(z=-2+t\). To find the points of intersection with the coordinate planes, we will solve these equations for when \(x\), \(y\), or \(z\) is zero.
02
Find Intersection with the XY-Plane
The XY-plane is defined by \(z = 0\). Set \(-2 + t = 0\) and solve for \(t\): \[ t = 2 \].Substitute \(t = 2\) back into the expressions for \(x\) and \(y\):\[ x = 3 + 2(2) = 7 \]\[ y = 7 + 8(2) = 23 \]Thus, the intersection point with the XY-plane is \((7, 23, 0)\).
03
Find Intersection with the XZ-Plane
The XZ-plane is defined by \(y = 0\). Set \(7 + 8t = 0\) and solve for \(t\): \[ t = -\frac{7}{8} \].Substitute \(t = -\frac{7}{8}\) back into the expressions for \(x\) and \(z\):\[ x = 3 + 2\left(-\frac{7}{8}\right) = \frac{5}{4} \]\[ z = -2 + \left(-\frac{7}{8}\right) = -\frac{23}{8} \]Thus, the intersection point with the XZ-plane is \(\left(\frac{5}{4}, 0, -\frac{23}{8}\right)\).
04
Find Intersection with the YZ-Plane
The YZ-plane is defined by \(x = 0\). Set \(3 + 2t = 0\) and solve for \(t\): \[ t = -\frac{3}{2} \].Substitute \(t = -\frac{3}{2}\) back into the expressions for \(y\) and \(z\):\[ y = 7 + 8\left(-\frac{3}{2}\right) = -5 \]\[ z = -2 + \left(-\frac{3}{2}\right) = -\frac{7}{2} \]Thus, the intersection point with the YZ-plane is \((0, -5, -\frac{7}{2})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Planes
In three-dimensional space, coordinate planes are the fundamental surfaces that help us understand the position of a point relative to the axes. There are three primary coordinate planes: the XY-plane, the XZ-plane, and the YZ-plane. Each of these divides the 3D space into different regions.
- The **XY-plane** is where the z-coordinate is always zero. This plane extends along the x and y axes and is crucial for examining how an object behaves when it is projected onto a flat 2D surface in 3D space.
- The **XZ-plane** has a constant y-coordinate of zero. In this plane, motion is only possible along the x and z directions. It plays a vital role in understanding vertical and horizontal movements relative to the y-axis.
- The **YZ-plane** features a constant x-coordinate of zero. Objects and movements constrained to this plane are only traceable along the y and z axes, presenting another layer of projection within 3D analyses.
Intersection Points
Intersection points between a line and coordinate planes help us determine where a line passes through these fundamental surfaces. To find these points, we usually set one of the coordinates (either x, y, or z) to zero and solve the corresponding parametric equations.
- For the **XY-plane**, set z = 0 because this defines the plane. Solve for the parameter that dictates the values of x and y when z is zero.
- For the **XZ-plane**, set y = 0 and solve the parametric equations to find x and z values.
- Similarly, for the **YZ-plane**, set x = 0 to uncover the intersection y and z values.
Line in 3D Space
A line in 3D space can be expressed using parametric equations, which define each point on the line as a function of a parameter, typically denoted as \(t\). Such equations are of the form:
- \(x = a + bt\)
- \(y = c + dt\)
- \(z = e + ft\)
- The **direction vector** \((b, d, f)\) suggests how the line moves in response to changes in \(t\).
- Understanding this setup allows for the exploration of how lines interact with various objects in space, such as determining where lines meet planes by substituting and solving equations.