Chapter 9: Problem 34
A description of a line is given. Find parametric equations for the line. The plane that contains the line \(| x=1-t, y=2+t\) \(z=-3 t\) and the point \(P(2,0,-6) .\) [Hint: A vector from any point on the line to \(P\) will lie in the plane.]
Short Answer
Step by step solution
Identify the Original Line
Calculate a Vector to Point P
Find the Normal Vector of the Plane
Calculate the Cross Product
Formulate Equation of the Plane
Simplify the Plane Equation
Rewrite the Plane Equation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Calculus
To better understand vector calculus, it's crucial to grasp how vectors work:
- Vectors can be represented as directed line segments in 2D or 3D space.
- They are often expressed as components along the axes, such as \(\vec{v} = (v_x, v_y, v_z)\) in 3D, where each component represents a direction in Cartesian coordinates.
- Vectors can be added, subtracted, or scaled (multiplied by a scalar) with ease.
- Operations like the dot product and cross product are fundamental, allowing us to delve deeper into vector relationships.
Cross Product
Here’s a quick rundown of the cross product:
- The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is denoted by \(\vec{a} \times \vec{b}\).
- It results in a new vector that is orthogonal (at a right angle) to both \(\vec{a}\) and \(\vec{b}\).
- The magnitude of the cross product is equal to the area of the parallelogram formed by \(\vec{a}\) and \(\vec{b}\).
- The formula for the cross product in terms of components is \(\vec{a} \times \vec{b} = (a_yb_z - a_zb_y, a_zb_x - a_xb_z, a_xb_y - a_yb_x)\).
Plane Equation
The basic formula for a plane in three dimensions is given by:
- \[ax + by + cz = d\]
Where \(a\), \(b\), and \(c\) are the components of the normal vector, and \(d\) is a constant found by substituting the point’s coordinates into the plane equation.
To construct this equation:
- Start with your normal vector, like \((-12, -9, 1)\) in this case.
- Use a point on the plane, for instance, \((1, 2, 0)\).
- Plug these into the point-normal form of the plane equation: \(-12(x - 1) - 9(y - 2) + 1(z - 0) = 0\).
- Simplify to get \(-12x - 9y + z + 30 = 0\) and then rearrange to the standard form \(12x + 9y - z = 30\).