👤

Find a vector equation and parametric equations for the line. (Use the parameter t.)
The line through the point (1, 0, 9) and perpendicular to the plane x + 2y + z = 6
r(t) = ?
(x(t), y(t), z(t)) = ?

Answer :

The line through the point (1, 0, 9) and perpendicular to the Plane                 x + 2y + z = 6. Therefore, the vector equation and parametric equations for the line is:

r(t) = (1,0,9) + t {( 1i^ + 2j^ + 1k^)}

(x(t), y(t), z(t)) = {( x -1) , (y/2), (9-z)}

Vector Equation:

Vector equations are used to represent lines or planes in a three-dimensional framework. A 3D plane requires 3 coordinates related to 3 axes. This is where vectors come in handy to easily express vector equations for lines or planes. In a 3-D frame, the unit vector along the x-axis is ^i, the unit vector along the y-axis is ^j, and the unit vector along the z-axis is ^k.

The vector equations are written ^i ,^j, ^k and can be represented geometrically in the 3D plane. The simplest form of the vector equation for a straight line is

r = a+λb and the vector equation for the plane is r. ^n = d.

Parametric Equation or Polar Equation:

A type of equation with an independent variable, called the parameter (often denoted by t), in which the dependent variable is defined as a continuous function of the parameter and does not depend on any other existing variables. Multiple parameters can be used if desired. For example, instead of the equation y=x² in Cartesian form, the same equation can be written as a pair of equations in parametric form for x=t and y=t². This conversion to parametric form, called parameterization, makes curve differentiation and integration very efficient.

According to the question:

Using the vector normal to the plane  AX +By + cz = D is :

       Ai^ + B j^ + C k^  

The vector x+2y + z = 6 perpendicular to the plane is Vector shape of straight line through points (1,0,9)

The vector perpendicular to the plane, x + 2y+ z = 6, is:

                      1 i^ +2 j^​ +1 k^

This allows us to write the point-vector form of the line passing through the point (1, 0,9)

(x, y, z) = (1, 0, 9) + t(1i^ +2j^ +1k^ )

From the point vector,

we have the three polar equations by observations:

 x = 1t+1

 y = 2t

 z = 1t + 9

Now,

To find the symmetric form, we solve each of the polar equations for t and then set them equal:

t = x -1

​t = y/2

​t = 9- z

​Putting them altogether , we get the equal gives us the symmetric form :

x−1​ = y/2 = 9-z

Learn more about Vector Equation:

https://brainly.com/question/18848864

#SPJ4

Go Teaching: Other Questions