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## Find Parametric Equations For The Line Through (2 4 6) That Is Perpendicular To The Plane X-Y+3Z=7

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## Answers ( 2 )

## Find Parametric Equations For The Line Through (2 4 6) That Is Perpendicular To The Plane X-Y+3Z=7

Solving parametric equations can be tricky, but understanding the concepts behind them is key to success. In this article, we’ll explore the basics of parametric equations and how to use them to find the line through (2 4 6) that is perpendicular to the plane x-y+3z=7. We’ll look at various methods for solving this equation, including using vector notation and point-slope form. We will also learn some tips and tricks for mastering parametric equations so that you can tackle any problem with ease.

## What are parametric equations?

Parametric equations are a set of equations that define a relationship between variables. In this case, the parametric equations define the relationship between the variables x, y, and z. The parametric equations for the line through ( ) that is perpendicular to the plane x-y+z=0 are:

x = t

y = -t

z = t

Where t is a real number.

## How to find parametric equations for a line

Given a point and a plane, we can find parametric equations for the line perpendicular to the plane that goes through the given point using the following steps:

1. Calculate the normal vector to the plane. This can be done by taking the cross product of any two non-parallel vectors in the plane.

2. Calculate a parametric equation for the line that goes through the given point and is parallel to the normal vector. This can be done by solving for t in the equation r = p + tn, where p is the given point, n is the normal vector, and r is a point on the desired line.

3. Find a value of t that makes the line perpendicular to the plane. This can be done by setting dot(r – p, v) = 0, where v is a vector in the plane and dot(a, b) denotes the dot product of vectors a and b.

4. Substitute this value of t into your parametric equation for the line to get parametric equations for a line perpendicular to the given plane that goes through your original point.

## What is the equation for a plane?

Assuming that you are working in three dimensional space, the equation for a plane is typically given as ax + by + cz = d, where a, b, and c are nonzero real numbers and d is any real number. The coefficients a, b, and c can be determined from any three noncollinear points that lie on the plane. Once you have these coefficients, you can plug in the coordinates of any other point to determine whether or not it lies on the plane.

## How to find the equation of a plane that is perpendicular to another plane

In order to find the equation of a plane that is perpendicular to another plane, you will need to first find the normal vector of the given plane. The normal vector of a plane is perpendicular to the plane itself. To find the normal vector of a plane, you will need to take the cross product of any two non-parallel vectors that are on the plane. Once you have found the normal vector, you can then use it to find the equation of a line that is perpendicular to the given plane.

## Conclusion

Problem solving is a key skill in mathematics and this article has given you an example of how to use parametric equations to find the line through (2 4 6) that is perpendicular to the plane x-y+3z=7. Knowing how to find solutions like these can help you tackle more complex problems, making it possible for you to solve even the most challenging math questions. With practice, your problem-solving skills will continue to improve and soon enough you’ll be able to conquer any equation with confidence!

Have you ever wanted to find the parametric equations for the line through (2 4 6) that is perpendicular to the plane x-y+3z=7? It can be a bit tricky, but with the right steps, you’ll be able to figure it out in no time.

Let’s start by breaking down the equation. The plane is x-y+3z=7, which means that the normal vector to the plane is (1, -1, 3). This is the vector that is perpendicular to the plane.

♂️ Since the vector (2, 4, 6) is on the line, the direction vector of the line is perpendicular to the normal vector of the plane. Therefore, the direction vector of the line is (3, 1, -1).

Now that we have the direction vector, we can find the parametric equations for the line. The parametric equations for the line through (2, 4, 6) with direction vector (3, 1, -1) are: x = 2 + 3t, y = 4 + t, z = 6 – t.

This means that the line through (2, 4, 6) that is perpendicular to the plane x-y+3z=7 can be expressed using the parametric equations x = 2 + 3t, y = 4 + t, z = 6 – t.

Now that you know how to find the parametric equations for the line through (2 4 6) that is perpendicular to the plane x-y+3z=7, you’ll be able to solve even more difficult problems with ease. Good luck!