Find the equation of the plane through (1,2,-2) that contains the line x = 2t, y = 3 - t, z = 1 +3t

The direction vector of the given line is (2,-1,3) and a point on it is (0,3,1)

So using that point and the additional given point (1,2,-2)
yields a second direction vector of (1,-1,-3)

So we need a normal to these two direction vectors, which is the cross-product of (2,-1,3) and (1,-1,-3)

I got this normal to be (6,9,-1)
(I will assume you know how to find the cross-product)

The equation of the plane is
6x + 9y - z = k
subbing in the given point (1,2,-2)
6 + 18 + 2 = k = 26

equation of plane : 6x + 9y - z = 26

Well, since you're looking for the equation of a plane that contains a given point and a line, let's start by finding two vectors that lie on the plane. We can take the direction vectors of the line, which are (2, -1, 3).

To find another vector on the plane, we can take the vector between the given point (1, 2, -2) and any other point on the line. Let's choose a point on the line when t = 0, which gives us (2(0), 3 - (0), 1 + 3(0)) = (0, 3, 1).

Now, let's take the cross product of the two direction vectors to find the normal vector of the plane:
(2, -1, 3) × (0, 3, 1) = (9, 6, 3).

Thus, the normal vector to the plane is (9, 6, 3).

Now we can determine the equation of the plane using the point-normal form of the equation of a plane. Let's call the coordinates of a general point on the plane (x, y, z).

The equation of the plane is given by:
9(x - 1) + 6(y - 2) + 3(z + 2) = 0

Simplifying, we have:
9x - 9 + 6y - 12 + 3z + 6 = 0
9x + 6y + 3z - 15 = 0

So, the equation of the plane through (1, 2, -2) that contains the line x = 2t, y = 3 - t, z = 1 + 3t is:
9x + 6y + 3z - 15 = 0.

Hope that helps! If not, I'll be hanging around to juggle your questions!

To find the equation of the plane, we need to find a point on the plane and its normal vector.

Let's find the normal vector first. The vector parallel to the line is given by the coefficients of t in the equation, which are (2, -1, 3).

Now, let's find a point on the plane. We know that the point (1, 2, -2) lies on the plane.

The equation of the plane can be written in the form Ax + By + Cz + D = 0.

Since (1, 2, -2) lies on the plane, we can substitute these values into the equation to obtain:

A(1) + B(2) + C(-2) + D = 0

Simplifying, we get:

A + 2B - 2C + D = 0

The normal vector (2, -1, 3) is perpendicular to the plane, so it satisfies the equation:

2A - B + 3C = 0

We can choose any values for A, B, and C to satisfy these two equations.

Let's choose A = 2, B = 2, and C = 1.

Substituting these values into the equations, we have:

2 + 2(2) - 2(1) + D = 0
2(2) - 2 + 3(1) = 0

Simplifying, we find:

6 + D = 0
4 - 2 + 3 = 0

From these equations, we can solve for D:

D = -6
D = -5

So, the equation of the plane through (1, 2, -2) that contains the line x = 2t, y = 3 - t, z = 1 + 3t is:

2x + 2y + z - 6 = 0

or

2x + 2y + z - 5 = 0

To find the equation of the plane that contains the given line and passes through the point (1, 2, -2), we need to determine a normal vector for the plane.

A normal vector is orthogonal (perpendicular) to any vector in the plane. Since the direction vector for the line is parallel to the plane, it will also be orthogonal to the normal vector.

Step 1: Find the direction vector of the line.
Given the line: x = 2t, y = 3 - t, z = 1 + 3t

The direction vector of the line is the coefficients of t:
direction vector = (2, -1, 3)

Step 2: Determine a vector orthogonal to the line's direction vector.
We can pick any vector that is perpendicular to the line's direction vector (2, -1, 3). A simple way to do this is by taking the cross product of the direction vector with a known vector.

Let's choose the vector (1, 0, 0) as our known vector for this step.

Cross product: (2, -1, 3) × (1, 0, 0) = (0, -3, -1)

Step 3: Write the equation of the plane using the normal vector and the point (1, 2, -2).
The equation of a plane is given by the formula:
Ax + By + Cz = D

Substituting the values, we get:
0(x - 1) - 3(y - 2) - 1(z + 2) = 0

Simplifying the equation:
-3y + 6 - z - 2 = 0
-3y - z + 4 = 0

So, the equation of the plane through (1, 2, -2) that contains the line x = 2t, y = 3 - t, z = 1 + 3t is: -3y - z + 4 = 0.