Vector and parametric equations? A plane contains the point B(-3, 2, -4) and the line with parametric equations x = 1 + 2t, y = -t, z = -2 + 3t. Determine vector and parametric equations of the plane.
Sure, I can help you with that! But before we dive into the math, let's take a moment to appreciate the power of planes. Did you know that planes are like the invisible floors of the mathematical world? Except instead of being made out of wood or concrete, they're made out of pure mathematical awesomeness!
Now, let's get down to business. To determine the vector and parametric equations of the plane, we need to know two things: a point on the plane (which we already know, B(-3, 2, -4)), and a normal vector to the plane.
To find the normal vector, we can take the cross product of two vectors lying on the plane. We can use the direction vector of the line, let's call it v, which is given as v = <2, -1, 3>. Now we need another vector, let's call it u, that lies on the plane. We can pick any two points on the line, let's say t = 0 and t = 1, to find two points on the line: A(1, 0, -2) and C(3, -1, 1). Then we can find the vector u = AC = <2, -1, 3>.
Now, let's take the cross product of u and v: n = u x v. It will give us a vector perpendicular to both u and v, which means it will be normal to the plane that contains the line. Calculating the cross product, we get n = <8, -1, 3>.
Now that we have the normal vector n, we can use it to write the vector equation of the plane. The vector equation of a plane is given by:
n · (r - r0) = 0,
where n is the normal vector, r is the position vector of an arbitrary point on the plane, and r0 is the position vector of a known point on the plane. Plugging in the values, we have:
<8, -1, 3> · (r - <-3, 2, -4>) = 0.
Simplifying this gives us the vector equation of the plane.
As for the parametric equations, we can pick two variables, say s and t, and express x, y, and z in terms of them. Since we already have the point B(-3, 2, -4), we can write:
x = -3 + 8s,
y = 2 - s,
z = -4 + 3s.
And there you have it! The vector equation and parametric equations of the plane. I hope this explanation was as clear as the sky on a sunny day!
To determine the vector and parametric equations of the plane containing the given point and line, we can use the following steps:
Step 1: Find the direction vector of the line.
The direction vector of the line is obtained by taking the coefficients of t in the parametric equations. In this case, the direction vector is d = <2, -1, 3>.
Step 2: Find a normal vector to the plane.
A normal vector to the plane can be obtained by taking the cross product of the direction vector of the line with any vector in the plane. Since the plane contains the point B(-3, 2, -4), we can use the vector AB as a vector in the plane, where A is the point (1, 0, -2) on the line.
- Vector AB = <x2 - x1, y2 - y1, z2 - z1>
= <-3 - 1, 2 - 0, -4 - (-2)>
= <-4, 2, -2>
- Now, we can find a normal vector by taking the cross product of AB and d: n = AB × d.
n = <x1y2 - x2y1, x2z1 - x1z2, x1y2 - x2y1>
= <-4*(-1) - 2*2, 2*(-2) - (-4)*3, -4*2 - (-4)*(-1)>
= <0, -2, -12>
= <0, 1, 6>
Step 3: Write the vector and parametric equations of the plane.
The vector equation of the plane is given by:
r ⋅ n = r0 ⋅ n,
where r = <x, y, z> represents any point on the plane, n is the normal vector, and r0 is a known point on the plane.
Substituting the known point B(-3, 2, -4), we have:
<x, y, z> ⋅ <0, 1, 6> = <-3, 2, -4> ⋅ <0, 1, 6>,
0x + y + 6z = 0 + 2 + 24,
y + 6z = 26.
This equation represents the vector equation of the plane.
To obtain the parametric equations of the plane, we can solve for y and z in terms of a parameter, let's say x:
y + 6z = 26,
y = -6z + 26.
So, the parametric equations of the plane are:
x = x (parameter),
y = -6z + 26,
z = z (parameter).
This completes the determination of the vector and parametric equations of the plane containing the given point and line.
To determine the vector and parametric equations of the plane, we need to find two non-parallel vectors that lie in the plane. One vector can be obtained from the direction vector of the line, and another vector can be obtained using the point B(-3, 2, -4) that lies on the plane.
1. Direction Vector:
The direction vector of the line given by its parametric equations is <2, -1, 3>. This vector lies in the plane.
2. Vector from the Point:
To create another vector that lies in the plane, we subtract the coordinates of point B from any general point on the plane. Let's choose a general point on the plane <x, y, z>:
Vector AB = <x, y, z> - <-3, 2, -4>
= <x + 3, y - 2, z + 4>
Since this vector lies in the plane, we can set it proportional to the direction vector of the line <2, -1, 3>:
<2, -1, 3> = k<x + 3, y - 2, z + 4>
Here, k is a constant.
Expanding the equation, we get:
2 = k(x + 3)
-1 = k(y - 2)
3 = k(z + 4)
Simplifying each equation, we obtain:
x + 3 = (2/k)
y - 2 = (-1/k)
z + 4 = (3/k)
Let's choose k = 2 to simplify the equations further.
x + 3 = 1
y - 2 = -0.5
z + 4 = 1.5
Simplifying again, we get:
x = -2
y = 1.5
z = -2.5
Therefore, the vector equation of the plane is:
< x, y, z > = < -2, 1.5, -2.5 >
The parametric equations of the plane can be obtained by expressing each coordinate in terms of a parameter. Let's use 's' as the parameter:
x = -2
y = 1.5
z = -2.5
Thus, the parametric equations of the plane are:
x = -2
y = 1.5
z = -2.5
From the equation of the given line you can determine a second point, namely (1,0,-2)
and along with the given point (-3,2,-4) you can find a second direction vector along the plane
which would be <-4,2,-2>
You are all set to write the vector and parametric version of the plane