Nohar C. answered • 09/24/23
Noharpatelsahabji
To find the equation of the plane that contains the point (1, 2, 3) and is also parallel to the line represented by the vector equation r(t) = <3t, 5 - 4t, 1 - 8t>, you can use the following approach:
1. First, find the direction vector of the line. The direction vector is the coefficient vector of t in the vector equation r(t). In this case, the direction vector is <3, -4, -8>.
2. Next, since the plane is parallel to the line, the normal vector of the plane will be perpendicular to the direction vector of the line. You can find the normal vector by taking the cross product of the direction vector of the line and any other vector. To keep it simple, let's take the cross product with the vector <1, 0, 0>. So, the normal vector of the plane is:
Normal Vector = <3, -4, -8> × <1, 0, 0> = <0, -8, 4>
3. Now that you have the normal vector, you can use the point-normal form of the equation of a plane. The point-normal form is:
Ax + By + Cz = D
Where A, B, and C are the components of the normal vector, and (x, y, z) are the coordinates of a point on the plane.
In this case, the normal vector is <0, -8, 4>, and the point (1, 2, 3) is on the plane. So, you can write the equation of the plane as:
0x - 8y + 4z = D
Now, plug in the coordinates of the point (1, 2, 3) to find the value of D:
0(1) - 8(2) + 4(3) = D
-16 + 12 = D
D = -4
So, the equation of the plane that contains the point (1, 2, 3) and is parallel to the line represented by the vector equation r(t) = <3t, 5 - 4t, 1 - 8t> is:
-8y + 4z = -4
You can simplify this equation further if needed.
