find an equation of a plane containing a point (0,6,-2) and parallel to vector v1= (0,-3,0), v2=(0,9,-1)
Help!
The cross-product of the two given vectors will give us the normal that we can use for the new plane.
You should have a method of your own to find that normal, I had (1,0,0)
(notice the dotproduct of (1,0,0) with each of the given vectors is zero)
so the equation of the new plane is
x + 0y + 0z = c
but the point (0,6,-2) lies on it, so
0 + 0 + 0 = c
c = 0
the equation of the plane is x = 0
find a line that passes through point (2,5,-3) and is perpendicular to the plane 2x-3y+4z+7=0
To find an equation of a plane containing a point and parallel to two vectors, we can use the point-normal form of the plane equation.
Step 1: Find the normal vector of the plane.
Since the plane is parallel to vector v1 = (0, -3, 0) and v2 = (0, 9, -1), the normal vector of the plane will be perpendicular to both v1 and v2. We can find the cross product of v1 and v2 to obtain the normal vector.
n = v1 x v2
= (0, -3, 0) x (0, 9, -1)
= (-3, 0, 0)
Therefore, the normal vector of the plane is (-3, 0, 0).
Step 2: Write the equation of the plane using the point-normal form.
The point-normal form of the plane equation is given by:
Ax + By + Cz = D
Substituting the values of the given point (0, 6, -2) and the normal vector (-3, 0, 0) into the equation, we get:
-3x + 0y + 0z = D
-3(0) + 0(6) + 0(-2) = D
0 = D
Therefore, the equation of the plane is:
-3x = 0
Simplifying further, we get:
x = 0
So, the equation of the plane containing the point (0, 6, -2) and parallel to the vectors v1 = (0, -3, 0) and v2 = (0, 9, -1) is x = 0.
To find the equation of a plane containing a point and parallel to two vectors, you can use the point-normal form of the equation of a plane.
1. Find the normal vector of the plane by taking the cross product of the given vectors. The normal vector (n) will be perpendicular to both v1 and v2.
n = v1 x v2
n = (0,-3,0) x (0,9,-1)
n = (-3, 0, 0)
2. Once you have the normal vector, you can write the equation of the plane using the point-normal form:
Ax + By + Cz = D
Let's use the point (0,6,-2) as the point on the plane and substitute its coordinates into the equation:
(-3)(0) + (0)(6) + (0)(-2) = D
0 + 0 + 0 = D
D = 0
So, the equation of the plane is:
-3x + 0y + 0z = 0
Simplifying, we get:
-3x = 0
x = 0
Therefore, the equation of the plane is:
x = 0
Note that x = 0 represents a vertical plane parallel to the yz-plane and passing through the point (0,6,-2).