Write a scalar equation of each plane, given its vector equation.
[x,y,z] = [3,7,-5] + s[1,2,-1] + t[1,-2,3]
follow the same method I showed you in your next post.
The two questions are of the same type.
To write a scalar equation of a plane given its vector equation, we need to find the normal vector of the plane and a point on the plane.
The normal vector of the plane can be found by taking the cross product of the direction vectors of two non-parallel vectors in the plane. In this case, the direction vectors are [1,2,-1] and [1,-2,3].
To find the normal vector, we take the cross product of [1,2,-1] and [1,-2,3]:
N = [1,2,-1] x [1,-2,3]
To compute the cross product, we use the formula:
N = [ (2)(3) - (-1)(-2), (-1)(1) - (3)(1), (1)(-2) - (1)(2) ]
N = [ 8, -4, -4 ]
Now that we have the normal vector N, we can find a point on the plane. In the vector equation of the plane, [3,7,-5] represents a point on the plane.
Now we can write the scalar equation of the plane using the formula of a plane:
Ax + By + Cz = D
Substituting the values, we get:
8x - 4y - 4z = D
To find the value of D, we substitute the coordinates of the point [3,7,-5] into the equation:
8(3) - 4(7) - 4(-5) = D
24 - 28 + 20 = D
D = 16
Therefore, the scalar equation of the plane is:
8x - 4y - 4z = 16