Final answer:
To find the required unit vector orthogonal to the plane, we calculate the cross product of two vectors within the plane to get the normal vector. We then normalize this vector and ensure the first coordinate is positive to obtain the desired unit vector, which is (√2/2, √2/2, 0).
Step-by-step explanation:
To find a unit vector with a positive first coordinate that is orthogonal to the plane through the points P(5,2,3), Q(6,3,4), and R(6,3,9), we first need to determine the normal vector to the plane. The normal vector can be found by taking the cross product of two vectors that lie in the plane.
Let's denote the vectors PQ and PR as follows:
- Vector PQ = Q - P = (6-5, 3-2, 4-3) = (1, 1, 1)
- Vector PR = R - P = (6-5, 3-2, 9-3) = (1, 1, 6)
The normal vector N to the plane is then N = PQ × PR. Calculating the cross product gives us:
- N = i(1×6 - 1×1) - j(1×6 - 1×1) + k(1×1 - 1×1)
- N = 5i - 5j + 0k = (5, -5, 0)
This vector is not a unit vector yet. To convert it into a unit vector, we divide it by its magnitude. The magnitude of N is |N| = √(5² + (-5)² + 0²) = √(50). Therefore, the unit vector in the direction of N is:
U = (1/√50) × (5, -5, 0) = (√2/2, -√2/2, 0)
However, we are asked for the unit vector with a positive first coordinate, so we simply take the positive version:
U = (√2/2, √2/2, 0)
What exactly does "orthogonal to the plane'' mean? Choose any two points P and Q in the plane, and consider the vector
PQ→. We say a vector
n⃗ is orthogonal to the plane if n⃗ is perpendicular to PQ→ for all choices of P and Q; that is, if n⃗ ⋅PQ→=0 for all P and Q
.
This gives us way of writing an equation describing the plane. Let P=(x0,y0,z0)
be a point in the plane and let n⃗ =⟨a,b,c⟩ be a normal vector
to the plane. A point Q=(x,y,z) lies in the plane defined by P and n⃗ if, and only if, PQ→ is orthogonal to n⃗ . Knowing PQ→=⟨x−x0,y−y0,z−z0⟩
, consider:
PQ→⋅n⃗ ⟨x−x0,y−y0,z−z0⟩⋅⟨a,b,c⟩a(x−x0)+b(y−y0)+c(z−z0)=0=0=0(10.6.1)