Final answer:
P's coordinates are (2,6). The magnitude of PR is √13, and the unit vector along PR is (2/√13, -3/√13).
Step-by-step explanation:
To solve the student's question about the parallelogram PQRS, we first find the vector components of PQ and SR. Since PQRS is a parallelogram, PQ and SR are parallel and have equal magnitudes.
Given the coordinates
P (x, y),
Q (5,7),
R (4,3),
S (1,2),
Let's calculate the components of PQ and SR:
- PQ = Q - P = (5, 7) - (x, y) = (5 - x, 7 - y)
- SR = R - S = (4, 3) - (1, 2) = (3,1)
For these vectors to be equal, the components must be equal:
- 5 - x = 3 -> x = 2
- 7 - y = 1 -> y = 6
So, the coordinates for P are (2, 6).
Next, we calculate the magnitude of vector PR using the Pythagorean theorem:
- PR = R - P = (4, 3) - (2, 6) = (2, -3)
- |PR| = √(2^2 + (-3)^2) = √(4 + 9) = √13
The unit vector along PR is the vector PR divided by its magnitude:
- unit vector = PR / |PR| = (2, -3) / √13
To find the unit vector in its simplest form, we commonly write it as:
- unit vector = (2/√13, -3/√13)