Let P=(1,2,-1),Q=(3,-1,4),and R=(2,6,2)
be three vertices of a parallelogram PQRS.
(a) Find the coordinates of S.
(b) Find the area of PQRS.
To find the coordinates of point S, we need to use the properties of a parallelogram.
Step 1: Find the vector representing the line segment PQ.
- Let's find vector PQ by subtracting the coordinates of P from the coordinates of Q.
- PQ = Q - P = (3, -1, 4) - (1, 2, -1) = (3 - 1, -1 - 2, 4 - (-1)) = (2, -3, 5).
Step 2: Find the vector representing the line segment PR.
- Let's find vector PR by subtracting the coordinates of P from the coordinates of R.
- PR = R - P = (2, 6, 2) - (1, 2, -1) = (2 - 1, 6 - 2, 2 - (-1)) = (1, 4, 3).
Step 3: Find the coordinates of S.
- The coordinates of S can be found by adding vector PR to point Q.
- S = Q + PR = (3, -1, 4) + (1, 4, 3) = (3 + 1, -1 + 4, 4 + 3) = (4, 3, 7).
Therefore, the coordinates of point S are (4, 3, 7).
Now let's calculate the area of parallelogram PQRS.
Step 1: Find the magnitude of vector PQ.
- The magnitude of vector PQ is given by ||PQ|| = √(PQ · PQ), where PQ · PQ is the dot product of PQ with itself.
- ||PQ|| = √(2^2 + (-3)^2 + 5^2) = √(4 + 9 + 25) = √38.
Step 2: Find the magnitude of vector PR.
- Similarly, the magnitude of vector PR is given by ||PR|| = √(PR · PR), where PR · PR is the dot product of PR with itself.
- ||PR|| = √(1^2 + 4^2 + 3^2) = √(1 + 16 + 9) = √26.
Step 3: Calculate the area of the parallelogram.
- The area of a parallelogram is given by the magnitude of the cross product of vectors PQ and PR.
- Area(PQRS) = ||PQ × PR|| = ||PQ|| × ||PR|| × sin(θ), where θ is the angle between vectors PQ and PR.
- The angle between vectors PQ and PR can be found using the dot product formula: PQ · PR = ||PQ|| × ||PR|| × cos(θ).
- The dot product of PQ and PR is (2 × 1) + (-3 × 4) + (5 × 3) = 2 - 12 + 15 = 5.
- Therefore, cos(θ) = PQ · PR / (||PQ|| × ||PR||) = 5 / (√38 × √26).
- The angle θ can be determined by taking the arccos of cos(θ): θ = arccos(5 / (√38 × √26)).
- Finally, the area of parallelogram PQRS is given by: Area(PQRS) = ||PQ|| × ||PR|| × sin(θ) = √38 × √26 × sin(θ).
You can evaluate the final expression to get the numerical value of the area.
To find the coordinates of vertex S, we can use the fact that a parallelogram has opposite sides that are equal in length and parallel.
Step 1: Find the vector formed by the vector subtraction PQ = Q - P:
PQ = (3, -1, 4) - (1, 2, -1) = (2, -3, 5)
Step 2: Add the vector PQ to the coordinate of vertex R to find the coordinates of S:
S = R + PQ = (2, 6, 2) + (2, -3, 5) = (4, 3, 7)
Therefore, the coordinates of vertex S are (4, 3, 7).
To find the area of the parallelogram PQRS, we can use the formula for the cross product of two sides of the parallelogram.
Step 3: Find the vectors formed by the vector subtraction PQ and PS:
PQ = (2, -3, 5)
PS = S - P = (4, 3, 7) - (1, 2, -1) = (3, 1, 8)
Step 4: Calculate the cross product of PQ and PS:
Cross product = PQ x PS = (2, -3, 5) x (3, 1, 8)
= ((-3*8 - 5*1), (2*8 - 5*3), (2*1 - (-3*3)))
= (-29, 14, -7)
Step 5: Calculate the magnitude (length) of the cross product vector:
Magnitude = sqrt((-29)^2 + 14^2 + (-7)^2)
= sqrt(841 + 196 + 49)
= sqrt(1086)
≈ 32.95
Therefore, the area of parallelogram PQRS is approximately 32.95 square units.
measuring from P,
(s-p)+(r-s) = (q-p)+(r-q) = r-p
= (2,6,2)-(1,2,-1) = (1,4,3)
the area is |(s-p)x(q-p)|
= |(0,2,4)x(4,-3,5)|
= |22,16,-8|
= √804