Use the cross product to find the area of the triangle with vertices:
P=(1,1,5)
Q=(3,4,3)
R=(1,5,7)
The area of the triangle is half of the area of a parallelogram having P, Q and R as three of its four vertices, and assume the fourth to be Q'.
The area of the parallogram PQRQ' is the magnitude of the cross product of the two vectors QP and QR.
Using P=(1,1,5), Q=(3,4,3), R=(1,5,7),
QP = (-2,-3,2) and
QR = (-2,1,4)
QPxQR=
|i j k|
|-2 -3 2|
|-2 1 4|
= (-14,4,-8)
Magnitude of the cross product
= |QPxQR|
= |(-14,4,-8)|
= √276
Area of the triangle
= (1/2)√276
=√69
To find the area of the triangle with vertices P=(1,1,5), Q=(3,4,3), and R=(1,5,7) using the cross product method, follow these steps:
Step 1: Calculate the vectors PQ and PR.
PQ = Q - P = (3,4,3) - (1,1,5) = (2,3,-2)
PR = R - P = (1,5,7) - (1,1,5) = (0,4,2)
Step 2: Compute the cross product of PQ and PR.
The cross product PQ x PR is given by:
| i j k |
| 2 3 -2 |
| 0 4 2 |
Cross product = (12 - 8)i - (0 + 0)j + (8 - 12)k
= 4i - 0j - 4k
= (4,0,-4)
Step 3: Calculate the magnitude of the cross product.
The magnitude of the cross product PQ x PR is given by:
|PQ x PR| = √(4^2 + 0^2 + (-4)^2)
= √(16 + 0 + 16)
= √32
= 4√2
Step 4: Calculate the area of the triangle.
The area of the triangle is equal to half the magnitude of the cross product.
Area = 1/2 * |PQ x PR|
= 1/2 * 4√2
= 2√2
Therefore, the area of the triangle with vertices P=(1,1,5), Q=(3,4,3), and R=(1,5,7) is 2√2 (square units).
To find the area of a triangle with vertices given as coordinates in 3D space, we can use the cross product of two vectors.
First, let's define the vectors PQ and PR using the given coordinates:
Vector PQ = Q - P = (3, 4, 3) - (1, 1, 5) = (2, 3, -2)
Vector PR = R - P = (1, 5, 7) - (1, 1, 5) = (0, 4, 2)
Next, calculate the cross product of PQ and PR:
Cross product = PQ × PR = (2, 3, -2) × (0, 4, 2)
The cross product of two vectors, (a, b, c) and (d, e, f), can be calculated as:
(i) a b c
(j) d e f
(k) aef - bdf - (ade - cdf) + (bde - cef)
Plugging in the values from PQ × PR:
Cross product = (2 * 2 - 3 * 4) i - (2 * 2 - (-2) * 0) j + (2 * 4 - 3 * 0) k
= (-2) i - 4 j + 8 k
The magnitude of the cross product vector gives us the area of the parallelogram formed by the two original vectors PQ and PR. Since we want the triangle's area, we can divide it by 2:
Area = |Cross product| / 2
Now, calculate the magnitude of the cross product vector:
|Cross product| = √((-2)^2 + (-4)^2 + 8^2)
= √(4 + 16 + 64)
= √84
Finally, divide the magnitude by 2 to get the area of the triangle:
Area = √84 / 2
= √21
Therefore, the area of the triangle with given vertices is √21 square units.