Find the area of the parallelogram with one corner at P1 and adjacent sides P1P2 and P1P3. NOTE: There is an arrow over P1P2 and P1P3. What does that arrow mean?
P1 = (0, 0, 0), P2 = (2, 3, 1),
P3 = (-2, 4, 1)
P1P2 with an arrow over it is the vector representation of the side that goes from P1 to P2. It would be written
2i + 3j + k
P1P3(with arrow), the adjacent side, would be written
-2 +4j + k
The vector whose magnitude is the parallelogram area is
|i j k|
|2 3 1| = -i +0j +2k
|-2 4 1|
(The 3x3 matrix above, surrounded by | | indicates a determinant).
The area is the magnitude of that vector, which is sqrt5 = 2.236
The arrow over P1P2 and P1P3 indicates that they are vectors. In this case, P1P2 and P1P3 represent vectors that originate from point P1 and point towards points P2 and P3, respectively. Vectors are commonly denoted with an arrow to indicate their direction and magnitude.
To find the area of the parallelogram, you can use the cross product of vectors P1P2 and P1P3. The cross product of two vectors gives a vector that is perpendicular to both of the original vectors. The magnitude of this vector represents the area of the parallelogram formed by the two original vectors.
Let's calculate the cross product of P1P2 and P1P3 using their coordinates:
P1P2 = (2, 3, 1)
P1P3 = (-2, 4, 1)
To compute the cross product, use the determinant method:
i j k
2 3 1
-2 4 1
= i[(3 * 1) - (4 * 1)] - j[(2 * 1) - (-2 * 1)] + k[(2 * 4) - (-2 * 3)]
= i(3 - 4) - j(2 + 2) + k(8 + 6)
= -i - 4j + 14k
The cross product of P1P2 and P1P3 is -i - 4j + 14k. To find the magnitude of this vector, which represents the area of the parallelogram, use the formula:
Area = √((-1)^2 + (-4)^2 + 14^2)
Area = √(1 + 16 + 196)
Area = √213
Therefore, the area of the parallelogram with one corner at P1 and adjacent sides P1P2 and P1P3 is √213.