Find the area of the parallelogram with vertices:. P(0,0,0), Q(4,-5,3), R(4,-7,1), S(8,-12,4).

97.98

Step-by-step explanation:

The area of the parallelogram PQR is the magnitude of the cross product of any two adjacent sides. Using PQ and PS as the adjacent sides;

Area of the parallelogram = |PQ×PS|

PQ = Q-P and PS = S-P

Given P(0,0,0), Q(4,-5,3), R(4,-7,1), S(8,-12,4)

PQ = (4,-5,3) - (0,0,0)

PQ = (4,-5,3)

Also, PS = S-P

PS = (8,-12,4)-(0,0,0)

PS = (8,-12,4)

Taking the cross product of both vectors i.e PQ×PS

(4,5,-3)×(8,-12,4)

PQ×PS = (20-36)i - (16-(-24))j + (-48-40)k

PQ×PS = -16i - 40j -88k

|PQ×PS| = √(-16)²+(-40)²+(-88)²

|PQ×PS| = √256+1600+7744

|PQ×PS| = √9600

|PQ×PS| ≈ 97.98

Hence the area of the parallelogram is 97.98

i pretty sure it is

uhhh idk u yolo?