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Find the area of the parallelogram with vertices:________.

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

User Ryan McCue
by
5.8k points

1 Answer

3 votes

Answer:

97.98

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

User Ali Lotfi
by
4.2k points
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