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High School Math Calculus PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

that the points are the vertices of a parallelogram and find the area of the parallelogram.

that the points are the vertices of a parallelogram and find the area of the parallelogram.

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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Question

Chapter 10, Problem 14CT

To determine

To show:that the points are the vertices of a parallelogram and find the area of the parallelogram.

Expert Solution & Answer

Explanation of Solution

Given information:

The points are A(2,−3,1),B(6,5,−1),C(3,−6,4),D(7,2,2) .

Proof:to show that the points are vertices of a parallelogram, find the sides of the quadrilateral vectors,

AB⇀=(6−2)i+(5+3)j+(−1−1)kAB⇀=4i+8j−2k [subtract and add]DC⇀=(7−3)i+(2+6)j+(2−4)kDC⇀=(4i+8j−2k) [subtract]AC⇀=(3−2)i+(−6+3)j+(4−1)kAC⇀=i−3j+3k [subtract and add]DB⇀=(6−7)i+(5−2)j+(−1−2)kDB⇀=−(i−3j+3k) [subtract]

From the above vectors AB⇀=−CD⇀ and AC⇀=−DB⇀ thus the AB⇀ is parallel to CD⇀ and AC⇀ is parallel to DB⇀ .

Thus, the points are the vertices of a parallelogram.

from the above part (a), it can be observed that the points are vertices of the parallelogram with AB⇀ and AC⇀ as adjacent sides.

To find the area of the parallelogram, find the cross product of the adjacent sides,

AB⇀×AC⇀=ijk48−21−33AB⇀×AC⇀=i(24+6)−j(12+2)+k(−12−8) [simplify]AB⇀×AC⇀=30i−14j−20k [subtract and add]

Now, the area of parallelogram is,

|AB⇀×AC⇀|=(30)2+(−14)2+(−20)2|AB⇀×AC⇀|=900+196+400 [square]|AB⇀×AC⇀|=1496 [add]|AB⇀×AC⇀|≈38.67 [simplify]

Thus, the area of parallelogram is 38.67 square units.

Chapter 10, Problem 13CT

Chapter 10, Problem 15CT

Chapter 10 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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