volume of pparallelepiped(急)

[匿名]

2012-10-13 5:12 pm

1.find vr such that the volume of the parallelepiped whos edges are vp=-4i+3k,vq=3i-5j-k and vr is maximum with |vr|=5

ans:+(5/sqroot 26)(3i+4j+4k) or -(5/sqroot 26)(3i+4j+4k)

回答 (1)

用戶10012

2012-10-14 3:22 am

✔ 最佳答案

Is vp correct ? Please confirm.

2012-10-13 19:22:23 補充：
vp x vq = | i j k |
| - 4 0 3 |
| 3 - 5 - 1 |
= 15 i + 5 j + 20 k
| vp x vq | = sqrt ( 15^2 + 5^2 + 20^2 ) = sqrt ( 225 + 25 + 400 ) = sqrt 650 = 5 sqrt 26
So the unit vector of vp x vq = (15 i + 5 j + 20 k)/5 sqrt 26 = 5(3 i + j + 4 k)/5 sqrt 26 = (3 i + j + 4 k)/sqrt 26
For volume of parallelpiped to be a maximum, vr must perpendicular to the base area, that means direction of vr must be the same as the unit vector of vp x vq or the reverse of it.
Since magnitude of vr is 5, so vr = +/- 5 ( 3 i + j + 4 k)/sqrt 26.

2012-10-13 19:23:05 補充：
Note : The first answer provided is wrong.

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