微積分(平面空間中的向量)

2010-08-19 2:23 am
1.
The following lines (u is a parameter) are known to be on one plane.

L1:r1(t)=<1+2t,-t,2+t>
L2:r2(t)=<ut,3+t,1-2t>

A)Find the value of u.
B)What is the scalar equation of the plane?


2.
Determine whether the vectors u=<1,-1,3>,v=<1,1,2> and w=<2,-4,7> lie in one plane.

-the vectors are in one plane since w(v * u) =0
-the vectors are in one plane since u(v * w) =0
-the vectors are in one plane since v(u * w) =0
-the vectors are in one plane since u(v * w) =1
-none of these
更新1:

3. Find the equation of the plane (if it exists) containing the two lines: a)r1(t)=<1-t,t,3+2t> r2(t)= 這題的方法?

更新2:

Find the equation of the plane (if it exists) containing the two lines: r1(t)=<1-t,t,3+2t> r2(t)= 這題的方法是?

更新3:

r1(t)=<1-t,t,3+2t> r2(t)=

更新4:

r2(t)=

更新5:

r2(t)=

更新6:

是r2的參數,不知為何沒顯示

回答 (1)

2010-08-25 8:54 am
✔ 最佳答案
Q1:
L1:r1(t)=<1+2t,-t,2+t>, 取點A(1, 0, 2), B(3, -1, 3)
L2:r2(t)=<ut,3+t,1-2t>, 取點C(0,3, 1), 方向為<u,1,-2>
過ABC三點的平面為 2x-y-5z+8=0
法向量<2, -1, -5>與<u, 1, -2>垂直, 故u= -9/2

Q2:
uxv∙w=<-5,1,2>∙<2,-4, 7>=0, 故三向量共平面
同理vxw∙u= uxw∙v=0,(二個外積,再與第3個向量內積=0即可)

Q3:題目不全


收錄日期: 2021-04-23 23:06:20
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