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Loking 發問時間： 科學數學 · 1 0 年前

# lines and planes in 3d

consider two planes

P1 : 3x-2y+4z=1

p2 2x-2y+z=3

and a line

L x=3+2t, y=1-5t, z=-2+6t

show that for any real number k, the plane

(2x-2y+4z-1)+k(2x-2y+z-3)=0 contains the line of intersection of P1 and P2

find an eqt of the plane containing the line of intersection of P1 and P2, and // L

find the shortest distance b/w L and the line of intersection of P1 and P2

P1 : 3x-2y+4z=1 no typo

2 個已更新項目:

may u explain a bit more on how do you find the distance between a line and a plane which is parallel to it?

3 個已更新項目:

what is the method in general?

### 2 個解答

• 1 0 年前
最佳解答

1. If A(x,y,z) belongs to P1 and P2, then

3x-2y+4z-1=0 and 2x-2y+z-3=0,

so, (3x-2y+4z-1)+k(2x-2y+z-3)=0+k*0=0

ie. A(x,y,z) satisfies (3x-2y+4z-1)+k(2x-2y+z-3)=0 for any k.

2. Let plane E: (3x-2y+4z-1)+k(2x-2y+z-3)=0 parallel L, then

the normal vector of E (3+2k, -2-2k, 4+k) is perpendicular to the

direction vector (2, -5, 6) of L, thus, 2(3+2k)-5(-2-2k)+6(4+k)=0, k=-2.

Hence, E : -x+2y+2z+5=0

3. The sortest distance = d(E, L) = | -3+2-4+5|/3=0.

2011-05-16 01:43:09 補充：

3. The shortest distance = d(E, L)= d(E, A(3, 1, -2))= | -3 + 2 -4 + 5 |/3=0.

題目: P1是 3x-2y+4z=1 or 2x-2y+4z=1?

2011-05-16 02:17:37 補充：

(2x-2y+4z-1)+k(2x-2y+z-3)=0 ????

2011-05-16 02:20:12 補充：

(2x-2y+4z-1)+k(2x-2y+z-3)=0 ????

2011-05-16 18:17:23 補充：

Because L//E, so the distance b/w E and L equals the distance b/w any point A of L and the plan E.

Taking A(3,1,-2), thus the distance d(L, E)= d(A, E)= | -3+2 -4 +5| /√(1+4+4)= 0.

In general, if E contains L2 and E parrelle to L1, then d(L1, L2)=d(E, L1)=d(E, A),

where A is any point of L1.

• Sam
Lv 6
1 0 年前

To : 煩惱即是菩提 ( 知識長)

第三題你是不是看錯題目?