# vectors!!! (20 pts)

1) With respect to right-handed Cartesian coordinates, let a = [1, -2, 0],

b = [-2, 3, 0], c = [2, -4, -1], and d = [3, -1, 5]. Find (a x b) x (c x d).

2) Let a = [1, 1, 0], b = [3, 2, 1], and c = [1, 0, 2] . Find the angle between a+c

and b+c.

3) For what c are 3x + z = 5 and 8x - y + cz = 9 orthogonal?

### 1 Answer

- 知足常樂Lv 76 years agoFavorite Answer
#3

Consider that the normal vectors of the planes are orthogonal.

(3, 0, 1)．(8, -1, c) = 0

24 + c = 0

c = -24

2014-01-26 23:08:38 補充：

#2

a + c = [2, 1, 2] and b + c = [4, 2, 3]

Let the required angle be θ.

(a + c)．(b + c) = |a + c| × |b + c| × cosθ

8 + 2 + 6 = √(2²+1²+2²) √(4²+2²+3²) cosθ

16 = √9 √29 cosθ

cosθ = 16/(3√29)

θ = 7.955800083°

2014-01-26 23:10:20 補充：

#1

I think you know how to do it.

Just tedious.

(a, b, c) × (x, y, z)

｜ｉ ｊ ｋ｜

＝｜ａ ｂ ｃ｜

｜ｘ ｙ ｚ｜

2014-02-05 04:10:49 補充：

[#1]

a x b = (1, -2, 0) × (-2, 3, 0)

｜ ｉ ｊ ｋ｜

＝｜ １－２ ０｜ ＝ －ｋ ＝ (0, 0, -1)

｜－２ ３ ０｜

c x d = (2, -4, -1) × (3, -1, 5)

｜ ｉ ｊ ｋ｜

＝｜ ２－４－１｜ ＝ －２１ｉ－１３ｊ＋１０ｋ ＝ (-21, -13, 10)

｜ ３－１ ５｜

(a x b) x (c x d) = (0, 0, -1) × (-21, -13, 10)

｜ ｉ ｊ ｋ｜

＝｜ ０ ０ －１｜ ＝ －１３ｉ＋２１ｊ ＝ (-13, 21, 0)

｜－２１－１３ １０｜

♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬

[#2]

a + c = [2, 1, 2] and b + c = [4, 2, 3]

Let the required angle be θ.

(a + c)．(b + c) = |a + c| × |b + c| × cosθ

8 + 2 + 6 = √(2²+1²+2²) √(4²+2²+3²) cosθ

16 = √9 √29 cosθ

cosθ = 16/(3√29)

θ = 7.955800083°

♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬

[#3]

Consider that the normal vectors of the planes are orthogonal.

(3, 0, 1)．(8, -1, c) = 0

24 + c = 0

c = -24