Question

a=3i-j+2k b=-i+3j+2k c= 4i+5j-2k &
d=i + 3j+5k then Compute the following
i)(axb)x(cxd) ii) (a×b).C- (axd).b​

Answer 1

Given :

a = 3i - j + 2k

b = -i + 3j + 2k

c = 4i + 5j - 2k

d = i + 3j + 5k

To find :

i) (a x b) x (c x d)

ii) (a × b).c - (a x d).b

Solution :

i × i = j × j = k × k = 0

i × j = k          j × i = -k

j × k = i          k × j = -i

k × i = j          i × k = -j  

i) a × b = (3i - j + 2k) × (-i + 3j + 2k)

a × b = (3)(-1)(i × i) + (3)(3)(i × j) + (3)(2)(i × k) + (-1)(-1)(j × i) + (-1)(3)(j × j) + (-1)(2)(j × k) + (2)(-1)(k × i) + (2)(3)(k × j) + (2)(2)(k × k)

a × b = 0 + 9k - 6j - k + 0 - 2i - 2j - 6i + 0

a × b = -8i - 8j + 8k

c x d = (4i + 5j - 2k) × (i + 3j + 5k)

c × d = (4)(1)(i × i) + (4)(3)(i × j) + (4)(5)(i × k) + (5)(1)(j × i) + (5)(3)(j × j) + (5)(5)(j × k) + (-2)(1)(k × i) + (-2)(3)(k × j) + (-2)(5)(k × k)

c × d = 0 + 12k - 20j - 5k + 0 + 25i - 2j + 6i + 0

c × d = 31i - 14j + 7k

(a × b) × (c × d) = (-8i - 8j + 8k) × (31i - 14j + 7k)

(a × b) × (c × d) = (-8)(31)(i × i) + (-8)(-14)(i × j) + (-8)(7)(i × k) + (-8)(31)(j × i) + (-8)(-14)(j × j) + (-8)(7)(j × k) + (8)(31)(k × i) + (8)(-14)(k × j) + (8)(7)(k × k)

(a × b) × (c × d) = 0 + 112k + 56j + 248k + 0 - 56i + 248j + 112i + 0

(a × b) × (c × d) = 56i + 304j + 360k

b) i.i = j.j = k.k =1

i.j = i.k = j.i = j.k = k.i = k.j =0

(a × b).c = (-8i - 8j + 8k).(4i + 5j - 2k)

(a × b).c = -32 - 40 - 16

(a × b).c = -88

(a × d) = (3i - j + 2k ) × (i + 3j + 5k)

(a × d) = 0 + 9k - 15j + k + 0 - 5i + 2j - 6i + 0

(a × d) = -11i - 13j + 10k

(a × d).b = (-11i - 13j + 10k).(-i + 3j + 2k)

(a × d).b = 11 - 39 + 20

(a × d).b = -8

(a × b).c - (a x d).b = -88 - (-8)

(a × b).c - (a x d).b = -80

Answer 2

The First Answer I've seen is wrong that's why I'm posting this. Hope it's helpful for you...