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If a,b,c,d are four vectors, then prove that
(a×b).(c×d)+(b×c).(a×d)+(c×a).(b×d)=0

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Solution

The value of (axb).(cxd) can be found by
(axb).(cxd)= (a.c)(b.d)-(a.d)(b.c)
So the value for given problem will become
(a×b).(c×d)+(b×c).(a×d)+(c×a).(b×d)=
(a.c)(b.d)-(a.d)(b.c)+(b.a)(c.d)-(b.d)(c.a)+(c.b)(a.d)-(c.d)(a.b)
If we consider a.c=c.a= scalar (number)
Then all the above expression will mutually cancel and we will get the value as zero

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Q. The value of

(a \times b) \cdot (c \times d) + (b \times c) \cdot (a \times d) + (c \times a) \cdot (b \times d)

\to a, \to b

\to c

be three unit vectors such that

\to a + \to b + \to c = \to 0

λ = \to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

\to d = \to a \times \to b + \to b \times \to c + \to c \times \to a

, then the ordered pair

(λ, \to d)

| \to a | = | \to b | = | \to c | = 2

\to a \cdot \to b = \to b \cdot \to c = \to c \cdot \to a = 2,

[\to a \to b \to c] cos 45^{\circ}

| a | = 2

, | b | = 3

| c | = 4

a + b + c = 0

then the value of

b \cdot c + c \cdot a + a \cdot b

(\to a \times \to b) \times \to c = \to a \times (\to b \times \to c)

\to a, \to b

\to c

are three non-zero vectors such that

\to a \cdot \to b \neq 0, \to b \cdot \to c \neq 0,

\to a

\to c

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