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Question

If a=^i+2^j+3^k, b=2^i^j+^k and c=3^i+2^j+^k and a×(b×c) is equal to xa+yb+zc, then x+y+z is equal to

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Solution

b×c=∣ ∣ ∣^i^j^k211321∣ ∣ ∣=^i(12)^j(23)+^k(4+3)=3^i+^j+7^ka×(b×c)=∣ ∣ ∣^i^j^k123317∣ ∣ ∣=11^i16^j+7^k

Given that

a×(b×c)=xa+yb+zc

x+2y+3z=11(1)2xy+2z=16(2)3x+y+z=7(3)

Adding (1) with 2×(2) we get,

5x+7z=21 ---- (4)

Adding (3) with (2)

5x+3z=9 ---- (5)

Subtracting (4) from (5), we get,

z=3

from (5)

5x+3z=9

5x+3×3=9

x=0

from (1)

x=2y+3z=11

0+2y+3×3=11

y=10

therefore,

x+y+z=7

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