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Question

(i) If a =3i^+4j^ and b =i^+j^+k^, find the value of a ×b .

(ii) If a =2i^+k^, b =i^+j^+k^, find the magnitude of a ×b .

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Solution

i Given: a=3i^+4j^+0k^ b=i^+j^+k^ a×b=i^j^k^340111 =4-0 i^ -3-0 j^ +3-4 k^ =4 i^-3j^-k^a×b=16+9+1 =26

ii Given:a=2i^+0j^+k^ b=i^+j^+k^a×b=i^j^k^201111 =0-1 i^ -2-1 j^ +2-0 k^ =- i^-j^+2k^a×b=-12+-12+22 =6

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