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Question : 35 of 160
Marks:
+1,
-0
Solution:
a=++,c=− Given,
a×b=c and
a⋅b=3 ∵‌a×b=c ⇒‌a×(a×b)=a×c ⇒‌(a⋅b)a−(a⋅a)b=a×c ⇒‌3a−|a|2b=a×c . . . (i)
‌∵‌a×b‌‌=c ⇒‌a×(a×b)‌‌=a×c ⇒(a⋅b)a−(a⋅a)b‌‌=a×c ⇒‌‌3a−|a|2b‌‌=a×c ‌‌=(−1−1)−(−1−0)+(1−0) ‌‌=−2++ |a|2=12+12+12=3 From Eq. (i),
3(++)‌‌−3b=−2++ 3+3+3+2−−‌‌=3b or
b=‌5+‌2+‌2
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