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Question : 33 of 160
Marks:
+1,
-0
Solution:
Given, |a|=|b|=|c|=1
‌a⋅b=0
‌(a−c)⋅(b+c)=0
‌a⋅b+a⋅c−c⋅b−|c|2=0
‌a⋅c−b⋅c=1
‌‌ Also, ‌c=lc+mb+n(a×b)
‌a⋅c=l(a)2+m(a⋅b)+n(a⋅(a×b)
‌a⋅c=l
Similarly, bâ‹…c=m
‌∴l−m=1⇒l2+m2−2lm=1
‌‌ Now, (c-la-mb) ‌=(n(a×b))2
‌⇒c2+l2a2+m2b−2mb⋅c+2lma⋅b−2la⋅c=n2|a|2|b|2
‌⇒1+l2+m2−2m2−2l2=n2
‌⇒1−l2−m2=n2
‌⇒n2=−2lm‌‌(∵−2lm=1−l2−m2)
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