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Question : 33 of 160
Marks:
+1,
-0
Solution:
We have,
(a+b)⋅p=(a+b)⋅‌ =‌‌‌[∵[bbc]=0] =1 Similarly,
(b+c)⋅q=(c+a)⋅r=1 ∴‌‌α=1+1+1=3 Now,
(a+b)⋅(b+c)×(a+b+c) =[a+bb+ca+b+c] =||[]. =[I(l−1)−1(0−1)][abc]=[abc]
and
b⋅(a×c)=[bac]=−[abc] ∴‌‌β=‌=−1 ∴‌‌α+β=3−1=2
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