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Question : 62 of 160
Marks:
+1,
-0
Solution:
Given,
B=3−2+ C=5+−3 BC=2+3−4 max{AB,BC,AC}=BC ∴BC is hypotenuse of
∆ABC
∠A=90∘ ∴‌‌AB⋅AC=0 BA⋅BC=|BA|BC|cos‌B CA⋅CB=|CA|CB|cos‌C ∴AB−AC+BA−BC+CA−CB =0+|BC|(|BA|‌cos‌B+|CB|cos‌C)
=0+|BC|BC| [ ∵ By projection formula]
=|BC|2−(√(2)2+32+42)2=4+9+16=29
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