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Question : 34 of 160
Marks:
+1,
-0
Solution:
From the given vectors,
A×B‌‌=|| ‌‌=i(2)−j(2)+k(2−1) ‌‌=2i−2j+k This implies,
(A×B)×C|‌‌=|(A×B)||C|sin‌30∘ ‌‌=√4+4+1|C|(‌) ‌‌=‌|C| |C−A|‌‌=2√2 This implies,
|C|2+|A|2−2C⋅A=8 |C|2+9−2|C|‌‌=8 (|C|−1)2‌‌=0 |C|‌‌=1 So
|(A×B)×C|=‌
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