c×a=b×c=−c×b, so c×(a+b)=0, meaning c parallel to a+b=5i^−2j^+2k^. Let c=t(a+b). 2a−3b=−5i^+21j^−26k^. Dot: t[(−25)+(−42)+(−52)]=−119t=2382, so t=−22. ∣c∣2=8⋅(25+4+4)=8⋅33=264.
Let a = 2 i + 3 j - 4 k and b = 3 i - 5 j + 6 k. Let be a vector such that × a = b × and ·(2 a-3 b) = 238√2 then | |^2 is equal to :
CUET UG 2022 — Mathematics Algebra
2022mcqhard
Let a=2i^+3j^−4k^ and b=3i^−5j^+6k^. Let c be a vector such that c×a=b×c and c⋅(2a−3b)=2382 then ∣c∣2 is equal to :
Official previous-year question
Held on 17 Aug 2022 · Verified 13 Jul 2026.
Options
- A
112
- B
164
- C
264
- D
260
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