AB=(1,0,1), AC=(−1,2,2). AB×AC=i^1−1j^02k^12=i^(0−2)−j^(2+1)+k^(2−0)=−2i^−3j^+2k^.
A vector perpendicular to a plane containing a triangle ABC having vertices as A(1,1,0), B(2,1,1) and C(0,3,2), is:
CUET UG 2022 — Mathematics Algebra
2022mcqmedium
A vector perpendicular to a plane containing a triangle ABC having vertices as A(1,1,0), B(2,1,1) and C(0,3,2), is:
Official previous-year question
Held on 23 Aug 2022 · Verified 13 Jul 2026.
Options
- A
−2i^−3j^+2k^
- B
i^+j^+k^
- C
2i^−j^−k^
- D
i^−2j^+2k^
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