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If vector a = vector(4 i + 3 j + 4 k) and vector b = vector(3 i + 2 k) then |vector(b x 2a)| is(A) 18 (B) 20 (C) 22 (D) 25 |
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Answer» Correct option: (C) 22 vec(b×2a)=2vec(3i+2k)×vec(4i+3j+4k) =2vec(12i×i+8k×i+9i×j+6k×j+12i×k+8k×k) =2vec(12.0+8j+9k-6i-12j+8*0) =2vec(-6i-4j+9k) Now |(b×2a)| =2√(6^2+4^2+9^2)=2√133 ~23 |
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