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Questions  

If A=2i^+4j^5k^ then the direction  cosines of the vector A are

a
25, 445   and -545
b
145, 245  and 345
c
445, 0 and 445
d
345, 245 and 545

detailed solution

Correct option is A

A→ = 2i^+4j^−5k^ ∴|A|→ = (2)2+(4)2+(−5)2   =  45∴   cosα = 245,    cosβ = 445, cosγ = -545

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