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Q.

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

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a

25, 445   and -545

b

145, 245  and 345

c

445, 0 and 445

d

345, 245 and 545

answer is A.

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Detailed Solution

A→ = 2i^+4j^−5k^ ∴|A|→ = (2)2+(4)2+(−5)2   =  45∴   cosα = 245,    cosβ = 445, cosγ = -545
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If A→ = 2i^+4j^−5k^ then the direction  cosines of the vector A→ are