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Q.

The direction cosines of a vector A are cosα=452,cosβ=12 and cosγ=352, then the vector A is

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a

4i^+j^+3k^

b

4i^+5j^+3k^

c

4i^-5j^-3k^

d

i^+j^-k^

answer is B.

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Detailed Solution

cosα=AxA=452cosβ=AyA=12=552cosγ=AzA=352A¯=Axi^+Ayj^+Azk^=4i^+5j^+3k^

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