Let →a,→b and →c be unit vectors, such that →a+→b+→c=→x,→a⋅→x=1,→b⋅→x=32,|→x|=2. Then the angle between →x and →c is
A
cos−1(37)
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B
cos−1(13)
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C
cos−1(34)
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D
cos−1(1√3)
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Solution
The correct option is Ccos−1(34) Let θ be the angle between →c and →x.
Given : →a+→b+→c=→x
Taking dot product of →x on both sides, we get →a⋅→x+→b⋅→x+→c⋅→x=|→x|2 ⇒1+32+|→c||→x|cosθ=4 ⇒2cosθ=32 ⇒θ=cos−1(34)