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Question

If a=i+j+k,b=4i+3j+4k and c=i+αj+βk are linearly dependent vectors and |c|=3, then

A
α=1,β=1
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B
α=1,β=±1
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C
α=1,β=±1
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D
α=±1,β=1
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Solution

The correct option is D α=±1,β=1
Since a,b,c are linearly dependent.

[abc]=0

∣ ∣1114341αβ∣ ∣=0

∣ ∣1004101α1β1∣ ∣=0

(β1)=0

β=1

Also, |c|=21+α2+β2=3

α2=1α=±1

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