Let →a,→b and →c be three non-zero vectors such that no two of these are collinear. If the vector →a+2→b is collinear with →c and →b+3→c is collinear with →a (λ being some non-zero scalar) then →a+2→b+6→c equals
A
λ¯¯¯a
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B
λ¯¯b
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C
λ¯¯c
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D
¯¯¯0
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Solution
The correct option is D¯¯¯0 →a+2→b is collinear with →c.
So we can write: →a+2→b=t1→c.......(i)
Similarly, →b+3→c is collinear with →a.
So we can write →b+3→c=t2→a.........(ii)
To eliminate →b we will perform (i)−2(ii) to get
→a−6→c=t1→c−2t2→a
(1+2t2)→a−(6+t1)→c=0
Since →a and →c are non-collinear, both 1+2t2 and 6+t1 must be zero to satisfy above equation. ∴t2=−12 and t1=−6
Substituting the values of t2 and t1 in (i) we get