Let a→, b→, c→ be three non-zero vectors which are pairwise non-collinear. If a→+3b→ is collinear with c→ and b→+2c→ is collinear with a→, then a→+b→+6c→ is :
Solutions · 1
Official solution
We are given that a→+3b→ is collinear with c→, and b→+2c→ is collinear with a→. This means we can write:
a→+3b→=λc→...(i)b→+2c→=μa→...(ii)
for some scalars λ and μ.
We are trying to find a→+b→+6c→ in terms of a→, b→, and c→. We can also express this as :
a→+3b→+6c→=(λ+6)c→...(iii)
by adding 6c→ to both sides of equation (i).
Now, from equation (ii), multiplying by 3 gives us :
3b→+6c→=3μa→...(iv)
Adding a→ to both sides of equation (iv) gives :
a→+3b→+6c→=(1+3μ)a→...(v)
Now, we have two expressions for a→+3b→+6c→, one in terms of c→ (from equation iii) and one in terms of a→ (from equation v). Setting these equal to each other gives :
(λ+6)c→=(1+3μ)a→...(vi)
Since a→ and c→ are not collinear, this equation can only hold if the coefficients on both sides are zero, hence :
λ+6=0 and 1+3μ=0
This gives λ=−6 and μ=−31.
Finally, substituting λ=−6 into equation (iii) gives :
a→+3b→+6c→=0
So, a→+b→+6c→=0→.
Therefore, the correct answer is Option D : 0→.