Single Choice

If $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} = a \vec{\delta}$$ and $$\vec{\beta} + \vec{\gamma} + \delta = b \vec{\alpha}, \vec{\alpha}$$ and $$\vec{\delta}$$ are non-collinear, then $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} + \vec{\delta}$$ equals

A$$a \vec{\alpha}$$
B$$b \vec{\delta}$$
C0
Correct Answer
D$$(a + b)\vec{\gamma}$$

Solution

Given $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} = a \vec{\delta}$$ (i)
$$\vec{\beta} + \vec{\gamma} + \delta = b \vec{\alpha}$$ (ii)
From (i), $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} + \vec{\delta} = (a + 1) \vec{\delta}$$ (iii)
From (ii), $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} + \vec{\delta} = (b + 1) \vec{\alpha}$$ (iv)
From (iii) and (iv),
$$(\alpha + 1) \vec{\delta} = (b + 1) \vec{\alpha}$$ (v)
Since $$\vec{\alpha}$$ is not parallel to $$\vec{\delta}$$,
From (v), $$a + 1 = 0$$ and $$b + 1 = 0$$
From (iii), $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} + \vec{\delta} = 0$$

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