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Selecting Consecutive Terms in A.P

The value of ...

Question

The value of $[\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}], where |\vec{a}| = 1, |\vec{b}| = 5, |\vec{c}| = 3, is$
(a) 0
(b) 1
(c) 6
(d) none of these

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Solution

(a) 0

$We have [\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}] = ((\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})) . (\vec{c} - \vec{a}) (By definition of scalar triple product) = ((\vec{a} - \vec{b}) \times \vec{b} - (\vec{a} - \vec{b}) \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b} - \vec{b} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b} - 0 - \vec{a} \times \vec{c} + \vec{b} \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b}) . (\vec{c} - \vec{a}) - (\vec{a} \times \vec{c}) . (\vec{c} - \vec{a}) + (\vec{b} \times \vec{c}) . (\vec{c} - \vec{a}) = (\vec{a} \times \vec{b}) . \vec{c} - (\vec{a} \times \vec{b}) . \vec{a} - (\vec{a} \times \vec{c}) . \vec{c} + (\vec{a} \times \vec{c}) . \vec{a} + (\vec{b} \times \vec{c}) . \vec{c} - (\vec{b} \times \vec{c}) . \vec{a} = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{a}] - [\vec{a} \vec{c} \vec{c}] + [\vec{a} \vec{c} \vec{a}] + [\vec{b} \vec{c} \vec{c}] - [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}] - 0 - 0 + 0 + 0 - [\vec{a} \vec{b} \vec{c}] (∵ [\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]) = 0$

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Q. The value of

[\to a - \to b \to b - \to c \to c - \to a]

| \to a | = 1, | \to b | = 5, | \to c | = 3

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[\to a - \to b \to b - \to c \to c - \to a]

| \to a | = 1, | \to b | = 5, | \to c | = 3

x = l o g_{b} a, y = l o g_{c} b, z = l o g_{a} c

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are three vectors such that

\vec{a} + \vec{b} + \vec{c} = \vec{0} and |\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5,

then the value of

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