A- -b- -c- are the position vectors of the vertices of a triangle A B C- If a-b-c-b-0- then the position vector of circumcentre is

The correct option is B (#160;a+#160;c)2 Given that #160; (#160;a #160; b).(#160;c #160; b) = 0 , we have that #160; ∠ B = 90^0. Thus #160 ...

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Applications of Dot Product

a⃗ ,b⃗ ,c⃗ ar...

Question

$\to a, \to b, \to c$ are the position vectors of the vertices of a triangle $A B C$ . If $(\to a - \to b) . (\to c - \to b)$ =0, then the position vector of circumcentre is

\frac{(\to b + \to c)}{2}

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\frac{\to a + \to b}{2}

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\frac{(\to a + \to c)}{2}

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\frac{\to a + \to b + \to c}{2}

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\to a, \to b, \to c

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\to r

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(| \to b - \to c | + | \to c - \to a | + | \to a - \to b |) \to r = | \to b - \to c | \to a + | \to c - \to a | \to b + ∣ ∣ \to a - \to b ∣ ∣ \to c

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\to a, \to b, \to c

| (\to a \times \to b) . \to c | = | \to a | | \to b | | \to c |

| \to a + \to b + \to c |^{2} =

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

(\to a \times \to b) \times \to c = \to a \times (\to b \times \to c)

\to a, \to b

\to c

\to a . \to b \neq 0, \to b . \to c \neq 0

\to a

\to c

\to a, \to b, \to c

(\to a + \to b) . \to c = (\to a - \to b) . \to c = 0

(\to a \times \to b) \times \to c

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