If a- b-c- are the position vectors of the vertices of a - ABC- then area of the triangle is

The correct option is C #10; 12|a⃗×b⃗+b⃗×c⃗+c⃗×a⃗| #10;Area of Δ =12|a⃗×b⃗+b⃗×c⃗+c⃗×a⃗| Where a⃗×b⃗+b⃗×c⃗+c⃗×a⃗=n⃗ where n⃗ is normal ...

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Applications of Cross Product

If a⃗, b⃗,c...

Question

If $\to a, \to b, \to c$ are the position vectors of the vertices of a $Δ A B C$ , then area of the triangle is

[\to a \to b \to c]

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\frac{1}{2} [\to a \to b \to c]

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[\to a \times \to b + \to b \times \to c + \to c \times \to a]

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\frac{1}{2} | \to a \times \to b + \to b \times \to c + \to c \times \to a |

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If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

\to a, \to b, \to c

[\to a \to b \to c] = 4

[\to a \times \to b

\to b \times \to c

\to c \times \to a] = ?

\to a, \to b, \to c

\to a, \to b, \to c

\to a \times \to b = \to b \times \to c = \to c \times \to a

[\to a \to b \to c] = 0.

\to a, \to b, \to c

[\to a \to b \to c] = 5

[\to a \times \to b, \to b \times \to c, \to c \times \to a]

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

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