Figure shows the three vectors a - b and c - where R is the midpoint of PQ- Then which of the following relations is correct -

The correct option is D a + b = 2 c Position vector of P is a⃗ 0⃗ = a⃗ Similarly, nbsp;Position vector of Q is b⃗ 0⃗ = b⃗ Thus, position v ...

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Perpendicular Distance of a Point from a Plane

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Question

Figure shows the three vectors $\to a, \to b$ and $\to c$ , where R is the midpoint of PQ. Then which of the following relations is correct ?

\to a + \to b = 2 \to c

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\to a + \to b = \to c

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\to a - \to b = 2 \to c

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\to a - \to b = \to c

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\to a, \to b, \to c

\to r

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

λ [\to a \to b \to c]

λ

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

\to a, \to b

\to c

(\to a + \to b + \to c) \cdot (\to p + \to q + \to r)

\to a, \to b, \to c

\to r = \frac{(\to r . \to a) (\to b \times \to c) + (\to r . \to b) (\to c \times \to a) + (\to r . \to c) (\to a \times \to b)}{λ},

λ

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

\to a, \to b

\to c

(\to a + \to b + \to c) \cdot (\to p + \to q + \to r)

\to a, \to b

\to c

\to p, \to q

\to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(\to a + \to b) \cdot \to p + (\to b + \to c) \cdot \to q + (\to c + \to a) \cdot \to r =

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