The position vectors of points A-B-C- are respectively a-b-c- If P divides A-B- in the ratio 3-4 and Q divides B-C- in the ratio 2-1 both externally then P-Q- is

The correct option is D 2(b⃗+c⃗ 2⃗a⃗) By section formula p⃗ = 3 b⃗ 4 a⃗3 4 = 3 b⃗ 4 a⃗ 1 and q⃗ = 2 c⃗ b⃗2 1 = 2 c⃗ b⃗1 Thus, P⃗ ...

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Position Vector

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The position vectors of points $\to A, \to B, \to C$ are respectively $\to a, \to b, \to c$ . If $P$ divides $\to A B$ in the ratio $3 : 4$ and $Q$ divides $\to B C$ in the ratio $2 : 1$ both externally then $\to P Q$ is

\to b + \to c - \to 2 a

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2 (\to b + \to c - \to 2 a)

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4 \to a - \to b - \to c

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

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\to a, \to b, \to c

2 \to a + \to b, \to a + 2 \to b + \to c, 4 \to a - 2 \to b - \to c

3 \to a + 4 \to b - 5 \to c

[\to a + 2 \to b - \to c, \to a - \to b, \to a - \to b - \to c] =

\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, \to b, \to c

_{1} = \to a - \to b + \to c,_{2} = \to b + \to c - \to a

_{3} = \to c + \to a - \to b, \to r = 2 \to a - 2 \to b + 4 \to c

\to r = x_{1}_{1} + x_{2}_{2} + x_{3}_{3}

\to a, \to b, \to c

A, B, C

Δ A B C

\to r

P

(| \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

P

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