If the position vectors of A- B- C- D are a- b- 2a-3b- a-2b- respectively- then A-C- D-B- B-A- D-A- are-

The correct option is A a⃗+3b⃗, 3b⃗ a⃗, a⃗ b⃗, 2b⃗ Given, #160; O⃗A⃗=a⃗ O⃗B⃗=b⃗ O⃗C⃗=2a⃗+3b⃗ O⃗D⃗=a⃗ 2b⃗ Then, #160; A⃗C⃗=O⃗C⃗ ...

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Addition of Vectors

If the positi...

Question

If the position vectors of A, B, C, D are $\to a, \to b . 2 \to a + 3 \to b, \to a - 2 \to b$ respectively, then $\to A C, \to D B, \to B A, \to D A$ are?

\to a + 3 \to b, 3 \to b - \to a, \to a - \to b, 2 \to b

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2 \to b, \to b - 2 \to a, 3 \to b + \to a, \to b - \to a

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\to a - 3 \to b, 3 \to b - \to a, \to a + \to b, 2 \to b

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- 2 \to b, \to b - 2 \to a, 3 \to b - \to a, \to b - \to a

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Solution

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A, B, C, D

\to a, \to b, 2 \to a + 3 \to b

\to a - 2 \to b

- - \to D B = 3 \to b - \to a

- - \to A C = \to a + 3 \to b

\to a, \to b, \to c

\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]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

2 \to a + 3 \to b - \to c, \to a - 2 \to b + 3 \to c, 3 \to a + 4 \to b - 2 \to c a n d \to a - 6 \to b + 6 \to c

\to a, \to b a n d \to c

\to A = (\to a - 2 \to b + 3 \to c)

\to B = (- 2 \to a + 3 \to b + 2 \to c)

\to C = (- 8 \to a + 13 \to b)

\to A B

\to A C

If $\to a a n d \to b$ are non-zero vectors such that $| \to a - 3 \to b | = | \to a + 3 \to b |$ , then the angle between $\to a a n d \to b$ is

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