The position vectors of A-B-C-D are a-b-2a- - 3b- and a- - 2b- respectively- Show that DB - 3b- - a- and AC - a- - 3b-

O⃗A⃗=a⃗ O⃗B⃗=b⃗ O⃗C⃗=2a⃗+3b⃗ O⃗D⃗=a⃗ 2b⃗ D⃗B⃗=O⃗B⃗ O⃗D⃗=b⃗ (a⃗ 2b⃗)=3b⃗ a⃗ A⃗C⃗=O⃗C⃗ O⃗A⃗=(2a⃗+3b⃗) (a⃗)=a⃗+3b⃗ ...

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Addition of Vectors

The position ...

Question

The position vectors of $A, B, C, D$ are $\to a, \to b, 2 \to a + 3 \to b$ and $\to a - 2 \to b$ respectively. Show that $- - \to D B = 3 \to b - \to a$ and $- - \to A C = \to a + 3 \to b$

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

\to A C, \to D B, \to B A, \to D A

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

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

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

2 \to a - \to b - \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

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