Nonzero vectors a- b- c- satisfy a-b-0-b-a- -b-c-0 and 2-b-c-b-a-If -a-b-4 c- then -

(b̅ a̅).(b̅+a̅ μb̅/4 )=0 ⇒ (4 μ)b^2=a^2 ⇒μ lt; 4 2|b⃗+a⃗ μb⃗/4|=|b̅ a̅| ⇒μ^2=5μ⇒μ = 0 or 5 ⇒μ = 0 ...

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Condition for Coplanarity of Four Points

Nonzero vecto...

Question

Nonzero vectors $\to a, \to b, \to c$ satisfy $\to a . \to b$ =0, $(\to b - \to a) . (\to b + \to c)$ = 0 and 2 $| \to b + \to c | = | \to b - \to a |$ .

If = $\to a = μ \to b + 4 \to c$ then $μ$ = ____

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\to a, \to b, \to c

\to a \cdot \to b = 0

(\to a - \to c) \cdot (\to b + \to c) = 0

\to c = λ \to a + μ \to b + ω (\to a \times \to b)

λ, μ, ω

\to a, \to b, \to c

\to a . \to b = 0

(\to a - \to c) . (\to b + \to c) = 0

\to c = λ \to a + μ \to b + ω (\to a \times \to b)

λ, μ, ω

\to a, \to b

\to c, \to a \cdot \to b = \to a \cdot \to c

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

\to b = \to c

\to a, \to b, \to c

\to a . \to b = 0

\to a + \to b = \to c

\to a, \to b, \to c

\to a \cdot \to b = 0 = \to a \cdot \to c

\to b

\to c

\frac{π}{3}

| \to a \times \to b - \to a \times \to c |

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