- a-b- - - a-b- - then angle between a- and b- is

The correct option is D π4 We have | a̅.b̅ |= | a̅ | | b̅ | | cosα #160; | | a̅×b̅ |= | a̅ | | b̅ | | sinα #160; | Equating | a̅ | | b̅ ...

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Vectors and Its Types

| a̅.b̅ |= | ...

Question

$∣ ∣ ¯ a . ¯ b ∣ ∣ = ∣ ∣ ¯ a \times ¯ b ∣ ∣$ , then angle between $¯ a$ and $¯ b$ is

0^{\circ}

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\frac{π}{6}

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\frac{π}{4}

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\frac{π}{2}

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| \begin{matrix} ¯ a \end{matrix} | = 4, | \begin{matrix} ¯ b \end{matrix} | = 2

¯ a

¯ b

\frac{π}{6}

(| ¯ a \times ¯ b ∣ ∣)^{2}

¯ a, ¯ b

¯ a ⊥ ¯ b

¯ r

| ¯ r | = 2, (¯ a, ¯ r) = \frac{π}{3}

(¯ a \times ¯ b) \times (¯ a \times ¯ r) = ¯ a

(¯ b, ¯ r) =

¯ a = x^i + y^j + 2^k, ¯ b =^i -^j +^k, ¯ c =^i + 2^j

(¯ a

¯ b) = π / 2, ¯ a . ¯ c = 4

¯ a = α_{1}^i + α_{2}^j + α_{3}^k

¯ b = β_{1}^i + β_{2}^j + β_{3}^k

¯ c = γ_{1}^i + γ_{2}^j + γ_{3}^k

| ¯ a | = 2 \sqrt{2}

¯ a

\frac{π}{3}

¯ b

¯ c

¯ b

¯ c

\frac{π}{3}

{∣ ∣ ∣ ∣ \begin{matrix} α_{1} & α_{2} & α_{3} β_{1} & β_{2} & β_{3} γ_{1} & γ_{2} & γ_{3} \end{matrix} ∣ ∣ ∣ ∣}^{n}

¯ A + B = ¯ A . ¯ B

¯ A . ¯ B = ¯ A + B

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