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Vectors and Its Types

Let the vecto...

Question

Let the vectors $\to a$ and $\to b$ be perpendicular to each other. Then a vector $\to v$ in terms of $\to a$ and $\to b$ satisfying the equation $\to v \cdot \to a = 0,$ $\to v \cdot \to b = 1$ and $[\to v \to a \to b] = 1$ is

A

\frac{\to b}{{∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}} + \frac{\to a \times \to b}{{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2}}

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B

\frac{\to b}{∣ ∣ ∣ \to b ∣ ∣ ∣} + \frac{\to a \times \to b}{{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2}}

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C

\frac{\to b}{{∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}} + \frac{\to a \times \to b}{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}

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D

\frac{\to b}{∣ ∣ ∣ \to b ∣ ∣ ∣} + \frac{\to a \times \to b}{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}

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Solution

The correct option is A $\frac{\to b}{{∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}} + \frac{\to a \times \to b}{{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2}}$
We have $\to a ⊥ \to b$
$\Rightarrow \to a, \to b, \to a \times \to b$ are linearly independent, $\to v$ can be expressed uniquely in terms of $\to a, \to b$ and $\to a \times \to b$

Let $\to v = x \to a + y \to b + z (\to a \times \to b) \dots (1)$
Given $\to a \cdot \to b = 0, \to v \cdot \to a = 0, \to v \cdot \to b = 1, [\to v \to a \to b] = 1$

From equation $(1),$
$\to v \cdot \to a = x {(\to a)}^{2} = 0$
$\Rightarrow x = 0 \to v \cdot \to b = x \to a \cdot \to b + y {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + z \to b \cdot (\to a \times \to b) = 1 \Rightarrow y {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} = 1$
$\Rightarrow y = \frac{1}{{∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}}$

$\to v \cdot (\to a \times \to b) = x \cdot 0 + y \cdot 0 + z {∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2} = 1 \Rightarrow z = \frac{1}{{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2}}$

$∴$ From equation $(1),$
$\to v = \frac{\to b}{{∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}} + \frac{\to a \times \to b}{{∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2}}$

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