If a→ is perpendicular to b→, then vector v→ in terms of a→ and b→ satisfying the condition v→⋅a→=0 and v→⋅b→=1 and [v→a→b→]=1 is

Let v→=xa→+yb→+z(a→×b→);v→⋅a→=0⇔x=0;v→⋅b→=1⇔y=1|b→|2v→⋅(a→×b→)=z|a→×b→|2=1;z=1|a→×b→|2;v→=b→|b→|2+a→×b→|a→×b→|2

If a→ is perpendicular to b→, then vector v→ in terms of a→ and b→ satisfying the condition v→⋅a→=0 and v→⋅b→=1 and [v→a→b→]=1 is

Let v→=xa→+yb→+z(a→×b→);v→⋅a→=0⇔x=0;v→⋅b→=1⇔y=1|b→|2v→⋅(a→×b→)=z|a→×b→|2=1;z=1|a→×b→|2;v→=b→|b→|2+a→×b→|a→×b→|2

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If $\vec{a}$ is perpendicular to $\vec{b}$ , then vector $\vec{v}$ in terms of $\vec{a}$ and $\vec{b}$ satisfying the condition $\vec{v} \cdot \vec{a} = 0 and \vec{v} \cdot \vec{b} = 1$ $and [\begin{array}{lll} \vec{v} & \vec{a} & \vec{b} \end{array}] = 1$ is

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$\frac{\vec{b}}{| \vec{b} |^{2}} + \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |^{2}}$

$\frac{\vec{b}}{| \vec{b} |^{2}} + \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |}$

$\vec{b} + \frac{(\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |}$

$\frac{\vec{b}}{| \vec{b} |} + \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |^{2}}$

answer is A.

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Detailed Solution

Let $\vec{v} = x \vec{a} + y \vec{b} + z (\vec{a} \times \vec{b}); \vec{v} \cdot \vec{a} = 0 \Leftrightarrow x = 0; \vec{v} \cdot \vec{b} = 1 \Leftrightarrow y = \frac{1}{| \vec{b} |^{2}}$

$\vec{v} \cdot (\vec{a} \times \vec{b}) = z | \vec{a} \times \vec{b} |^{2} = 1; z = \frac{1}{| \vec{a} \times \vec{b} |^{2}}; \vec{v} = \frac{\vec{b}}{| \vec{b} |^{2}} + \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |^{2}}$