Let a→=iˆ-jˆ+2kˆ and let b→ be a vector such that a→×b→=2iˆ-kˆ and a→⋅b→=3 . Then the projection of b→ on the vector a→-b→ is___

a‾×(a‾×b‾)=3a→-6b‾=iˆjˆkˆ1-1220-1=iˆ+5jˆ+2kˆ 6b‾=2iˆ-8jˆ⇒b‾=iˆ3-4jˆ3+2kˆ3Projection =b‾⋅(a‾-b‾)|a‾-b‾|=3-2196+219-6=6/9219=221

Let a→=iˆ-jˆ+2kˆ and let b→ be a vector such that a→×b→=2iˆ-kˆ and a→⋅b→=3 . Then the projection of b→ on the vector a→-b→ is___

a‾×(a‾×b‾)=3a→-6b‾=iˆjˆkˆ1-1220-1=iˆ+5jˆ+2kˆ 6b‾=2iˆ-8jˆ⇒b‾=iˆ3-4jˆ3+2kˆ3Projection =b‾⋅(a‾-b‾)|a‾-b‾|=3-2196+219-6=6/9219=221

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Let $\vec{a} = \hat{i} - \hat{j} + 2 \hat{k}$ and let $\vec{b}$ be a vector such that $\vec{a} \times \vec{b} = 2 \hat{i} - \hat{k}$ and $\vec{a} \cdot \vec{b} = 3$ . Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b}$ is___

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$\frac{2}{\sqrt{21}}$

$2 \sqrt{\frac{3}{7}}$

$\frac{2}{3} \sqrt{\frac{7}{3}}$

$\frac{2}{3}$

answer is A.

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Detailed Solution

$\begin{array}{r} \overline{a} \times (\overline{a} \times \overline{b}) = 3 \vec{a} - 6 \overline{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 2 \\ 2 & 0 & - 1 \end{matrix}| = \hat{i} + 5 \hat{j} + 2 \hat{k} \\ 6 \overline{b} = 2 \hat{i} - 8 \hat{j} \Rightarrow \overline{b} = \frac{\hat{i}}{3} - \frac{4 \hat{j}}{3} + \frac{2 \hat{k}}{3} \end{array}$

Projection $= \frac{\overline{b} \cdot (\overline{a} - \overline{b})}{| \overline{a} - \overline{b} |} = \frac{3 - \frac{21}{9}}{\sqrt{6 + \frac{21}{9} - 6}} = \frac{6 / 9}{\sqrt{\frac{21}{9}}} = \frac{2}{\sqrt{21}}$