Let a→=3i^−j^+2k^,b→=a→×(i^−2k^) and c→=b→×k^. Then the projection of c→−2j^ on a→ is :

b→=a→×(i^−3k^)=i^j^k^3−1210−2=2i^+8j^+k^c→=b→×k^=8i^−2j^c→−2j^=8i^−4j^Projection of (i^−2j^) on a→(c→−2j^)⋅a^=⟨8,−4,0⟩⋅⟨3,−1,2⟩14=2814=214

Let a→=3i^−j^+2k^,b→=a→×(i^−2k^) and c→=b→×k^. Then the projection of c→−2j^ on a→ is :

b→=a→×(i^−3k^)=i^j^k^3−1210−2=2i^+8j^+k^c→=b→×k^=8i^−2j^c→−2j^=8i^−4j^Projection of (i^−2j^) on a→(c→−2j^)⋅a^=⟨8,−4,0⟩⋅⟨3,−1,2⟩14=2814=214

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Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}, \vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k}) and \vec{c} = \vec{b} \times \hat{k} .$ Then the projection of $\vec{c} - 2 \hat{j} on \vec{a}$ is :

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$3 \sqrt{7}$

$\sqrt{14}$

$2 \sqrt{14}$

$2 \sqrt{7}$

answer is C.

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Detailed Solution

$\vec{b} = \vec{a} \times (\hat{i} - 3 \hat{k})$
$\begin{array}{l} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 2 \\ 1 & 0 & - 2 \end{array}| = 2 \hat{i} + 8 \hat{j} + \hat{k} \\ \vec{c} = \vec{b} \times \hat{k} = 8 \hat{i} - 2 \hat{j} \\ \vec{c} - 2 \hat{j} = 8 \hat{i} - 4 \hat{j} \end{array}$
Projection of $(\hat{i} - 2 \hat{j}) on \vec{a}$
$\begin{array}{l} (\vec{c} - 2 \hat{j}) \cdot \hat{a} = \frac{⟨ 8, - 4, 0 ⟩ \cdot ⟨ 3, - 1, 2 ⟩}{\sqrt{14}} \\ = \frac{28}{\sqrt{14}} = 2 \sqrt{14} \end{array}$