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Let $\vec{a}=6 \vec{i}-3 \vec{j}-6 \vec{k}$ and $\vec{d}=\vec{i}+\vec{j}+\vec{k}$. Suppose that $\vec{a}=\vec{b}+\vec{c}$ where $\vec{b}$ is parallel to $\vec{d}$ and $\vec{c}$ is perpendicular to $\vec{d}$. Then $\vec{c}$ is
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The correct answer is:
$7 \vec{i}-2 \vec{j}-5 \vec{k}$
$\vec{a}=6 \vec{i}-3 \vec{j}-6 \vec{k}$
$\vec{d}=\vec{i}+\vec{j}+\vec{k}$ $\vec{a}=\vec{b}+\vec{c}$
$\vec{b}=\lambda \vec{d} \quad \& \quad \vec{c} \cdot \vec{d}=0$ $\vec{b}=\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}$ From $(1)$ $6 \hat{i}-3 \hat{j}-6 \hat{k}=\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}+\vec{c}$ $\vec{c}=(6-\lambda) \hat{i}-(3+\lambda) \hat{j}-(6+\lambda) \hat{k}$ $\vec{c} \cdot \vec{d}=0$ $\Rightarrow \lambda=-1$ $\vec{c}=7 \hat{i}-2 \hat{j}-5 \hat{k}$
$\vec{d}=\vec{i}+\vec{j}+\vec{k}$ $\vec{a}=\vec{b}+\vec{c}$
$\vec{b}=\lambda \vec{d} \quad \& \quad \vec{c} \cdot \vec{d}=0$ $\vec{b}=\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}$ From $(1)$ $6 \hat{i}-3 \hat{j}-6 \hat{k}=\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}+\vec{c}$ $\vec{c}=(6-\lambda) \hat{i}-(3+\lambda) \hat{j}-(6+\lambda) \hat{k}$ $\vec{c} \cdot \vec{d}=0$ $\Rightarrow \lambda=-1$ $\vec{c}=7 \hat{i}-2 \hat{j}-5 \hat{k}$
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