Let vec{a} = 6 hat{i} - 3 hat{j} - 6 hat{k} a | vedclass

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(B) Given $\vec{a} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$.Since $\vec{b}$ is parallel to $\vec{d}$,we can wr

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$7 \hat{i} - 2 \hat{j} - 5 \hat{k}$

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(B) Given $\vec{a} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$.Since $\vec{b}$ is parallel to $\vec{d}$,we can write $\vec{b} = \lambda \vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k})$.Given $\vec{a} = \vec{b} + \vec{c}$,we have $\vec{c} = \vec{a} - \vec{b} = (6 - \lambda) \hat{i} - (3 + \lambda) \hat{j} - (6 + \lambda) \hat{k}$.Since $\vec{c}$ is perpendicular to $\vec{d}$,their dot product is zero: $\vec{c} \cdot \vec{d} = 0$.$(6 - \lambda)(1) + (-3 - \lambda)(1) + (-6 - \lambda)(1) = 0$.$6 - \lambda - 3 - \lambda - 6 - \lambda = 0$.$-3 - 3\lambda = 0 \Rightarrow \lambda = -1$.Substituting $\lambda = -1$ into the expression for $\vec{c}$:$\vec{c} = (6 - (-1)) \hat{i} - (3 + (-1)) \hat{j} - (6 + (-1)) \hat{k} = 7 \hat{i} - 2 \hat{j} - 5 \hat{k}$.

Let $\vec{a} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$. 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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