Let vec{a}=2 hat{i}-hat{j}+hat{k} and vec{b}= | vedclass

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(D) Given: $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$.Since $\vec{a} \parallel \vec{c}$,we can write $\vec{c} = k \vec{a}$

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$-\frac{1}{3}(5 \vec{a}+3 \vec{b})$

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(D) Given: $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$.Since $\vec{a} \parallel \vec{c}$,we can write $\vec{c} = k \vec{a}$ for some scalar $k$.Given $\vec{b} = \vec{c} - \vec{d}$,we have $\vec{d} = \vec{c} - \vec{b} = k \vec{a} - \vec{b}$.Since $\vec{a} \perp \vec{d}$,their dot product is zero: $\vec{a} \cdot \vec{d} = 0$.$\vec{a} \cdot (k \vec{a} - \vec{b}) = 0 \Rightarrow k |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0$.Calculate $|\vec{a}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$.Calculate $\vec{a} \cdot \vec{b} = (2)(0) + (-1)(2) + (1)(-3) = 0 - 2 - 3 = -5$.Substituting these values: $k(6) - (-5) = 0 \Rightarrow 6k = -5 \Rightarrow k = -\frac{5}{6}$.Thus,$\vec{c} = -\frac{5}{6} \vec{a}$.Then $\vec{d} = \vec{c} - \vec{b} = -\frac{5}{6} \vec{a} - \vec{b}$.Finally,$\vec{c} + \vec{d} = (-\frac{5}{6} \vec{a}) + (-\frac{5}{6} \vec{a} - \vec{b}) = -\frac{10}{6} \vec{a} - \vec{b} = -\frac{5}{3} \vec{a} - \vec{b} = -\frac{1}{3}(5 \vec{a} + 3 \vec{b})$.

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$. If $\vec{b}=\vec{c}-\vec{d}$,$\vec{a}$ is parallel to $\vec{c}$,and $\vec{a}$ is perpendicular to $\vec{d}$,then $\vec{c}+\vec{d}=$

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