Let vec{alpha}=hat{i}+hat{j}+hat{k}, vec{beta | vedclass

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(C) Since $\vec{\delta}$ lies in the plane of $\vec{\alpha}$ and $\vec{\beta}$, we can write $\vec{\delta} = \vec{\alpha} + \lambda \vec{\beta}$ for s

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$-\hat{i}+3\hat{j}+3\hat{k}$

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(C) Since $\vec{\delta}$ lies in the plane of $\vec{\alpha}$ and $\vec{\beta}$, we can write $\vec{\delta} = \vec{\alpha} + \lambda \vec{\beta}$ for some scalar $\lambda$.Substituting the given vectors, we get $\vec{\delta} = (\hat{i}+\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}-\hat{k}) = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1-\lambda)\hat{k}$.The projection of $\vec{\delta}$ on $\vec{\gamma}$ is given by $\frac{\vec{\delta} \cdot \vec{\gamma}}{|\vec{\gamma}|} = \frac{1}{\sqrt{3}}$.Calculating the dot product: $\vec{\delta} \cdot \vec{\gamma} = (1+\lambda)(-1) + (1-\lambda)(1) + (1-\lambda)(-1) = -1 - \lambda + 1 - \lambda - 1 + \lambda = -1 - \lambda$.The magnitude $|\vec{\gamma}| = \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{3}$.Thus, $\frac{-1-\lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow -1-\lambda = 1 \Rightarrow \lambda = -2$.Substituting $\lambda = -2$ into the expression for $\vec{\delta}$, we get $\vec{\delta} = (1-2)\hat{i} + (1-(-2))\hat{j} + (1-(-2))\hat{k} = -\hat{i} + 3\hat{j} + 3\hat{k}$.

Let $\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}$, $\vec{\beta}=\hat{i}-\hat{j}-\hat{k}$ and $\vec{\gamma}=-\hat{i}+\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{\delta}$, in the plane of $\vec{\alpha}$ and $\vec{\beta}$, whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}}$, is given by

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