Let vec{a} = hat{i} + hat{j} + hat{k},vec{b} | vedclass

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(C) vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$ is given by $\vec{v} = m\vec{a} + n\vec{b}$.Substituting the given vectors: $\vec{v} = m(

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

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(C) vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$ is given by $\vec{v} = m\vec{a} + n\vec{b}$.Substituting the given vectors: $\vec{v} = m(\hat{i} + \hat{j} + \hat{k}) + n(\hat{i} - \hat{j} + \hat{k}) = (m+n)\hat{i} + (m-n)\hat{j} + (m+n)\hat{k}$.The projection of $\vec{v}$ on $\vec{c}$ is $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.Calculating the dot product: $\vec{v} \cdot \vec{c} = (m+n)(1) + (m-n)(-1) + (m+n)(-1) = m+n - m+n - m-n = n-m$.The magnitude $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.Thus,$\frac{n-m}{\sqrt{3}} = \frac{1}{\sqrt{3}} \implies n-m = 1$,or $n = m+1$.Substituting $n = m+1$ into $\vec{v}$: $\vec{v} = (2m+1)\hat{i} - \hat{j} + (2m+1)\hat{k}$.For $m=0$,$\vec{v} = \hat{i} - \hat{j} + \hat{k}$. For $m=1$,$\vec{v} = 3\hat{i} - \hat{j} + 3\hat{k}$.Comparing with the options,option $C$ is $3\hat{i} - \hat{j} + 3\hat{k}$.

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three vectors. $A$ vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $1/\sqrt{3}$,is given by:

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