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

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(B) Any vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$ can be expressed as a linear combination: $\vec{V} = x\vec{a} + y\vec{b}$. For simpli

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

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(B) Any vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$ can be expressed as a linear combination: $\vec{V} = x\vec{a} + y\vec{b}$. For simplicity,we can write $\vec{V} = \vec{a} + \lambda\vec{b}$ (assuming the coefficient of $\vec{a}$ is non-zero).$\vec{V} = (\hat{i}+\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k}) = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1+\lambda)\hat{k}$.The projection of $\vec{V}$ on $\vec{c}$ is given by $\frac{\vec{V} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.Given $\vec{c} = \hat{i}-\hat{j}-\hat{k}$,we have $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.So,$\frac{\vec{V} \cdot \vec{c}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \vec{V} \cdot \vec{c} = 1$.$((1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1+\lambda)\hat{k}) \cdot (\hat{i}-\hat{j}-\hat{k}) = 1$.$(1+\lambda) - (1-\lambda) - (1+\lambda) = 1$.$1 + \lambda - 1 + \lambda - 1 - \lambda = 1$.$\lambda - 1 = 1 \Rightarrow \lambda = 2$.Substituting $\lambda = 2$ into the expression for $\vec{V}$:$\vec{V} = (1+2)\hat{i} + (1-2)\hat{j} + (1+2)\hat{k} = 3\hat{i} - \hat{j} + 3\hat{k}$.

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 $\frac{1}{\sqrt{3}}$,is given by:

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