Given vec{A}=(2 hat{i}-3 hat{j}+hat{k}),vec{B | vedclass

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(A) First,calculate the sum of vectors $\vec{A}$ and $\vec{B}$:$\vec{A}+\vec{B} = (2 \hat{i}-3 \hat{j}+\hat{k}) + (3 \hat{i}+\hat{j}-2 \hat{k})$$\vec{

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$10$

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(A) First,calculate the sum of vectors $\vec{A}$ and $\vec{B}$:$\vec{A}+\vec{B} = (2 \hat{i}-3 \hat{j}+\hat{k}) + (3 \hat{i}+\hat{j}-2 \hat{k})$$\vec{A}+\vec{B} = (2+3) \hat{i} + (-3+1) \hat{j} + (1-2) \hat{k}$$\vec{A}+\vec{B} = 5 \hat{i} - 2 \hat{j} - \hat{k}$Now,calculate the dot product of $(\vec{A}+\vec{B})$ with $\vec{C}$:$(\vec{A}+\vec{B}) \cdot \vec{C} = (5 \hat{i} - 2 \hat{j} - \hat{k}) \cdot (3 \hat{i} + 2 \hat{j} + \hat{k})$Using the property $\hat{i} \cdot \hat{i} = 1$,$\hat{j} \cdot \hat{j} = 1$,$\hat{k} \cdot \hat{k} = 1$ and cross terms are $0$:$(\vec{A}+\vec{B}) \cdot \vec{C} = (5)(3) + (-2)(2) + (-1)(1)$$(\vec{A}+\vec{B}) \cdot \vec{C} = 15 - 4 - 1 = 10$Therefore,the correct option is $A$.

Given $\vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k})$,$\vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k})$. The value of $(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

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