Let vec{p}=2 hat{i}+3 hat{j}+hat{k} and vec{q | vedclass

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(A) Given vectors are $\vec{p}=2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2 \hat{j}+\hat{k}$.First,calculate the sum and difference of the vect

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$3$

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(A) Given vectors are $\vec{p}=2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2 \hat{j}+\hat{k}$.First,calculate the sum and difference of the vectors:$\vec{p}+\vec{q} = (2+1)\hat{i} + (3+2)\hat{j} + (1+1)\hat{k} = 3\hat{i} + 5\hat{j} + 2\hat{k}$$\vec{p}-\vec{q} = (2-1)\hat{i} + (3-2)\hat{j} + (1-1)\hat{k} = \hat{i} + \hat{j} + 0\hat{k}$Since $\vec{r}$ is perpendicular to both $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$,$\vec{r}$ must be parallel to their cross product:$(\vec{p}+\vec{q}) 	imes (\vec{p}-\vec{q}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 5 & 2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-2) - \hat{j}(0-2) + \hat{k}(3-5) = -2\hat{i} + 2\hat{j} - 2\hat{k}$.Let $\vec{v} = -2\hat{i} + 2\hat{j} - 2\hat{k}$. The magnitude is $|\vec{v}| = \sqrt{(-2)^2 + 2^2 + (-2)^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.The vector $\vec{r}$ is given by $\vec{r} = \pm |\vec{r}| \frac{\vec{v}}{|\vec{v}|} = \pm \sqrt{3} \frac{-2\hat{i} + 2\hat{j} - 2\hat{k}}{2\sqrt{3}} = \pm (-\hat{i} + \hat{j} - \hat{k})$.Thus,$\vec{r} = \mp \hat{i} \pm \hat{j} \mp \hat{k}$.Comparing with $\vec{r} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,we get $|\alpha|=1, |\beta|=1, |\gamma|=1$.Therefore,$|\alpha|+|\beta|+|\gamma| = 1+1+1 = 3$.

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