Two position vectors are given by overrightar | vedclass

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(A) Given vectors are $\overrightarrow{r_1} = \hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{r_2} = \hat{i} - \hat{j} + \hat{k}$.First,we calculate

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$\frac{\hat{i}}{\sqrt{2}} - \frac{\hat{k}}{\sqrt{2}}$

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(A) Given vectors are $\overrightarrow{r_1} = \hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{r_2} = \hat{i} - \hat{j} + \hat{k}$.First,we calculate the cross product $\overrightarrow{r_1} 	imes \overrightarrow{r_2}$ using the determinant method:$\overrightarrow{r_1} 	imes \overrightarrow{r_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -1 & 1 \end{vmatrix}$$= \hat{i}(1 - (-1)) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$$= \hat{i}(2) - \hat{j}(0) + \hat{k}(-2) = 2\hat{i} - 2\hat{k}$.Now,the magnitude of this vector is $|\overrightarrow{r_1} 	imes \overrightarrow{r_2}| = \sqrt{2^2 + 0^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.The unit vector $\hat{n}$ in the direction of $\overrightarrow{r_1} 	imes \overrightarrow{r_2}$ is given by $\frac{\overrightarrow{r_1} 	imes \overrightarrow{r_2}}{|\overrightarrow{r_1} 	imes \overrightarrow{r_2}|} = \frac{2\hat{i} - 2\hat{k}}{2\sqrt{2}} = \frac{\hat{i}}{\sqrt{2}} - \frac{\hat{k}}{\sqrt{2}}$.

Two position vectors are given by $\overrightarrow{r_1} = (1, 1, 1)$ and $\overrightarrow{r_2} = (1, -1, 1)$. Find the unit vector in the direction of $\overrightarrow{r_1} \times \overrightarrow{r_2}$.

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