Let overrightarrow{a}=2 hat{i}-hat{j}+hat{k}, | vedclass

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Question

(A) Any vector $\overrightarrow{r}$ in the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$ can be written as $\overrightarrow{r} = m\overrighta

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$2 \hat{i}+3 \hat{j}-3 \hat{k}$

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(A) Any vector $\overrightarrow{r}$ in the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$ can be written as $\overrightarrow{r} = m\overrightarrow{b} + n\overrightarrow{c}$. For simplicity,consider $\overrightarrow{r} = m\overrightarrow{b} + \overrightarrow{c}$ (assuming the vector is not parallel to $\overrightarrow{c}$).$\overrightarrow{r} = m(\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} + \hat{j} - 2\hat{k}) = (m+1)\hat{i} + (2m+1)\hat{j} + (-m-2)\hat{k}$.The projection of $\overrightarrow{r}$ on $\overrightarrow{a}$ is given by $\frac{|\overrightarrow{r} \cdot \overrightarrow{a}|}{|\overrightarrow{a}|} = \sqrt{\frac{2}{3}}$.$\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$,so $|\overrightarrow{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.$\overrightarrow{r} \cdot \overrightarrow{a} = 2(m+1) - 1(2m+1) + 1(-m-2) = 2m + 2 - 2m - 1 - m - 2 = -m - 1$.Thus,$\frac{|-m-1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{2}{\sqrt{6}}$.$|-m-1| = 2$,which implies $-m-1 = 2$ or $-m-1 = -2$.If $-m-1 = 2$,then $m = -3$. Substituting $m = -3$ into $\overrightarrow{r}$: $\overrightarrow{r} = -2\hat{i} - 5\hat{j} + \hat{k}$.If $-m-1 = -2$,then $m = 1$. Substituting $m = 1$ into $\overrightarrow{r}$: $\overrightarrow{r} = 2\hat{i} + 3\hat{j} - 3\hat{k}$.Comparing with the options,$2\hat{i} + 3\hat{j} - 3\hat{k}$ is present.

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