$\bar{a}-\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{j}-\hat{k}$, then vector $\bar{r}$ satisfying $\overline{\mathrm{a}} imes \overline{\mathrm{r}}=\overline{\mathrm{b}}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$ is

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$\bar{a}-\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{j}-\hat{k}$, then vector $\bar{r}$ satisfying $\overline{\mathrm{a}} 	imes \overline{\mathrm{r}}=\overline{\mathrm{b}}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$ is

Aicharya · Accepted Answer

Given $\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$ $\overline{\mathrm{a}} 	imes \overline{\mathrm{r}}=\overline{\mathrm{b}}$ Let $\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$ $\begin{aligned} \overline{\mathrm{a}} 	imes \overline{\mathrm{r}} & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 1 & 1 & 1 \ x & y & \mathrm{z} \end{array}ight| \ & =(\mathrm{z}-y) \hat{\mathrm{i}}-\hat{\mathrm{j}}(z-x)+\hat{\mathrm{k}}(y-x) \end{aligned}$ Given $\overline{\mathrm{a}} 	imes \overline{\mathrm{r}}=\overline{\mathrm{b}}$ $	herefore \quad(z-y) \hat{\mathrm{i}}-(\mathrm{z}-x) \hat{\mathrm{j}}+(y-x) \hat{\mathrm{k}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ Comparing $\begin{aligned} & \mathrm{z}-y=0 ... (i)\ & \mathrm{z}-x=-1 ... (ii)\ & y-x=-1 ... (iii) \end{aligned}$ Also, $\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$ $\begin{aligned} & (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})(x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}})=3 \ & x+y+z=3 \end{aligned}$ Solving equations (i), (ii), (iii) and (iv), we get $\begin{aligned} & x=\frac{5}{3}, y=\frac{2}{3}, \mathrm{z}=\frac{2}{3} \ 	herefore \quad \overline{\mathrm{r}} & =\frac{5}{3} \hat{\mathrm{i}}+\frac{2}{3} \hat{\mathrm{j}}+\frac{2}{3} \hat{\mathrm{k}} \end{aligned}$

$\bar{a}-\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{j}-\hat{k}$, then vector $\bar{r}$ satisfying…

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