Question

If a vector $\overline{r}$ satisfies the equation $\overline{r}\times \left(\overline{i}+2\overline{j}+\overline{k}\right)=\overline{i}-\overline{k},$ then $\overline{r}$ is equal to

• A. $\overline{i}+3\overline{j}+\overline{k}$
• B. $3\overline{i}-7\overline{j}-3\overline{k}$
• C. $\overline{k}+t\left(\overline{i}+2\overline{j}+\overline{k}\right)$ where $t$ is any scalar
• D. $2\overline{i}+\left(t+3\right)\overline{j}-5\overline{k}$ where $t$ is any scalar

Answer 1

The correct option is A $\overline{i}+3\overline{j}+\overline{k}$

$\begin{array}{c}\overrightarrow{C}=\overrightarrow{C}.\hat{i}=\overrightarrow{C}.\hat{j}=\overrightarrow{C}.\hat{k}and\left|\overrightarrow{C}\right|=100\\ Let\overrightarrow{C}=x\hat{i}+y\hat{j}+z\hat{k}\\ x^2+y^2+z^2=100---\left(i\right)\\ Now,\\ \overrightarrow{C}\widehat{.i}=\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\hat{i}=x\\ \overrightarrow{C}.\hat{j}=y\\ \overrightarrow{C}.\hat{k}=z\\ \therefore x=y=z\\ Puttingthesevalueineq{u}^n\left(i\right)weget\\ 3x^2=100\\ x^2=\frac{100}{3}\\ \therefore x=\pm \frac{10}{\sqrt{3}}\\ Requiredvector=\pm \frac{10}{\sqrt{3}}\left(\hat{i}+\hat{j}+\hat{k}\right)\\ \overrightarrow{r}\times \left(\hat{i}+2\hat{j}+\hat{k}\right)=\hat{i}+\hat{k}\\ Let\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ \left(x\hat{i}+y\hat{j}+z\hat{k}\right)\left(\hat{i}+\hat{j}+\hat{k}\right)=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ 1& 2& 1\end{array}\right]\\ \hat{i}\left(y-2z\right)-\hat{j}\left(x-z\right)+\hat{k}\left(x-y\right)=\hat{i}-\hat{j}\\ Oncomparing,weget\\ \overrightarrow{r}=\hat{i}+3\hat{j}+\hat{k}\end{array}$

Hence,option $A$ is the correct answer.