Question

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

• A. $\hat{i}+3\hat{j}+\hat{k}$
• B. $3\hat{i}+7\hat{j}+3\hat{k}$
• C. $\hat{j}+t(\hat{i}+2\hat{j}+\hat{k})$
• D. $\hat{i}+(t+3)\hat{j}+\hat{k}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $3\hat{i}+7\hat{j}+3\hat{k}$
B $\hat{j}+t(\hat{i}+2\hat{j}+\hat{k})$
C $\hat{i}+3\hat{j}+\hat{k}$

Let, $\overline{r}=x\hat{i}+y\hat{j}+z\hat{k}$
But, $\overline{r}$ satisfies the equation
$\overline{r}\times (\hat{i}+2\hat{j}+\hat{k})=\hat{i}-\hat{k}$
$\therefore (x\hat{i}+y\hat{j}+z\hat{k})\times (\hat{i}+2\hat{j}+\hat{k})=\hat{i}-\hat{k}$
$\Rightarrow (y-2z)\hat{i}-(x-z)\hat{j}+(2x-y)\hat{k}=\hat{i}-\hat{k}$
On comparing both sides, we get $z=x,y=2x+1$
$\therefore \overline{r}=x\hat{i}+(2x+1)\hat{j}+x\hat{k}$
which can be written as,
$\overline{r}=\hat{j}+x(\hat{i}+2\hat{j}+\hat{k})$
Replace $x$ with $t$, ($\because x$ is a scaler).
$\therefore \overline{r}=\hat{j}+t(\hat{i}+2\hat{j}+\hat{k})$
For $t=1\Rightarrow \overline{r}=\hat{i}+3\hat{j}+\hat{k}$
and at $t=3\Rightarrow \overline{r}=\hat{i}+7\hat{j}+3\hat{k}$
Hence, option $A,B,C$.