Question

Let $\overrightarrow{a}=2\hat{i}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{c}=4\hat{i}-3\hat{j}+7\hat{k}$. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r}x\overrightarrow{b}=\overrightarrow{c}x\overrightarrow{b}$ and $\overrightarrow{r}.\overrightarrow{a}=0,$ then $\overrightarrow{r}$ is:

• A. $-\hat{i}-8\hat{j}+2\hat{k}$
• B. $\hat{i}-8\hat{j}+2\hat{k}$
• C. $\hat{i}+8\hat{j}+2\hat{k}$
• D. None of these

Answer 1

The correct option is A

$-\hat{i}-8\hat{j}+2\hat{k}$

Explanation for the correct option:

Step 1: Equating $\overrightarrow{r}x\overrightarrow{b}=\overrightarrow{c}x\overrightarrow{b}$

Let,$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ .

$$\begin{gathered} \overrightarrow{r}.\overrightarrow{a}=0\mathbf{}\mathbf{\left[}\mathbf{\text{Given}}\mathbf{\right]} \\ \Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(2\hat{i}+\hat{k}\right)=0\mathbf{}\mathbf{[}\mathbf{\because }\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{2}\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{k}}\mathbf{]} \\ \Rightarrow 2x+z=0...\left(i\right) \end{gathered}$$

Given,

$\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{c}=4\hat{i}-3\hat{j}+7\hat{k}$

$\overrightarrow{r}x\overrightarrow{b}=\overrightarrow{c}x\overrightarrow{b}$

$$\begin{gathered} \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ 1& 1& 1\end{array}\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -3& 7\\ 1& 1& 1\end{array}\right| \\ \Rightarrow \hat{i}\left(y-z\right)-\hat{j}\left(x-z\right)+\hat{k}\left(x-y\right)=\hat{i}\left(-3-7\right)-\hat{j}\left(4-7\right)+\hat{k}\left(4-\left(-3\right)\right) \\ \Rightarrow \hat{i}\left(y-z\right)-\hat{j}\left(x-z\right)+\hat{k}\left(x-y\right)=\hat{i}\left(-10\right)-\hat{j}\left(-3\right)+\hat{k}\left(7\right) \\ \Rightarrow \hat{i}\left(y-z\right)-\hat{j}\left(x-z\right)+\hat{k}\left(x-y\right)=-10\hat{i}+3\hat{j}+7\hat{k} \end{gathered}$$

From the above we get

$$\begin{gathered} y-z=-10...\left(ii\right) \\ z-x=3...\left(iii\right) \\ x-y=7...\left(iv\right) \end{gathered}$$

Step 2: Finding the value of $\overrightarrow{r}$:

Solving equations $\left(i\right)$ and $\left(ii\right)$ we get,

$2x+y=-10...\left(v\right)$

Solving equations $\left(iv\right)$ and $\left(v\right)$ we get,

$$\begin{gathered} 3x=-3 \\ \Rightarrow x=-1 \end{gathered}$$

Substituting $x=-1$ in equation $\left(iii\right)$ we get

$$\begin{gathered} \Rightarrow z-(-1)=3 \\ \Rightarrow z=2 \end{gathered}$$

Substituting $x=-1$ in equation $\left(ii\right)$ we get

$\Rightarrow y=-8$

Therefore,$\overrightarrow{r}=-1\hat{i}-8\hat{j}+2\hat{k}$

Hence, option (A) is the correct answer.