If $\vec a =\hat i+\hat k,\vec b=\hat j-\hat k$ and $\vec c=\hat i+\hat j+\hat k$ such that $\vec r ×\vec b =\vec c×\vec b$ and $\vec r.\vec a = 0$, then $\vec r$ is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\hat i+3\hat j-\hat k$
$\vec{a} = \hat{i} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$, $\vec{c} = \hat{i} + \hat{j} + \hat{k}$
Given $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$
$\vec{r} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0} \ \Rightarrow \ (\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$
Thus $\vec{r} - \vec{c}$ is parallel to $\vec{b}$: $\vec{r} - \vec{c} = \lambda \vec{b}$
$\vec{r} = \vec{c} + \lambda \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + \lambda (\hat{j} - \hat{k})$
$\vec{r} = \hat{i} + (1+\lambda)\hat{j} + (1-\lambda)\hat{k}$
Also given $\vec{r} \cdot \vec{a} = 0$:
$(\hat{i} + (1+\lambda)\hat{j} + (1-\lambda)\hat{k}) \cdot (\hat{i} + \hat{k}) = 1 + (1-\lambda) = 0$
$2 - \lambda = 0 \ \Rightarrow \ \lambda = 2$
Substitute: $\vec{r} = \hat{i} + (1+2)\hat{j} + (1-2)\hat{k} = \hat{i} + 3\hat{j} - \hat{k}$