If \(\vec{a}=\hat{\imath}+\hat{k},\ \vec{b}=\hat{\jmath}-\hat{k}\) and \(\vec{c}=\hat{\imath}+\hat{\jmath}+\hat{k}\) such that \(\vec{r}\times\vec{b}=\vec{c}\times\vec{b}\) and \(\vec{r}\cdot\vec{a}=0\), then \(\vec{r}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\hat{\imath}+3\hat{\jmath}-\hat{k}\)
From \(\vec{r}\times\vec{b}=\vec{c}\times\vec{b}\):
\((\vec{r} - \vec{c})\times\vec{b} = \vec{0}\)
A zero cross product means the vectors are parallel, so \(\vec{r} - \vec{c} = \lambda\vec{b}\) for some scalar \(\lambda\):
\(\vec{r} = (\hat{\imath}+\hat{\jmath}+\hat{k}) + \lambda(\hat{\jmath}-\hat{k})\)
\(\vec{r} = \hat{\imath} + (1+\lambda)\hat{\jmath} + (1-\lambda)\hat{k}\)
Since \(\vec{r}\cdot\vec{a} = 0\) with \(\vec{a} = \hat{\imath} + \hat{k}\):
\(1 + (1 - \lambda) = 0\)
\(\lambda = 2\)
\(\vec{r} = \hat{\imath} + 3\hat{\jmath} - \hat{k}\)
Shortcut: Only one option satisfies \(\vec{r}\cdot\vec{a} = 0\), i.e. (coefficient of \(\hat{\imath}\)) + (coefficient of \(\hat{k}\)) = 0:
Option (1): \(1 + (-1) = 0\)
Option (2): \(1 + 1 = 2\)
Option (3): \(3 + (-1) = 2\)
Option (4): \(1 + 3 = 4\)