A vector \(\vec{r}\) satisfies \(\vec{r}\cdot\hat{\imath} = 2\), \(\vec{r}\cdot\hat{\jmath} = -1\) and \(\vec{r}\) is perpendicular to \(\hat{\imath}+\hat{\jmath}+\hat{k}\). Then \(\vec{r}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\hat{\imath}-\hat{\jmath}-\hat{k}\)
Let \(\vec{r} = x\hat{\imath}+y\hat{\jmath}+z\hat{k}\). The scalar product with a unit vector along an axis picks out that component:
\(\vec{r}\cdot\hat{\imath} = x = 2\)
\(\vec{r}\cdot\hat{\jmath} = y = -1\)
Since \(\vec{r}\) is perpendicular to \(\hat{\imath}+\hat{\jmath}+\hat{k}\):
\(x+y+z = 0\)
\(2-1+z = 0\)
\(z = -1\)
So \(\vec{r} = 2\hat{\imath}-\hat{\jmath}-\hat{k}\).