If \(\vec{r}\) is a vector with \(|\vec{r}| = 2\), then the greatest possible value of \(\vec{r}\cdot(2\hat{\imath}-\hat{\jmath}+2\hat{k})\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
Let \(\vec{b} = 2\hat{\imath}-\hat{\jmath}+2\hat{k}\) and let \(\theta\) be the angle between \(\vec{r}\) and \(\vec{b}\).
\(|\vec{b}| = \sqrt{4+1+4} = 3\)
\(\vec{r}\cdot\vec{b} = |\vec{r}|\,|\vec{b}|\cos\theta = 6\cos\theta\)
Since \(\cos\theta \le 1\), the greatest value is 6. It is reached when \(\theta = 0\), that is, when \(\vec{r}\) points along \(\vec{b}\):
\( \displaystyle \vec{r} = \frac{2}{3}(2\hat{\imath}-\hat{\jmath}+2\hat{k}) \)
\( \displaystyle \vec{r}\cdot\vec{b} = \frac{2}{3}\times 9 = 6 \)