Step 1: {Find the angle between \( \vec{a} \) and \( \vec{b} \)}
The dot product formula gives: \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \implies \sqrt{3} = (1)(2)\cos\theta \implies \cos\theta = \frac{\sqrt{3}}{2}. \] Thus, \( \theta = \frac{\pi}{6} \).
Step 2: {Angle between \( 2\vec{a} \) and \( -\vec{b} \)}
Since \( 2\vec{a} \) and \( -\vec{b} \) involve a scalar multiplication, the angle becomes: \[ \pi - \frac{\pi}{6} = \frac{5\pi}{6}. \]
Step 3: {Verify the options}
The correct angle is \( \frac{5\pi}{6} \), matching option (C).
Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $, $ \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} $, $ \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $ and $ \vec{d} $ be a vector such that $ \vec{b} \times \vec{d} = \vec{c} \times \vec{d} $ and $ \vec{a} \cdot \vec{d} = 4 $. Then $ |\vec{a} \times \vec{d}|^2 $ is equal to _______
Consider two vectors $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$, $\lambda>0$. The angle between them is given by $\cos^{-1} \left( \frac{\sqrt{5}}{2\sqrt{7}} \right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: