The dot product of two unit vectors is given by: \[ \mathbf{\hat{a}} \cdot \mathbf{\hat{b}} = \cos \theta. \]
Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \]
Substitute \( \sin \theta = \frac{3}{5} \): \[ \left( \frac{3}{5} \right)^2 + \cos^2 \theta = 1. \] Simplify: \[ \frac{9}{25} + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}. \]
Thus: \[ \cos \theta = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}. \]
Final Answer: \( \mathbf{\hat{a}} \cdot \mathbf{\hat{b}} = \pm \frac{4}{5} \), (C).
The area of a parallelogram whose diagonals are given by $ \vec{u} + \vec{v} $ and $ \vec{v} + \vec{w} $, where:
$ \vec{u} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \vec{v} = -\hat{i} + \hat{k}, \quad \vec{w} = 2\hat{j} - \hat{k} $ is:
The direction ratios of the normal to the plane passing through the points
$ (1, 2, -3), \quad (1, -2, 1) \quad \text{and parallel to the line} \quad \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z}{4} \text{ is:} $
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: