Step 1: The given vector equation \( \mathbf{a} \times \mathbf{c} = \mathbf{a} \times \mathbf{b} \) implies that \( \mathbf{c} \) lies in the plane formed by \( \mathbf{a} \) and \( \mathbf{b} \). We will use this condition to express \( \mathbf{c} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \).
Step 2: The dot product equation \( (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) = 168 \) provides a second condition to find \( \mathbf{c} \). Expand the dot product and solve for \( | \mathbf{c} |^2 \).
Step 3: By solving these equations, we find that the maximum value of \( | \mathbf{c} |^2 \) is 308. Thus, the correct answer is (3).
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.