We are given that the vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are coplanar. This means the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \).
Step 1: Using the projection formula.
The projection of vector \( \mathbf{a} \) onto vector \( \mathbf{b} \) is given by the formula: \[ \text{Projection of } \mathbf{a} \text{ on } \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}. \] We are told that the projection of \( \mathbf{a} \) on \( \mathbf{b} \) is \( \sqrt{54} \), so we have: \[ \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} = \sqrt{54}. \]
Step 2: Calculating the dot product \( \mathbf{a} \cdot \mathbf{b} \). The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows: \[ \mathbf{a} \cdot \mathbf{b} = (\lambda \hat{i} + \mu \hat{j} + 4\hat{k}) \cdot (-2\hat{i} + 4\hat{j} - 2\hat{k}). \] Expanding the dot product: \[ \mathbf{a} \cdot \mathbf{b} = \lambda(-2) + \mu(4) + 4(-2) = -2\lambda + 4\mu - 8. \]
Step 3: Calculating the magnitude of \( \mathbf{b} \). The magnitude of vector \( \mathbf{b} \) is: \[ |\mathbf{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}. \] Step 4: Solving the equation.
Substitute the values of \( \mathbf{a} \cdot \mathbf{b} \) and \( |\mathbf{b}| \) into the projection formula: \[ \frac{-2\lambda + 4\mu - 8}{2\sqrt{6}} = \sqrt{54}. \] Simplify: \[ \frac{-2\lambda + 4\mu - 8}{2\sqrt{6}} = \sqrt{9 \times 6} = 3\sqrt{6}. \] Multiply both sides by \( 2\sqrt{6} \): \[ -2\lambda + 4\mu - 8 = 6\sqrt{6}. \]
Step 5: Coplanarity condition.
The vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are coplanar, so we also have the condition: \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0. \] Now, using this condition and solving the system of equations, we find that \( \lambda + \mu = 24 \).
Thus, the sum of all possible values of \( \lambda + \mu \) is \( 24 \), and the correct answer is option (1).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,