Solution:
The given lines are in symmetric form. Two lines intersect if they satisfy:
\[ \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \] \[ \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}. \]For the lines to intersect, the determinant condition must hold:
\[ \begin{vmatrix} 2 & -3 & \alpha \\ 5 & 2 & \beta \\ (4-1) & (1-3) & (1-3) \end{vmatrix} = 0. \]Expanding the determinant:
\[ 2(2 \cdot (-2) - \beta \cdot (-2)) - (-3)(5 \cdot (-2) - \beta \cdot 3) + \alpha (5 \cdot (-2) - 2 \cdot 3) = 0. \]Simplifying, we get the relation between \( \alpha \) and \( \beta \).
Using optimization, the minimum value of \( 8\alpha \beta \) is found to be:
\[ 18. \]Final Answer: \( \mathbf{18} \).
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
Let α, β(α > β) be the roots of the quadratic equation x2 – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.
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 : \(A_2 \rightleftharpoons 2A\)