Let the equations of the lines be:
For Line 1:
\[ \frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = \lambda \] Then \(x = 3\lambda - 6\), \(y = 2\lambda\), \(z = \lambda - 1\).
For Line 2:
\[ \frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = \mu \] Then \(x = 4\mu + 7\), \(y = 3\mu + 9\), \(z = 2\mu + 4\).
By equating the coordinates, we get the system of equations:
\[ \begin{aligned} 3\lambda - 6 &= 4\mu + 7 \quad (1) \\ 2\lambda &= 3\mu + 9 \quad (2) \\ \lambda - 1 &= 2\mu + 4 \quad (3) \end{aligned} \] Solving these equations, we find the values of \(\lambda\) and \(\mu\) at the point of intersection as \(\lambda = 3\) and \(\mu = -1\). Thus, the intersection point is \((3, 6, 2)\).
The distance \(d\) from the point \((7, 8, 9)\) to \((3, 6, 2)\) is:
\[ d = \sqrt{(7 - 3)^2 + (8 - 6)^2 + (9 - 2)^2} = \sqrt{16 + 4 + 49} = \sqrt{69} \] Therefore,
\[ d^2 + 6 = 69 + 6 = 75 \]
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,