We are given that line \( PQ \) is the line of shortest distance between the lines \( L_1 \) and \( L_2 \), where:
The direction ratios of \( PQ \) are given as: \[ \text{Direction Ratios of } PQ = (3\mu - 2\lambda + 1, 4\mu - 3\lambda + 2, 5\mu - 4\lambda + 2) \] The condition \( PQ \perp L_2 \) (i.e., \( PQ \) is perpendicular to \( L_2 \)) implies that the dot product of the direction ratios of \( PQ \) and \( L_2 \) is zero.
The dot product of the direction ratios of \( PQ \) with those of \( L_2 \) gives the following equation: \[ (3\mu - 2\lambda + 1) \times 2 + (4\mu - 3\lambda + 2) \times 3 + (5\mu - 4\lambda + 2) \times 4 = 0 \] Simplifying the equation: \[ 2(3\mu - 2\lambda + 1) + 3(4\mu - 3\lambda + 2) + 4(5\mu - 4\lambda + 2) = 0 \] \[ 6\mu - 4\lambda + 2 + 12\mu - 9\lambda + 6 + 20\mu - 16\lambda + 8 = 0 \] \[ 38\mu - 29\lambda + 16 = 0 \quad \text{...(1)} \]
Similarly, using the condition \( PQ \perp L_2 \), we get the second equation: \[ (3\mu - 2\lambda + 1) \times 3 + (4\mu - 3\lambda + 2) \times 4 + (5\mu - 4\lambda + 2) \times 5 = 0 \] Simplifying: \[ 3(3\mu - 2\lambda + 1) + 4(4\mu - 3\lambda + 2) + 5(5\mu - 4\lambda + 2) = 0 \] \[ 9\mu - 6\lambda + 3 + 16\mu - 12\lambda + 8 + 25\mu - 20\lambda + 10 = 0 \] \[ 50\mu - 38\lambda + 21 = 0 \quad \text{...(2)} \]
Now, we solve the system of equations: \[ 38\mu - 29\lambda + 16 = 0 \quad \text{...(1)} \] \[ 50\mu - 38\lambda + 21 = 0 \quad \text{...(2)} \] Using equation (1) and (2), solve for \( \lambda \) and \( \mu \): \[ \lambda = \frac{1}{3}, \quad \mu = \frac{-1}{6} \]
Substituting \( \lambda = \frac{1}{3} \) and \( \mu = \frac{-1}{6} \) into the equations for \( P \) and \( Q \): \[ P\left( 2\left(\frac{1}{3}\right) + 1, 3\left(\frac{1}{3}\right) + 2, 4\left(\frac{1}{3}\right) + 3 \right) = \left( \frac{5}{3}, 3, \frac{13}{3} \right) \] \[ Q\left( 3\left(\frac{-1}{6}\right) + 2, 4\left(\frac{-1}{6}\right) + 4, 5\left(\frac{-1}{6}\right) + 5 \right) = \left( \frac{3}{2}, \frac{10}{3}, \frac{25}{6} \right) \]
The parametric form of the line \( PQ \) is given by: \[ \frac{x - \frac{5}{3}}{\frac{1}{6}} = \frac{y - 3}{\frac{-1}{3}} = \frac{z - \frac{13}{3}}{\frac{1}{6}} \] Simplifying this, we get: \[ \frac{x - \frac{5}{3}}{1} = \frac{y - 3}{-2} = \frac{z - \frac{13}{3}}{1} \]
Substituting the coordinates \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) into the parametric equations: \[ \frac{x - \frac{5}{3}}{1} = \frac{14/3 - 5/3}{1} = \frac{9}{3} = 3 \] \[ \frac{y - 3}{-2} = \frac{-3 - 3}{-2} = \frac{-6}{-2} = 3 \] \[ \frac{z - \frac{13}{3}}{1} = \frac{22/3 - 13/3}{1} = \frac{9}{3} = 3 \] Since all equations are equal, the point \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) lies on the line \( PQ \).
Therefore, the point \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) lies on the line \( PQ \), confirming that our calculations are correct.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).