Question

Let a line passing through the point \((-1, 2, 3)\) intersect the lines\( L_1 : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{-2} \) at \( M(\alpha, \beta, \gamma) \) and\( L_2 : \frac{x + 2}{-3} = \frac{y - 2}{-2} = \frac{z - 1}{4} \) at \( N(a, b, c) \).Then the value of \( \frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} \) equals ____.

Answer 1

Correct Answer: 196

Write the equation of the line passing through \( (-1, 2, 3) \) with direction ratios \( (l, m, n) \):

\[ x = -1 + \lambda l, \quad y = 2 + \lambda m, \quad z = 3 + \lambda n. \]

Intersection with \( L_1 \): For intersection, equate:

\[ -1 + \lambda l = 1 + 3\mu, \quad 2 + \lambda m = 2 + 2\mu, \quad 3 + \lambda n = -1 - 2\mu. \]

Intersection with \( L_2 \): For intersection, equate:

\[ -1 + \lambda l = -2 - 3\nu, \quad 2 + \lambda m = 2 + 4\nu, \quad 3 + \lambda n = 1 - 2\nu. \]

Solve for \( \alpha, \beta, \gamma, a, b, \) and \( c \).

Calculate:

\[ \frac{(\alpha + \beta + \gamma)^2}{a + b + c} = 196. \]

Answer 2

This problem involves finding the intersection points, M and N, of a line passing through a given point P with two other lines, L₁ and L₂. Since the points P, M, and N lie on the same line, they are collinear. We will use this property to find the coordinates of M and N and then evaluate the given expression.

Concept Used:

1. Parametric Equation of a Line: Any point on a line \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda \) can be represented in terms of a parameter \( \lambda \) as \( (x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda) \).

2. Collinearity of Three Points: If three points P, M, and N are collinear, the direction ratios of the vector \( \vec{PM} \) are proportional to the direction ratios of the vector \( \vec{PN} \).

\[ \frac{x_M - x_P}{x_N - x_P} = \frac{y_M - y_P}{y_N - y_P} = \frac{z_M - z_P}{z_N - z_P} \]

Step-by-Step Solution:

Step 1: Write the parametric coordinates for points M on line L₁ and N on line L₂.

The line \( L_1 \) is given by \( \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{-2} \). Let this ratio be equal to a parameter \( \lambda \). Any point \( M(\alpha, \beta, \gamma) \) on \( L_1 \) can be written as:

\[ \alpha = 3\lambda + 1, \quad \beta = 2\lambda + 2, \quad \gamma = -2\lambda - 1 \]

The line \( L_2 \) is given by \( \frac{x + 2}{-3} = \frac{y - 2}{-2} = \frac{z - 1}{4} \). Let this ratio be equal to a parameter \( \mu \). Any point \( N(a, b, c) \) on \( L_2 \) can be written as:

\[ a = -3\mu - 2, \quad b = -2\mu + 2, \quad c = 4\mu + 1 \]

Step 2: Use the collinearity of points P, M, and N.

The given point is \( P(-1, 2, 3) \). Since P, M, and N lie on the same line, they are collinear. This means the direction ratios of the vector \( \vec{PM} \) are proportional to the direction ratios of \( \vec{PN} \).

Direction ratios of \( \vec{PM} \) are \( (\alpha - (-1), \beta - 2, \gamma - 3) \):

\[ ( (3\lambda + 1) + 1, (2\lambda + 2) - 2, (-2\lambda - 1) - 3 ) = (3\lambda + 2, 2\lambda, -2\lambda - 4) \]

Direction ratios of \( \vec{PN} \) are \( (a - (-1), b - 2, c - 3) \):

\[ ( (-3\mu - 2) + 1, (-2\mu + 2) - 2, (4\mu + 1) - 3 ) = (-3\mu - 1, -2\mu, 4\mu - 2) \]

Step 3: Set up and solve the equations from the proportionality of direction ratios.

\[ \frac{3\lambda + 2}{-3\mu - 1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda - 4}{4\mu - 2} \]

From the second part of the equation, we get:

\[ \frac{2\lambda}{-2\mu} = \frac{\lambda}{-\mu} \]

Equating the first two parts:

\[ \frac{3\lambda + 2}{-3\mu - 1} = \frac{\lambda}{-\mu} \implies -\mu(3\lambda + 2) = \lambda(-3\mu - 1) \] \[ -3\lambda\mu - 2\mu = -3\lambda\mu - \lambda \implies -2\mu = -\lambda \implies \lambda = 2\mu \]

Now, equate the second and third parts and substitute \( \lambda = 2\mu \):

\[ \frac{\lambda}{-\mu} = \frac{-2\lambda - 4}{4\mu - 2} \implies \frac{2\mu}{-\mu} = \frac{-2(2\mu) - 4}{4\mu - 2} \] \[ -2 = \frac{-4\mu - 4}{4\mu - 2} \implies -2(4\mu - 2) = -4\mu - 4 \] \[ -8\mu + 4 = -4\mu - 4 \implies 8 = 4\mu \implies \mu = 2 \]

Now find \( \lambda \):

\[ \lambda = 2\mu = 2(2) = 4 \]

Step 4: Determine the coordinates of M and N.

For point \( M(\alpha, \beta, \gamma) \), substitute \( \lambda = 4 \):

\[ \alpha = 3(4) + 1 = 13 \] \[ \beta = 2(4) + 2 = 10 \] \[ \gamma = -2(4) - 1 = -9 \]

For point \( N(a, b, c) \), substitute \( \mu = 2 \):

\[ a = -3(2) - 2 = -8 \] \[ b = -2(2) + 2 = -2 \] \[ c = 4(2) + 1 = 9 \]

Final Computation & Result:

Calculate the values of \( \alpha + \beta + \gamma \) and \( a + b + c \).

\[ \alpha + \beta + \gamma = 13 + 10 + (-9) = 14 \] \[ a + b + c = -8 + (-2) + 9 = -1 \]

Finally, compute the required ratio:

\[ \frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} = \frac{(14)^2}{(-1)^2} = \frac{196}{1} \]

The value of the expression is 196.