Question

The lines $L_1: \frac{x-1}{3} = \frac{y}{0} = \frac{z-3}{-1}$ and $L_2: \frac{x-2}{1} = \frac{y}{2} = \frac{z-\frac{8}{3}}{-1}$ lies in plane $\Omega$. If $\Omega$ passes through (0, 3, 1), then line $x - 1 = 1 - y = 3z - 1$ cuts the plane $\Omega$ at $(\alpha, \beta, \gamma)$. The value of $2\alpha + \beta + \gamma$ is equal to

• A. $\frac{22}{3}$
• B. $\frac{13}{3}$
• C. $\frac{25}{3}$
• D. $\frac{29}{3}$

Answer 1

Answer: (A)

$\frac{22}{3}$

Answer 2

1. Plane Determination: The direction vectors of $L_1$ ($\vec{v_1} = \langle 3, 0, -1 \rangle$) and $L_2$ ($\vec{v_2} = \langle 1, 2, -1 \rangle$) are parallel to the plane $\Omega$. Their cross product $\vec{v_1} \times \vec{v_2} = \langle 2, 2, 6 \rangle$, which simplifies to $\langle 1, 1, 3 \rangle$, gives the normal vector to $\Omega$. The plane $\Omega$ passes through the point $(0, 3, 1)$. Using the point-normal form of a plane equation $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$, we get $1(x-0) + 1(y-3) + 3(z-1) = 0$, which simplifies to $x + y + 3z = 6$.

2. Line Parameterization: The third line $x - 1 = 1 - y = 3z - 1$ is parameterized. Setting $t = x-1 = 1-y = 3z-1$, we get the parametric equations: $x = 1+t$, $y = 1-t$, $z = \frac{1+t}{3}$.

3. Intersection Point: The intersection of the parameterized line with plane $\Omega$ is found by substituting the parametric equations into the plane equation: $(1+t) + (1-t) + 3(\frac{1+t}{3}) = 6$. Solving for $t$ yields $t=3$.

4. Coordinates of Intersection: Substituting $t=3$ back into the parametric equations gives the intersection point $(\alpha, \beta, \gamma) = (1+3, 1-3, \frac{1+3}{3}) = (4, -2, \frac{4}{3})$.

5. Final Calculation: The required value is $2\alpha + \beta + \gamma = 2(4) + (-2) + \frac{4}{3} = 8 - 2 + \frac{4}{3} = 6 + \frac{4}{3} = \frac{18}{3} + \frac{4}{3} = \frac{22}{3}$.