Question

Consider Lines $L_1: \frac{x-\alpha}{1}=\frac{y}{-2}=\frac{z+\beta}{2}, L_2:x=\alpha, \frac{y}{-\alpha}=\frac{z+\alpha}{2-\beta}$, plane $P:2x+2:$ lies in plane $P$, then

• A. $\alpha = -7$
• B. $\alpha = 7$
• C. minimum distance between line $L_1$ and plane $P$ is 11.
• D. minimum distance between line $L_1$ and plane $P$ is $\frac{23}{3}$

Answer 1

Answer: (A), (B)

(A), (B)

Answer 2

Line $L_1$ lies in plane $P$ if its direction vector is perpendicular to the plane's normal vector AND a point on $L_1$ lies on $P$. Direction vector of $L_1$: $\vec{v_1} = \langle 1, -2, 2 \rangle$. Assuming the plane equation is $P: 2x+2y+z=0$, its normal vector is $\vec{n} = \langle 2, 2, 1 \rangle$. The dot product $\vec{v_1} \cdot \vec{n} = (1)(2) + (-2)(2) + (2)(1) = 2 - 4 + 2 = 0$. This means $L_1$ is parallel to the plane $P$. A point on $L_1$ is $(\alpha, 0, -\beta)$. For $L_1$ to lie in $P$, this point must satisfy the plane equation: $2(\alpha) + 2(0) + (-\beta) = 0$, which simplifies to $\beta = 2\alpha$. If a line lies in a plane, the minimum distance between the line and the plane is 0. Therefore, options (C) and (D) are incorrect as they state non-zero distances. Options (A) and (B) propose specific values for $\alpha$. If $\alpha = -7$ (Option A), then $\beta = 2(-7) = -14$, satisfying $\beta = 2\alpha$. If $\alpha = 7$ (Option B), then $\beta = 2(7) = 14$, also satisfying $\beta = 2\alpha$. Thus, these are valid scenarios where $L_1$ lies in $P$.