Question

Let the plane $P : 8 x+\alpha_1 y+\alpha_2 z+12=0$ be parallel to the line $L : \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $P$ on the $y$-axis is 1 , then the distance between $P$ and $L$ is :

• A. $\sqrt{\frac{2}{7}}$
• B. $\frac{6}{\sqrt{14}}$
• C. $\sqrt{\frac{7}{2}}$
• D. $\sqrt{14}$

Hint:

When solving distance problems involving planes and lines, first find the equation of the plane, then use the appropriate distance formula to calculate the shortest distance between the plane and the line or point.

Answer 1

The Correct Option is D

To find the distance between the plane \( P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0 \) and the line \( L : \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5} \), we follow these steps:

1. Understand the relationship between the plane and the line. The plane \( P \) is parallel to the line \( L \). A line parallel to a plane means that the direction ratios of the line satisfy an important condition: they must be orthogonal to the normal vector of the plane.
2. Write the normal vector of the plane \( P \) as \( \langle 8, \alpha_1, \alpha_2 \rangle \).
3. The direction ratios of the line \( L \) are 2, 3, and 5. Since the line is parallel to the plane, the normal vector of the plane is orthogonal to the direction vector of the line. Thus, the dot product is zero:
4. Substitute the condition of the y-intercept of 1 for the given plane:
5. With \( \alpha_1 = -12 \), substitute into the orthogonal condition:
6. Rewrite the plane equation with the known coefficients:
7. Use the plane's point and the line's direction ratios, as well as the formula to calculate the distance \( d \) from the line to the plane, which is:
8. The point on the line when \( t = 0 \) (to find a point for line) is \((-2, 3, -4)\) as it lies on the line.
9. Plug the xi, yi, and zi into the distance formula for line (-2,3,-4) to plane equation:

Thus, the distance from the plane to the line is \(\sqrt{14}\), corresponding to the correct answer choice:

$\sqrt{14}$