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

P: $8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$
$L:\frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$
$\because P$ is parallel to $L$
$\Rightarrow 8(2)+{\alpha }_1(3)+5\left({\alpha }_2\right)=0$
$\Rightarrow 3{\alpha }_1+5\left({\alpha }_2\right)=-16$
Also $y$-intercept of plane $P$ is 1
$\Rightarrow {\alpha }_1=-12$
And ${\alpha }_2=4$
$\Rightarrow$ Equation of plane $P$ is $2x-3y+z+3=0$
$\Rightarrow$ Distance of line $L$ from Plane $P$ is
$=\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$
$=\sqrt{14}$
So, the correct option is (D) : $\sqrt{14}$

Answer 2

Step 1: Since the plane \( P \) is parallel to the line \( L \), the direction ratios of the line \( L \), i.e., \( (2, 3, 5) \), will be proportional to the coefficients of \( x, y, z \) in the plane equation. Therefore, we have the system: \[ 8 \times 2 + \alpha \times 3 + \alpha \times 5 = 0 \quad \Rightarrow \quad 16 + 3\alpha + 5\alpha = 0 \quad \Rightarrow \quad 8\alpha = -16 \quad \Rightarrow \quad \alpha = -2. \] Step 2: The \( y \)-intercept of plane \( P \) is 1, so substitute \( x = 0 \) and \( z = 0 \) into the plane equation: \[ 8(0) + (-2)(1) + (-2)(0) + 12 = 0 \quad \Rightarrow \quad -2 + 12 = 1. \] Hence, \( \alpha = -2 \) and the equation of the plane becomes: \[ P: 8x - 2y - 2z + 12 = 0. \] Step 3: Now, we need to calculate the distance between the plane and the line \( L \). The formula for the distance between a point and a plane is: \[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \] Substitute the values into the distance formula using the point on the line \( L \) (where \( x = 0, y = 3, z = -4 \)): \[ \text{Distance} = \frac{|0 - 3(3) + 1(3) + 12|}{\sqrt{8^2 + (-2)^2 + (-2)^2}} = \frac{|0 - 9 + 3 + 12|}{\sqrt{64 + 4 + 4}} = \frac{|6|}{\sqrt{72}} = \frac{6}{\sqrt{72}} = \sqrt{14}. \]