Let the plane P: 8x+alpha_1y+alpha_2z+12=0 be parallel to the line L:(x+2)/(2)=(y-3)/(3)=(z+4)/(5) . If the intercept of P on the y -axis is 1 , then the distance between P and L is

3D Geometry: Let the plane P: 8x+alpha_1y+alpha_2z+12=0 be parallel to the line… | JEE Main 2023

JEE Main 2023 (Online) 31st January Evening ShiftQuestion 1

Let the plane $P: 8x+\alpha_1y+\alpha_2z+12=0$ 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

3D Geometry

$\sqrt{\frac27}$$\frac{6}{\sqrt{14}}$$\sqrt{\frac72}$$\sqrt{14}$✓ Correct answer

Solution

$y$-intercept $-\frac{12}{\alpha_1}=1$ gives $\alpha_1=-12$; parallel to $L$: $16+3\alpha_1+5\alpha_2=0$ gives $\alpha_2=4$. Plane $2x-3y+z+3=0$; distance from $(-2,3,-4)$: $\frac{14}{\sqrt{14}}=\sqrt{14}$.

3D Geometry - JEE Mathematics question (JEE Main 2023 (Online) 31st January Evening Shift)

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