Question

The x-intercept of a plane $\pi$ passing through the point $(1, 1, 1)$ is $\dfrac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\dfrac{5}{7}$. If the y-intercept of the plane $\pi$ is negative and the z-intercept is positive, then its y-intercept is

• A. $-\dfrac{5}{3}$
• B. $-\dfrac{5}{6}$
• C. $-\dfrac{3}{2}$
• D. $-\dfrac{5}{2}$

Hint:

Use intercept form and plug the given point into the plane equation to relate intercepts.

Answer 1

The Correct Option is A

Let plane equation in intercept form be: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$
Given $a = \dfrac{5}{2}$ and plane passes through $(1,1,1)$, so:
$\dfrac{1}{5/2} + \dfrac{1}{b} + \dfrac{1}{c} = 1 \Rightarrow \dfrac{2}{5} + \dfrac{1}{b} + \dfrac{1}{c} = 1$
$\Rightarrow \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{3}{5}$
Also, perpendicular from origin to plane = $\dfrac{|D|}{\sqrt{A^2 + B^2 + C^2}} = \dfrac{5}{7}$
From geometry and sign constraints, solving gives $b = -\dfrac{5}{3}$