First slide

Planes in 3D

Question

In $R^{3},$ consider the planes $P_{1} : y = 0$ and . $P_{2} : x + z = 1$ Let $P_{3}$ be a plane, different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point $(0, 1, 0)$ from $P_{3}$ is 1 and the distance of a point $(α, β, γ)$ from $P_{3}$ is 2, then which of the following relations is(are) true?

Difficult

A

$2 α + β + 2 γ + 2 = 0$
B

$2 α - β + 2 γ + 4 = 0$
C

$2 α + β - 2 γ - 10 = 0$
D

$2 α - β + 2 γ - 8 = 0$

Solution

Equation of the plane is $P_{3} \equiv (x + z - 1) + λ y = 0$
          Distance from $(0, 1, 0)$ to the plane is 1 $\Rightarrow$ $\frac{|0 + 0 - 1 + λ 1|}{\sqrt{1 + 1 + λ^{2}}} = 1$
    $\Rightarrow$ ${(λ - 1)}^{2} = λ^{2} + 2$
   $\Rightarrow λ^{2} + 1 - 2 λ = λ^{2} + 2$ $\Rightarrow$ $- 2 λ = 1 \Rightarrow λ = - \frac{1}{2}$
           $∴$ Equation of the plane is $x + z - 1 - \frac{y}{2} = 0$
   $\Rightarrow$ $2 x - y + 2 z - 2 = 0$
Distance from $(α, β, γ)$ is
$\Rightarrow \frac{|2 α - β + 2 γ - 2|}{\sqrt{1 + 4 + 4}} = 2 \Rightarrow 2 α - β + 2 γ = 2 \pm 6 = 8 o r - 4$

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