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 point 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 $(\alpha ,\beta ,\gamma )$ from ${{P}_{3}}$ is 2, then which of the following are true
$\begin{aligned} & \text{a)}\text{ 2}\alpha +\beta +2\gamma +2=0 \\\ & \text{b) 2}\alpha -\beta +2\gamma +4=0 \\\ & \text{c) 2}\alpha +\beta -2\gamma -10=0 \\\ & \text{d) 2}\alpha -\beta +2\gamma -8=0 \\\ \end{aligned}$

Answer 1

Now we are given equation of two planes ${{P}_{1}},{{P}_{2}}$ and we know that the plane ${{P}_{3}}$ passes through intersection of ${{P}_{1}}$ and ${{P}_{2}}$. Now we know equation of plane passing through intersection of planes ${{P}_{1}}=0\text{ and }{{\text{P}}_{2}}=0$ is ${{P}_{2}}+\lambda {{P}_{1}}=0$ . Hence we find the equation of ${{P}_{3}}$ in terms of λ. Now we will use the Condition that the plane is at a distance 1 from point $(0,1,0)$to find the value of λ. Now we know that the distance between point $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and the plane $ax+by+cz+d=0$ is given by the formula $\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$. Hence using the condition we will find the value of λ and Now we have the equation of plane ${{P}_{3}}$. Now we will use our next condition which is the distance of a point is $(\alpha ,\beta ,\gamma )$ from ${{P}_{3}}$ is 2. Hence again using the formula of distance of point from plane we can find the conditions equation in α, β, γ.

Complete step by step answer:
Now we are given equations of planes ${{P}_{1}}$ and ${{P}_{2}}$. ${{P}_{1}}:y=0$ and ${{P}_{2}}:x+z=1$
Hence we have ${{P}_{1}}:y=0$ and ${{P}_{2}}:x+z-1=0$ . \
Now we know that equation of plane passing through intersection of planes ${{P}_{1}}=0\text{ and }{{\text{P}}_{2}}=0$ is ${{P}_{2}}+\lambda {{P}_{1}}=0$ and we are given that the plane ${{P}_{3}}$ passes through the intersection of ${{P}_{1}}=0\text{ and }{{\text{P}}_{2}}=0$ .
Hence we have equation of ${{P}_{3}}$
${{P}_{3}}:x+z-1+\lambda y=0................(1)$
It is given that the distance of the point $(0,1,0)$ from this plane is 1.
Now we know that the distance between point $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and the plane $ax+by+cz+d=0$ is given by the formula $\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$.
Hence applying this we get

& \left| \dfrac{1(0)+1(0)-1+\lambda (1)}{\sqrt{{{1}^{2}}+{{\lambda }^{2}}+{{1}^{2}}}} \right|=1 \\\ & \Rightarrow \left| \dfrac{\lambda -1}{\sqrt{{{\lambda }^{2}}+2}} \right|=1 \\\ \end{aligned}$$ Now we will solve further by taking square on both sides. $$\begin{aligned} & \dfrac{{{(\lambda -1)}^{2}}}{{{\left( \sqrt{{{\lambda }^{2}}+2} \right)}^{2}}}=1 \\\ & \Rightarrow \dfrac{{{\lambda }^{2}}+1-2\lambda }{{{\lambda }^{2}}+2}=1 \\\ & \Rightarrow {{\lambda }^{2}}+1-2\lambda ={{\lambda }^{2}}+2 \\\ & \Rightarrow 1-2\lambda =2 \\\ & \Rightarrow -2\lambda =2-1 \\\ & \Rightarrow \lambda =-\dfrac{1}{2} \\\ \end{aligned}$$ Now we have found that $\lambda =-\dfrac{1}{2}$ Substituting the value of λ in equation (1) we get the equation for plane ${{P}_{3}}$ ${{P}_{3}}:x+z-1-\dfrac{y}{2}=0$ Now multiplying the whole equation by 2 we get ${{P}_{3}}:2x+2z-2-y=0$ Now we will use the second condition given to us that is distance of a point is $(\alpha ,\beta ,\gamma )$ from ${{P}_{3}}$ is 2 Hence again using the formula distance between point $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and the plane $ax+by+cz+d=0$ is given by $$\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$$ Now using this we get $$\begin{aligned} & \left| \dfrac{2\alpha +2\gamma -2-\beta }{\sqrt{{{2}^{2}}+{{2}^{2}}+{{(-1)}^{2}}}} \right|=2 \\\ & \Rightarrow \left| \dfrac{2\alpha +2\gamma -2-\beta }{\sqrt{4+4+1}} \right|=2 \\\ & \Rightarrow \left| \dfrac{2\alpha +2\gamma -2-\beta }{3} \right|=2 \\\ & \Rightarrow |2\alpha +2\gamma -2-\beta |=6 \\\ & \Rightarrow 2\alpha +2\gamma -2-\beta =\pm 6 \\\ \end{aligned}$$ Let us first take $$2\alpha +2\gamma -2-\beta =6$$ $$\begin{aligned} & \Rightarrow 2\alpha +2\gamma -2-\beta -6=0 \\\ & \Rightarrow 2\alpha +2\gamma -\beta -8=0 \\\ \end{aligned}$$ And also $$2\alpha +2\gamma -2-\beta =-6$$ $$\begin{aligned} & \Rightarrow 2\alpha +2\gamma -2-\beta +6=0 \\\ & \Rightarrow 2\alpha +2\gamma -\beta +4=0 \\\ \end{aligned}$$ **So, the correct answer is “Option B and D”.** **Note:** The formula for distance between point $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and the plane $ax+by+cz+d=0$ is given by $$\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$$. Not confuse it with $$\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}}} \right|$$ . Also the equation of plane passing through intersection of planes ${{P}_{1}}=0\text{ and }{{\text{P}}_{2}}=0$ can be taken as ${{P}_{1}}+\lambda {{P}_{2}}=0$ .