Question

Find the equation of the plane passing through the line of intersection of the planes
$\overrightarrow{r}.(\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{j}}\,)-6=0$ and $\overrightarrow{r}.(3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-4\overset{\wedge }{\mathop{k}}\,)=0$whose perpendicular distance from the origin is unity.

Answer 1

To solve the above question, we use the formulas of the equation of a plane passing through the intersection of the other two planes and the distance of a plane from origin. We will use ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ this formula to solve this question compare these two with the given question and then solve it.

Complete step-by-step solution:
Given planes are $\overrightarrow{r}.(\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{j}}\,)-6=0$ and $\overrightarrow{r}.(3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-4\overset{\wedge }{\mathop{k}}\,)=0$;

We have to find the equation of the plane passing through the line of intersection of these two planes.
We know that: equation of plane passing through the line of intersection the planes
${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is
${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda ({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}})=0$
Here given equations of the planes are $\overrightarrow{r}.(\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{j}}\,)-6=0$ and $\overrightarrow{r}.(3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-4\overset{\wedge }{\mathop{k}}\,)=0$ where $r=ax+by+cz+d$
Thus, we can rewrite the planes as;
$x+3y-6=0$ and $3x-y-4k=0$
Hence the required equation is: $x+3y-6+\lambda (3x-y-4z)=0$
By simplifying we get;
$x(3\lambda +1)+y(3-\lambda )+z(-4\lambda )-6=0.......(1)$
And here we have perpendicular distance of this plane from origin is given as 1;
As we know perpendicular distance of the plane $ax+by+cz+d=0$ from origin is $\dfrac{\left| d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
Where,

& a=3\lambda +1 \\\ & b=3-\lambda \\\ & c=-4\lambda \\\ & d=-6 \\\ \end{aligned}$$ On substituting we get; $$\begin{aligned} & \Rightarrow \left| \dfrac{-6}{\sqrt{{{(3\lambda +1)}^{2}}+{{(3-\lambda )}^{2}}+{{(-4\lambda )}^{2}}}} \right|=\left| \dfrac{6}{\sqrt{9{{\lambda }^{2}}+6\lambda +1+9+{{\lambda }^{2}}-6\lambda +16{{\lambda }^{2}}}} \right| \\\ & \\\ & which\,\,is\,\,given\,\,as\,\,unity\,\, \\\ & \\\ & \Rightarrow \left| \dfrac{6}{\sqrt{9{{\lambda }^{2}}+6\lambda +1+9+{{\lambda }^{2}}-6\lambda +16{{\lambda }^{2}}}} \right|=1 \\\ & \\\ \end{aligned}$$ $$\Rightarrow 6=\sqrt{9{{\lambda }^{2}}+6\lambda +1+9+{{\lambda }^{2}}-6\lambda +16{{\lambda }^{2}}}$$ On squaring on both sides of the above equation and simplifying we get; $$36=26{{\lambda }^{2}}+10$$ $$\Rightarrow {{\lambda }^{2}}=1$$ $$\therefore \lambda =\pm 1$$ $$Now\,\,substitute\,\,the\,\,value\,\,\lambda \,\,in\,\,(1)\,\,we\,\,get:$$ $$If\,\,\lambda =1$$ $$x(3+1)+y(3-1)+z(-4)-6=0$$ $$\begin{aligned} & \Rightarrow 4x+2y-4z-6=0 \\\ & \Rightarrow 2x+y-2z-3=0 \\\ \end{aligned}$$ $$\begin{aligned} & If\,\,\lambda =-1\,\, \\\ & x(-2)+y(4)+4z-6=0 \\\ & \Rightarrow -x+2y+2z-3=0 \\\ \end{aligned}$$ Hence the equations of required planes are $$2x+y-2z-3=0\,\,\,\,\,and\,\,\,\,-x+2y+2z-3=0$$ Hence the equations of the plane passing through line of intersection of the planes $$\overrightarrow{r}.(\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{j}}\,)-6=0$$ and $$\overrightarrow{r}.(3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-4\overset{\wedge }{\mathop{k}}\,)=0$$ whose perpendicular distance from the origin is unity are $$2x+y-2z-3=0\,\,\,\,\,and\,\,\,\,-x+2y+2z-3=0$$. **Note:** In this context of planes, we have to be familiar with the general form of lines and planes in both doing theoretically and thinking graphically. If we are good at the general forms we can compare the given problem and can solve it easily by comparing. And inn this type of question we need to solve it by taking the constant term which helps us to reduce the complexity of the solution. In this question, we have taken the $\lambda $ as the constant