Question

What are the equations of the planes that are parallel to the plane $x+2y-2z=1$ and two units away from it?

Answer 1

In this problem, we have to find the equations of the planes that are parallel to the plane $x+2y-2z=1$ and two units away from it. We can first find the normal vector from the given equation from which we can find the equation of the plane parallel to the original one passing through the point $P\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$, we will get an equation with an unknown value d, we can then find the value of d using the distance formula to complete the required equation.

Complete step by step answer:
Here we have to find the equations of the planes that are parallel to the plane $x+2y-2z=1$ and two units away from it.
We know that the given equation is,
$x+2y-2z=1$……. (1)
We can now find the normal vector for the equation $x+2y-2z=1$, we get
$\Rightarrow \overset{\to }{\mathop{n}}\,=<1,2,-2>$ ……. (2)
We can now find the equation of the plane parallel to the original one passing through the point $P\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$.
$\Rightarrow \overset{\to }{\mathop{n}}\,\times < x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}} >$
$\Rightarrow <1,2,-2>\times < x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}} >$
We can now simplify the above step, we get
$\Rightarrow x+2y-2z-{{x}_{0}}-2{{y}_{0}}+2{{z}_{0}}=0$
We can now write the above step as,
$\Rightarrow x+2y-2z+d=0$……. (3)
Where a = 1, b = 2, c = -2.
We can now choose a point from the original plane (1),
We can now take x = 0 and y = 0, from (1) we get

& \Rightarrow 0+-2z=1 \\\ & \Rightarrow z=-\dfrac{1}{2} \\\ \end{aligned}$$ The point $${{P}_{1}}\left( 0,0,-\dfrac{1}{2} \right)$$….. (4) We can now find the value of d. We are given that the distance between the point and the plane is 2 units. We know that the formula for the distance between the plane and the point is, $$D=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$$ We can now substitute the point (4), the value of a, b , c, the normal vector points (2) and the distance value, we get $$\Rightarrow 2=\dfrac{\left| \left( 1 \right)\left( 0 \right)+\left( 2 \right)\left( 0 \right)+\left( -2 \right)\left( \dfrac{-1}{2} \right)+d \right|}{\sqrt{1+4+4}}$$ We can now simplify the above step, $$\begin{aligned} & \Rightarrow 2\times 3=\left| 1+d \right| \\\ & \Rightarrow d=6-1=5 \\\ \end{aligned}$$ We can now substitute the value of d in (3), we get $$\Rightarrow x+2y-2z+5=0$$ We will have another solution, $$\begin{aligned} & \Rightarrow \left| 1+d \right|=-6 \\\ & \Rightarrow d=-7 \\\ \end{aligned}$$ We can now substitute the value of d in (3), we get $$\Rightarrow x+2y-2z-7=0$$ Therefore, the required equations are $$x+2y-2z-7=0$$ and $$x+2y-2z+5=0$$. **Note:** We should always remember the formula for the distance between the plane and the point is,$$D=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$$. We should also remember that we can find the normal vector from the given equation from which we can find the equation of the plane parallel to the original one passing through the point $$P\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$$.