Question

Find the perpendicular distance of p(1,1,1) from the plane 2x+2y-z=6​

Answer 1

Let the point (x, y, z) be the foot of the perpendicular to the plane drawn from (1, 1, 1). So the vector $\left<x-1,\ y-1,\ z-1\right>$ is perpendicular to the plane.

The equation of the plane is,

$\longrightarrow 2x+2y-z=6$

$\longrightarrow\left<x,\ y,\ z\right>\cdot\left<2,\ 2,\ -1\right>=6$

from which it is clear that the vector $\left<2,\ 2,\ -1\right>$ is perpendicular to the plane.

[Recall that the equation of the plane is $\vec{r}\cdot\vec{n}=d$ where $\vec{r}=\left<x,\ y,\ z\right>$ and $\vec{n}$ is a vector normal to the plane.]

Now the vectors $\left<x-1,\ y-1,\ z-1\right>$ and $\left<2,\ 2,\ -1\right>$ are parallel to each other, since they're perpendicular to the plane. So we can write,

$\longrightarrow\left<x-1,\ y-1,\ z-1\right>=k\left<2,\ 2,\ -1\right>$

where $k$ is a real number. Then we get,

• $x=2k+1$

• $y=2k+1$

• $z=1-k$

Since the point (x, y, z) is on the plane, considering the equation of the plane,

$\longrightarrow 2x+2y-z=6$

$\longrightarrow 2(2k+1)+2(2k+1)-(1-k)=6$

$\longrightarrow 9k+3=6$

$\longrightarrow k=\dfrac{1}{3}$

Then,

• $x=\dfrac{5}{3}$

• $y=\dfrac{5}{3}$

• $z=\dfrac{2}{3}$

Now, the perpendicular distance is,

$\longrightarrow d=\sqrt{\left(1-\dfrac{5}{3}\right)^2+\left(1-\dfrac{5}{3}\right)^2+\left(1-\dfrac{2}{3}\right)^2}$

$\longrightarrow d=\sqrt{\dfrac{4}{9}+\dfrac{4}{9}+\dfrac{1}{9}}$

$\longrightarrow\underline{\underline{d=1}}$