Question

6.
Find the perpendicular distance of the plane *. (2î – Î + 2k) = 12.
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Answer 1

We're asked to find the perpendicular distance of the plane $2x-y+2z=12$ from the origin.

From the equation of the plane it is clear that the vector $\left<2,\ -1,\ 2\right>$ is normal to the plane.

Let (x, y, z) be a point on the plane, then it forms a vector $\left<x,\ y,\ z\right>$ with origin, which is assumed to be perpendicular to the plane.

That means the vectors $\left<x,\ y,\ z\right>$ and $\left<2,\ -1,\ 2\right>$ are parallel vectors. Thus,

$\longrightarrow \left<x,\ y,\ z\right>=t\left<2,\ -1,\ 2\right>,\quad t\in\mathbb{R}-\{0\}$

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

So the perpendicular distance of the plane from the origin is,

$\longrightarrow d=\sqrt{x^2+y^2+z^2}$

$\longrightarrow d=\sqrt{(2t)^2+(-t)^2+(2t)^2}$

$\longrightarrow d=\sqrt{4t^2+t^2+4t^2}$

$\longrightarrow d=\sqrt{9t^2}$

$\longrightarrow d=3t\quad\quad\dots(1)$

This implies,

• $x=2t$

• $y=-t$

• $z=2t$

Putting these values in the plane equation,

$\longrightarrow2(2t)-(-t)+2(2t)=12$

$\longrightarrow4t+t+4t=12$

$\longrightarrow9t=12$

$\longrightarrow t=\dfrac{4}{3}$

Then (1) becomes,

$\longrightarrow\underline{\underline{d=4}}$

Hence the perpendicular distance is 4 units.