Question

3.
Show that the distance of any point (x, y, z) on the plane 2x+3y-2 =12, from the origin is
given by ! = )x2 + y2 + (2x+3y - 12). Hence find a point on the plane that is nearest to the
origin
nuors​

Answer 1

The given equation of the plane is,

$\longrightarrow 2x+3y-z=12$

(from which it is clear that the vector $\left<2,\ 3,\ -1\right>$ is perpendicular to the plane)

$\longrightarrow z=2x+3y-12$

Then the distance of the point (x, y, z) on the plane from the origin is,

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

$\longrightarrow\underline{\underline{d=\sqrt{x^2+y^2+(2x+3y-12)^2}}}$

The position vector of the point on the plane, that is the nearest to the origin, should be perpendicular to the plane, because the distance of that point from the origin is nothing but the perpendicular distance of the plane.

Let (x, y, z) be the nearest point, then its position vector is $\left<x,\ y,\ z\right>.$

Now this vector is perpendicular to the plane, also the vector $\left<2,\ 3,\ -1\right>$ is perpendicular to the plane as said earlier, then these two vectors should have same direction ratio.

[Two vectors perpendicular to a plane have same direction ratio.]

Thus we can say that,

$\longrightarrow\left<x,\ y,\ z\right>=\lambda\left<2,\ 3,\ -1\right>$

Then,

• $x=2\lambda$

• $y=3\lambda$

• $z=-\lambda$

Putting these values in equation of the plane,

$\longrightarrow 2(2\lambda)+3(3\lambda)-(-\lambda)=12$

$\longrightarrow14\lambda=12$

$\longrightarrow\lambda=\dfrac{6}{7}$

Then,

• $x=\dfrac{12}{7}$

• $y=\dfrac{18}{7}$

• $z=-\dfrac{6}{7}$

Hence the nearest point is $\bf{\left(\dfrac{12}{7},\ \dfrac{18}{7},\ -\dfrac{6}{7}\right).}$