Troubles with finding closest point on a plane to the origin

Question

I have a trouble with the coordinates of a point on the plane that is closest to the origin

The equation of the plane is $5x-7y+z=-21$

The book I am using uses the formula $\vec{OR}=\vec{OQ}+\vec{QP}_\vec{n_u}$ where Q is the point outside the plane, P is a point belongs to the plane and $\vec{n_u}$ is the unitary normal vector of the plane.

As a result, I calculated as following:

• P(0,0,-21) and unitary normal vector $\vec{n_u}=[\frac{5}{\sqrt{75}},\frac{-7}{\sqrt{75}},\frac{1}{\sqrt{75}}]$

• $\vec{QP}$ = (0,0,-21)-(0,0,0)= (0,0,-21)

• $\vec{QP}_\vec{n_u} = \frac{|\vec{QP}\cdotp\vec{n_u}|}{1}\vec{n_u}$

• $(\frac{7}{5},\frac{-49}{25},\frac{7}{25})$

However, first of all the result I obtained does not belong to plane $\pi$, secondly, the answer key gave the answer of $(\frac{-7}{5},\frac{49}{25},\frac{-7}{25})$ which belongs to the plane as well

The only way, as far as I am concerned, is if $\vec{n} = [-5,7,-1]$, which in the case the equation would be totally different, and result in a different plane

I am not sure where I did wrong, because I have checked again that I am using the right formula so I assume I made mistakes in calculations?

Answer 1

Note, the line that passes through origin and is normal to the plane is

$$x=\frac{5t}{\sqrt{75}},\>\>\>y=\frac{-7t}{\sqrt{75}},\>\>\>z=\frac{t}{\sqrt{75}}\tag 1$$

Substitute above in the equation of the plane $5x-7y+z=-21$ to obtain $t=-\frac{21}{\sqrt{75}}$. Plug it into (1) to obtain the point on the plane,

$$(-\frac{7}{5},\frac{49}{25},-\frac{7}{25})$$

Answer 2

The line connecting a point off the plane to the nearest point on the plane will be perpendicular to the plane. So the direction vector of that line will be parallel to the normal vector of the plane.

$$ \vec{v}=\lambda\mathbf{N} $$

So the equation of a line through the point $(x_0,y_0,z_0)$ off the plane and perpendicular to the plane $ax+by+cz+d=0$ will be

$$ (x,y,z)=(x_0,y_0,z_0)+ \lambda(a,b,c)t$$

Since $(0,0,0)$ is not on the given plane, we know that the line containing it and the near point $(x_1,y_1,z_1)$ on the plane satisfies

\begin{eqnarray} (x_1,y_1,z_1)&=&(\lambda at,\lambda bt,\lambda ct)\\ &=&(5\lambda t,-7\lambda t,\lambda t) \end{eqnarray}

and that

\begin{eqnarray} 5x_1-7y_1+z_1&=&-21\\ 5(5\lambda t)-7(-7\lambda t)+\lambda t&=&-21\\ 75\lambda t&=&-21\\ \lambda t&=&-\frac{7}{25} \end{eqnarray}

So

$$ (x_1,y_1,z_1)=(-\frac{7}{5},-\frac{49}{25},\frac{7}{25} )$$

Notice that the value of $\lambda$ is inconsequential. You should get the same answer regardless of the value of $\lambda$.