Distance from a point to a plane $\pi$

Question

Let $x=(1,2,3)$, $u=(-2,3,1)$. let $\pi$ the plane through the origin which is orthogonal to the vector $u$.

Compute the distance from $x$ to $\pi$.

The formula is: $d=\dfrac {\left| Ax_{1}+Bx_{2}+Cx_{3}+D\right| }{\sqrt {A^{2}+B^{2}+C^{2}}}$ .

Initially, I need to find a equation from point of $x$ to point of $u$, right?

How can I compute, can you help?

Answer 1

Given a normal vector $u=(a, b, c)$ for the plane and a point $O(x_0, y_0, z_0)$ on the plane, then the plane equation is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Your $u=(-2, 3, 1)$ and $O=(0, 0, 0)$ so the plane equation is $-2x+3y+z=0$. Now get the coefficients $A, B, C, D$ from here and plug $x$ in your distance formula to find the answer.

Answer 2

In the formula $d=\dfrac {\left| Ax_{1}+Bx_{2}+Cx_{3}+D\right| }{\sqrt {A^{2}+B^{2}+C^{2}}}$ are $A,B,C$ the coordinates of $u$ since this vector is orthogonal to $\pi,$ and $D=0$ because $\pi$ passes through origin. Thus $$d=\dfrac {\left| 1\times(-2)+2\times3+3\times1\right| }{\sqrt {(-2)^{2}+3^{2}+1^{2}}}=\frac{7}{\sqrt{14}}=\frac{\sqrt{14}}{2}$$