tough matrix problem!! can you help me solve it?

wildwildwest568

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Mar 20, 2006
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Given: The point A(x0, y0, z0) lies outside a plane ? with equation ax+by+cz+d=0. The point P(x, y, z) lies on the plane ?. AB is the distance between the point A and the plane ?.

Derive: The equation: (AB)= | ax0+by0+cz0+d | / v(a^2+b^2+c^2)

(the bars are an absolute value and the v is a square root sign)

ANY input (advice, strategies, comments, etc) is much appreciated

THANKS!!
 
Hello, wildwildwest568!

Here's a game plan for you . . .

Given: The point \(\displaystyle A(x_o, y_o, z_o)\) lies outside a plane \(\displaystyle \pi\) with equation \(\displaystyle ax\,+\,by\,+\,cz\,+\,d\:=\:0\).

The point \(\displaystyle P(x, y, z)\) lies on the plane \(\displaystyle \pi\).

\(\displaystyle d(AB)\) is the distance between the point \(\displaystyle A\) and the plane \(\displaystyle \pi\).

Derive the equation: \(\displaystyle \L\; d(AB)\;=\;\frac{\left|ax_o\,+\,by_o\,+\,cz_o\,+\,d\right|}{\sqrt{a^2\,+\,b^2\,+\,c^2}}\)
We want line \(\displaystyle L\) through point \(\displaystyle A\) and perpendicular to the plane.

The plane has normal vector: \(\displaystyle \,\vec{n}\:=\:\langle a,b,c\rangle\)

Hence, line \(\displaystyle L\) has parametric equations: \(\displaystyle \:\begin{array}{cc}x\:=\:x_o\,+\,at \\ y\:=\:y_o\,+\,bt \\ z\:=\:z_o\,+\,ct\end{array}\)

Find point \(\displaystyle B\), the intersection of line \(\displaystyle L\) and plane \(\displaystyle \pi\).
\(\displaystyle \;\;\)(Substitute the parametric equation of \(\displaystyle L\) into the equation of the plane and solve for \(\displaystyle t\).)

Then use the Distance Formula on points \(\displaystyle A\) and \(\displaystyle B\).
 
wildwildwest568 said:
how did you get it all formatted right?
The tutor used LaTeX. To learn the basic commands, try the links in the "Forum Help" pull-down menu at the very top of the page.

Eliz.
 
wildwildwest568 said:
does anyone else have ideas about solving?
Where are you stuck?

Please reply showing all of the steps you have tried.

Thank you.

Eliz.
 
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