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 A(xo,yo,zo)\displaystyle A(x_o, y_o, z_o) lies outside a plane π\displaystyle \pi with equation ax+by+cz+d=0\displaystyle ax\,+\,by\,+\,cz\,+\,d\:=\:0.

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

d(AB)\displaystyle d(AB) is the distance between the point A\displaystyle A and the plane π\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 L\displaystyle L through point A\displaystyle A and perpendicular to the plane.

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

Hence, line L\displaystyle L has parametric equations: x=xo+aty=yo+btz=zo+ct\displaystyle \:\begin{array}{cc}x\:=\:x_o\,+\,at \\ y\:=\:y_o\,+\,bt \\ z\:=\:z_o\,+\,ct\end{array}

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

Then use the Distance Formula on points A\displaystyle A and B\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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