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enter image description here

I have to find intersection of two lines ($AH$ & $CD$)

$A(3.42,-1.84,8.56) $

$B(-3.42,3.84,-8.56) $

$C(0.00,16.25,0.00)$

$AH$ is the perpendicular; $CD$ is the median

I tried so:

Firstly I found coordinates of the point $H$:

$x_H=(3.42-3.42)/2=0$

$y_H=(-1.84+3.84)/2=1$

$z_H=(8.56-8.56)/2=0$

So $H=(0,1,0)$

After, I found the median by formula:

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\frac{x-0}0=\frac{y-16.25}{1-16.25}=\frac{z-0}0$

-> it's mean median equation: $y=16.25$

All this is correct?

Please help me, I do not know what to do next ...

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  • 1
    $\begingroup$ It seems that what you call the coordinates of $H$ are the coordinates of the midpoint between $A$ and $B$, which, according to the sketch, would be $D$, not $H$? $\endgroup$ – joriki Dec 14 '11 at 22:34
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As joriki says in the comments, you have found $D$, not $H$. To find $H$, note that $(A-H)\cdot(C-B)=0$, and $H=C+(C-B)t$ for some real number $t$. You should be able to find $t$, and thus $H$, from these two equations.

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  • $\begingroup$ Thanks to all, my answer is 0;5;0 and it's right! $\endgroup$ – Lu Vue Dec 16 '11 at 6:29

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