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Consider the triangle ABC in $\mathbb{R}^3$ formed by the point $A(3,2,1)$, $B(4,4,2)$, $C(6,1,0)$.

Find the coordinates of the point $D$ on $BC$ such that $AD$ is perpendicular to $BC$.

I believe this uses projections, but I can't seem to get started. I tried the projection of $AC$ onto $BD$ and $AB$ onto $BC$, but to no avail.

Any help is loved! Thanks.

user73229
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2 Answers2

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Here is one way, presumably unsuitable.

The direction vector of $BC$ is $(2,-3,-2)$.

A generic point $D_t$ on $BC$ is given by $(4,4,2)+t(2,-3,-2)$. The direction vector of $AD_t$ is $(1,2,1)+t(2,-3,-2)$. The dot product of this with $(2,-3,-2)$ is $-6+17t$. This dot product must be $0$.

We end up with $D=\left(\frac{80}{17},\frac{50}{17},\frac{22}{17} \right)$.

André Nicolas
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Edited:

There are two ideas used is this.

(1),$AD$ must be perpendicular to $BC$, so their dot product must equal 0.

(2),$D$ is on $BC$, so we must find an eqn. for line $BC$.

Let's start with (2). We know that the direction vector of the line $BC$ is parallel to the vector $\overrightarrow BC$ which is $<2,-3,-2>$ and the line passes through $B(4,4,2)$. Therefore the parametric eqn. for line $BC$ must be $$x = 2t+4$$ $$y = -3t+4$$ $$z=-2t+2$$

This means that all points on the line $BC$, in particular, $D$, will have coordinates of the form $$D(2t+4,-3t+4,-2t+2)$$

Now let's move on to (1).

The vectors $\overrightarrow {AD}$ and $\overrightarrow{BC}$ are perpendicular to each other, so their dot product must be 0, i.e $$<3-(2t+4),2-(-3t+4),1-(-2t+2)> \cdot <2,-3,-2> = 0$$

All you have to do is to find $t$ and plug it back in.

hyg17
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  • sorry, could you elaborate a little further (as in, show the next steps?) – user73229 May 25 '13 at 03:45
  • Long story short, the cross product of $\overrightarrow {AB}$ and $\overrightarrow {AC}$ gives you the normal vector. – hyg17 May 25 '13 at 03:47
  • I don't understand your idea either. We don't want the normal to that plane, since the desired point $D$, which lies on side $BC$, is in the plane. – MJD May 25 '13 at 04:17
  • Oops, sorry misread the problem. I will edit the answer so bear with me. – hyg17 May 25 '13 at 04:31