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Given:

$ A=(1,-1,-4)$ $ B=(3,-2,-1)$ $ C=(-1,0,m-4) $

Find the point $D$ which satisfies the conditions:

  • $ \vec{AC\:}$ and $\vec{AD\:}$ are orthogonal.
  • The projection of the vector $\vec{AD\:}$ onto $\vec{AB\:}$ is $\frac{1}{\sqrt{14}}$
  • The volume of the Parallelepiped constructed from the vectors $\vec{AD\:}$, $\vec{AB\:}$ and $ \vec{AC\:}$ is $4$.

Discuss the solutions base of on the parameter $m$.

Looking at the problem at first I felt enthusiastic, as it felt very basic, but going in it became a bit of a mess.

Let's say that $D=(x,y,z)$.

We have $\vec{AC\:}=(-2,1,m)$ and $\vec{AD\:}=(x-1,y+1,z+4)$

Their dot product gives: $mz+4m+y-2x+3=0$ as the first condition.

Onto the next one:

$\vec{AB\:}=(2,-1,3)$ and it's magnitude is $\sqrt{14}$.

The magnitude of $\vec{AD\:}$ is $\sqrt{\left(x+1\right)^2+\left(y+1\right)^2+\left(z+4\right)^2}$

Now we find that: $\cos \left(θ\right)=\frac{2x+3z-y+9}{\left(\sqrt{14}\right)\left(\sqrt{x^2-2x+y^2+2y+z^2+8z+18}\right)}$

Which when "plugged" in into the projection formula gives that: $2x+3z-y+9=14$ which is our second condition and also second equation.

The last contidion is the mixed product of these three vectors:

$V=-2my-mx-m-6y-3x-3=4$ Third and final condition.

Now we have a system of equations that has to be discussed in respect to $m$

Is what I did so far correct?

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    $\vec {AB\:} =(2,-1,-5) $. – 2017-01-29

1 Answers 1

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Is what I did so far correct?

No, the second condition and the third condition you've got are not correct.

For the second :

Let $D'(2k+1,-k-1,3k-4)$ where $k\not=0\in\mathbb R$ be the point on $AB$ such that $D'D\perp D'A$.

Now $$\begin{align}D'D\perp D'A&\iff -2k(x-2k-1)+k(y+k+1)-3k(z-3k+4)=0\\\\&\iff k=\frac{2x-y+3z+9}{14}\end{align}$$

So, the second condition is $$\begin{align}AD'=\frac{1}{\sqrt{14}}&\iff \sqrt{14}|k|=\frac{1}{\sqrt{14}}\\\\&\iff \sqrt{14}\left|\frac{2x-y+3z+9}{14}\right|=\frac{1}{\sqrt{14}}\\\\&\iff \color{red}{|2x-y+3z+9|=1}\end{align}$$

For the third :

The third condition is $$\color{red}{|-2my-mx-m-6y-3x-3|=4}$$