Linear algebra vector and lines in r3

[alingy1]

Let $$A(2,-1,1)$$, $$B$$ and $$C$$ be the vertices of a triangle where $$\overrightarrow{AB}$$ is parallel to $$\vec{v}=(2,0,-1), $$$$\overrightarrow{BC}$$ is parallel to $$\vec{w}=(1,-1,1)$$ and $$\angle(BAC)=90°$$. Find the equation of the line through $$\(A\)$$ and $$\(C\)$$ in vector and parametric forms.

Well, there is not much I can do here. I could find the equation of the plane:
$$-x-3y-2z=-1$$ by finding the normal of the two vectors given.

 

[alingy1]

I only wish to have a hint.

 

[LCKurtz]

Let $$A(2,-1,1)$$, $$B$$ and $$C$$ be the vertices of a triangle where $$\overrightarrow{AB}$$ is parallel to $$\vec{v}=(2,0,-1), $$$$\overrightarrow{BC}$$ is parallel to $$\vec{w}=(1,-1,1)$$ and $$\angle(BAC)=90°$$. Find the equation of the line through $$\(A\)$$ and $$\(C\)$$ in vector and parametric forms.

Well, there is not much I can do here. I could find the equation of the plane:
$$-x-3y-2z=-1$$ by finding the normal of the two vectors given.

First, I suggest you don't use the same notation for points as vectors. I would suggest angle brackets for vectors, i.e., $\vec v = \langle 2,0,-1\rangle$. Second, are you sure you have stated the problem correctly? As it is, there is more than one solution. For example, you could just take side AB equal to your $\vec v$, go the right distance along the $\vec w$ direction to make AC perpendicular to AB. Once you have one solution you could extend both legs of the right triangle proportionally to a larger similar triangle. Infinitely many answers.

 

[HallsofIvy]

(On this board use "itex" and "/itex" for formulas you want to be "inline". That would not give that strange spacing!)

You are told that line $\overline{AB}$ is parallel to vector <2, 0, -1>. That should tell you immediately that line $\overline{AB}$ can be written in parametric form $x= 2t+ x_0$, $y= y_0$, $z= -t+ z_0$ where $(x_0, y_0, z_0)$ can be any point on the line. Since the point A itself is given as (2, -1, 1), the line is $x= 2t+ 2$, $y= -1$, $z= -t+ 1$.

Similarly, we can write line $\overline{BC}$ as $x= t+ x_1$, $y= -t+ y_1$, $z= t+ z_1$ but we are not given a point on
$\overline{BC}$.

We do know, however, that $\overline{AC}$ is perpendicular to $\overline{AB}$ so must be in the plane through A normal to $\overline{AB}$: $2(x- 2)- (z- 1)= 0$ as well as lying in the same plane that $\overline{BC}$ is in.