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.