Equation of 3d vector question

[canadian_beef]

Homework Statement

Find an equation of the line through the point A (4,5,5) which meets the line (x-11)/3 = (y+8)/-1 = z-4 at right angles.

Homework Equations

The Attempt at a Solution

well i know that the direction of the given line is d(3,-1,1) and the original line crosses through the point (11,-8,4)

The direction of the new line must be able to dot with the old line to get 0 (thus right angles).

I don't know how to find a direction that i know will pass through A.

Help or hint please

 

[AlephZero]

You can write any point P on the given line in parametric form, e.g. (x = t, y = a function of t, z = another function of t). Then you can write the direction of AP in terms of t.

 

[tim_lou]

So, you have point A and point P (P on the line). what can you do with the orthogonal projection of AP onto the given line? from there, how can you find a vector that points from the line to A and that is orthogonal to <3,-1,1> ?

 

[benorin]

There are two common ways to write a vector equation of a line:

i. Given two points, say P and Q, an equation of the line through P and Q (in that order as t ranges over [0,1]) is given by

$$\vec{r}(t)=\vec{Q}t+\vec{P}(1-t)$$ or $$\vec{r}(t)=\vec{P}+\left(\vec{Q}-\vec{P}\right) t$$​

notice that $$\vec{Q}-\vec{P}$$ is a vector in the direction of the line, hence

ii. Given a point and a vector in the direction of a line, say point P and vector $$\vec{v}$$, we have an equation of the line through P and parallel to $$\vec{v}$$ is given by

$$\vec{r}(t)=\vec{P}+\vec{v} t$$​

 

[HallsofIvy]

Any vector, <a,b,c>, perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0 or c= b- 3a. Such a line passing through (4, 5, 5) must be of the form (x-4)/a= (y- 5)/b= (z-5)/(b- 3a). Now, what are the conditions on a and b, in order to ensure that (x-4)/a= (y- 5)/b= (z-5)/(b- 3a) intersects (x-11)/3 = (y+8)/-1 = z-4 ?

 

[canadian_beef]

Any vector, <a,b,c>, perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0 or c= b- 3a. Such a line passing through (4, 5, 5) must be of the form (x-4)/a= (y- 5)/b= (z-5)/(b- 3a). Now, what are the conditions on a and b, in order to ensure that (x-4)/a= (y- 5)/b= (z-5)/(b- 3a) intersects (x-11)/3 = (y+8)/-1 = z-4 ?

couple questions

Any vector, <a,b,c>, perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0

shouldnt there be a D value such that 3a-b+c+d=0

and how do i make sure they intersect

 

[canadian_beef]

couple questions

Any vector, <a,b,c>, perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0

shouldnt there be a D value such that 3a-b+c+d=0

and how do i make sure they intersect

cant I just make up any A and B values and it will work?