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Equation of 3d vector question

1. 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.

2. Homework Equations



3. 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 dont know how to find a direction that i know will pass through A.

Help or hint please
 

AlephZero

Science Advisor
Homework Helper
6,953
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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.
 
682
1
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

Homework Helper
1,066
13
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

[tex]\vec{r}(t)=\vec{Q}t+\vec{P}(1-t)[/tex] or [tex]\vec{r}(t)=\vec{P}+\left(\vec{Q}-\vec{P}\right) t [/tex]​

notice that [tex]\vec{Q}-\vec{P}[/tex] 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 [tex]\vec{v}[/tex], we have an equation of the line through P and parallel to [tex]\vec{v}[/tex] is given by

[tex]\vec{r}(t)=\vec{P}+\vec{v} t [/tex]​
 

HallsofIvy

Science Advisor
Homework Helper
41,728
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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 ?
 
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
 
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?
 

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