# Intersection of 2 lines

## Homework Statement

Find the vector equation of the line through 4,5,5 that meets (x-11)/3=7+8=z+4 at right angles

## The Attempt at a Solution

i have no idea where to start.

Dick
Homework Helper
Come on. You must have SOME idea. For example, what does the equation represent? A line, a plane, or what? Try and figure that out first. I'm not totally sure you posted it correctly.

it was a question from a test that i got 0 on, the exact wording is as follows:

find a vector equ ation of the line through the point (4,5,5) that meets the line (x-11)/3=(y+8)/1=(z-4)/1 at right angles

i understand that the dot product of the vector i have to find and of the line given will be equal to zero. other than that, im completely lost.

i dont know if this makes sense, but i tried to first came up with an expression for the vector i am trying to find

4-(11+3t), 5-(-8+t), 5-(4+t) (starting point of the vector subtract a point on the line)

then using dot product equation, i isolated t to -7/11

is there anything flawed in this logic?

Dick
Homework Helper
Yeah, I don't think it's quite right. A general point on the line is, fixing z=t. y=t-4 and x=3t+23. i.e. (3t+23,t-4,t). Or did I do that wrong? I would then take the difference between that and (4,5,5) and minimize the distance wrt to t. That would give you two points on the line, yes?

Well, from the given question it is evident that the equation which to be found is of a line. Now, the equation,
$$\frac{x-11}{3}=\frac{y+8}{1}=\frac{z-4}{1}$$
directly gives you the direction ratios of the required line, isnt it? What would they be?

it does not give me the vectors to the line...
there are an infinite number of possible normals to the original line

Well, (3,1,1) are the direction ratios! And if this line passes through (4,5,5), substitue it in the general equation,
$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$

where, (a,b,c) are the direction ratios and $$(x_1,y_1,z_1)$$ is the point through which the line passes!