Finding the Vector Equation of a Line at Right Angles: A Homework Challenge

AI Thread Summary
The discussion revolves around finding the vector equation of a line that passes through the point (4,5,5) and intersects another line at right angles. Participants clarify that the given equation represents a line, providing its direction ratios as (3,1,1). The dot product of the sought vector and the direction vector must equal zero to ensure orthogonality. There is some confusion regarding the setup of the problem and the correct application of the distance minimization technique. Ultimately, the correct approach involves substituting the point into the general equation of the line using the identified direction ratios.
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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


Homework Equations





The Attempt at a Solution


i have no idea where to start.
 
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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, I am completely lost.
 
i don't 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?
 
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, isn't 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,
<br /> \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!
 
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