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

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SUMMARY

The discussion focuses on finding the vector equation of a line through the point (4,5,5) that is perpendicular to the line defined by the equation (x-11)/3=(y+8)/1=(z-4)/1. The direction ratios of the given line are established as (3,1,1). The solution involves using the dot product to ensure orthogonality between the sought vector and the direction vector of the given line. The final approach requires substituting the point (4,5,5) into the general equation of a line to derive the vector equation.

PREREQUISITES
  • Understanding of vector equations in three-dimensional space
  • Knowledge of the dot product and its properties
  • Familiarity with direction ratios of lines
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study vector equations of lines in three-dimensional space
  • Learn about the properties of the dot product and orthogonality
  • Explore the concept of direction ratios and their applications
  • Practice solving problems involving perpendicular lines in vector calculus
USEFUL FOR

Students studying vector calculus, particularly those tackling problems involving lines and planes in three-dimensional geometry. This discussion is especially beneficial for those preparing for exams or homework in advanced mathematics courses.

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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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